No electrostatic forces of repulsion ndash metal is deformed (malleable)
Electrostatic forces
of repulsion
Force
Force
broken crystal
>
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Chemical ChangeChemical Change
Any change involving a rearrangement of atomsAny change involving a
rearrangement of atoms
During aldquochemical reactionrdquonew materials are
formed by a change in the way atoms are bonded together
During aldquochemical reactionrdquonew materials are
formed by a change in the way atoms are bonded together
Physical and Chemical PropertiesExamples of Physical Properties
Boiling point Color Slipperiness Electrical conductivity
Melting point Taste Odor Dissolves in water
Shininess (luster) Softness Ductility Viscosity (resistance to flow)
Volatility Hardness Malleability Density (mass volume ratio)
Examples of Chemical Properties
Burns in air Reacts with certain acids Decomposes when heated
Explodes Reacts with certain metals Reacts with certain nonmetals
Tarnishes Reacts with water Is toxic
Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction
No electrostatic forces of repulsion ndash metal is deformed (malleable)
Electrostatic forces
of repulsion
Force
Force
broken crystal
>
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
During aldquochemical reactionrdquonew materials are
formed by a change in the way atoms are bonded together
During aldquochemical reactionrdquonew materials are
formed by a change in the way atoms are bonded together
Physical and Chemical PropertiesExamples of Physical Properties
Boiling point Color Slipperiness Electrical conductivity
Melting point Taste Odor Dissolves in water
Shininess (luster) Softness Ductility Viscosity (resistance to flow)
Volatility Hardness Malleability Density (mass volume ratio)
Examples of Chemical Properties
Burns in air Reacts with certain acids Decomposes when heated
Explodes Reacts with certain metals Reacts with certain nonmetals
Tarnishes Reacts with water Is toxic
Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction
No electrostatic forces of repulsion ndash metal is deformed (malleable)
Electrostatic forces
of repulsion
Force
Force
broken crystal
>
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Physical and Chemical PropertiesExamples of Physical Properties
Boiling point Color Slipperiness Electrical conductivity
Melting point Taste Odor Dissolves in water
Shininess (luster) Softness Ductility Viscosity (resistance to flow)
Volatility Hardness Malleability Density (mass volume ratio)
Examples of Chemical Properties
Burns in air Reacts with certain acids Decomposes when heated
Explodes Reacts with certain metals Reacts with certain nonmetals
Tarnishes Reacts with water Is toxic
Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction
No electrostatic forces of repulsion ndash metal is deformed (malleable)
Electrostatic forces
of repulsion
Force
Force
broken crystal
>
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
No electrostatic forces of repulsion ndash metal is deformed (malleable)
Electrostatic forces
of repulsion
Force
Force
broken crystal
>
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
No electrostatic forces of repulsion ndash metal is deformed (malleable)
Electrostatic forces
of repulsion
Force
Force
broken crystal
>
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Chemical Bonds
Increasing ionic character
Nonpolar covalent
Electrons are sharedequally
Cl Cl
Polar covalent
Electrons are sharedunequally
ClH
Ionic bonding
Electrons are transferred
Cl1-Na1+
Ralph A Burns Fundamentals of Chemistry 1999 page 229
bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Temperature vs Heat
Measuredwith a
Calorimeter
TotalKineticEnergy
Joules(calories)
Measuredwith a
Thermometer
AverageKineticEnergy
oCelcius(or Kelvin)
Alike Different
A Propertyof
Matter
HaveKineticEnergy
Heat
Different
Topic Topic
Temperature
thermometercalorimeter
Conservation of Matter
Reactants yield Products
Densitybull
Density is an INTENSIVE property of matter
- does NOT depend on quantity of matter - color melting point boiling point odor density
bull Contrast with
EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Conservation of Matter
Reactants yield Products
Densitybull
Density is an INTENSIVE property of matter
- does NOT depend on quantity of matter - color melting point boiling point odor density
bull Contrast with
EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Densitybull
Density is an INTENSIVE property of matter
- does NOT depend on quantity of matter - color melting point boiling point odor density
bull Contrast with
EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances
Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)
Lithium
Water
Aluminum
Lead
10 19 053
10 10 10
10 37 27
10 058 114
Density
D
M
Vensity
ass
olume
D = M V
M = D x V
V =M D
Two ways of viewing density
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71
Equal volumeshellip
hellipbut unequal masses
The more massive object(the gold cube) has thegreater density
aluminum gold
(A)
Equal masseshelliphellipbut unequal volumes
(B)
gold
aluminumThe object with the larger volume (aluminum cube) has the smaller density
Specific Gravity
Jaffe New World of Chemistry 1955 page 66
09025
water 10
ice
cork
aluminum
27
>
>
>
Archimedes Principle
Vfinal = 985 cm3
- Vinitial = 445 cm3
Vfishing sinker = 540 cm3
Before immersion
Water
445 cm3
After immersion
Fishing sinker
985 cm3
Thread
>
Dissolving of Salt in Water
NaCl(s) + H2O Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Density
D
M
Vensity
ass
olume
D = M V
M = D x V
V =M D
Two ways of viewing density
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71
Equal volumeshellip
hellipbut unequal masses
The more massive object(the gold cube) has thegreater density
aluminum gold
(A)
Equal masseshelliphellipbut unequal volumes
(B)
gold
aluminumThe object with the larger volume (aluminum cube) has the smaller density
Specific Gravity
Jaffe New World of Chemistry 1955 page 66
09025
water 10
ice
cork
aluminum
27
>
>
>
Archimedes Principle
Vfinal = 985 cm3
- Vinitial = 445 cm3
Vfishing sinker = 540 cm3
Before immersion
Water
445 cm3
After immersion
Fishing sinker
985 cm3
Thread
>
Dissolving of Salt in Water
NaCl(s) + H2O Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Two ways of viewing density
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71
Equal volumeshellip
hellipbut unequal masses
The more massive object(the gold cube) has thegreater density
aluminum gold
(A)
Equal masseshelliphellipbut unequal volumes
(B)
gold
aluminumThe object with the larger volume (aluminum cube) has the smaller density
Specific Gravity
Jaffe New World of Chemistry 1955 page 66
09025
water 10
ice
cork
aluminum
27
>
>
>
Archimedes Principle
Vfinal = 985 cm3
- Vinitial = 445 cm3
Vfishing sinker = 540 cm3
Before immersion
Water
445 cm3
After immersion
Fishing sinker
985 cm3
Thread
>
Dissolving of Salt in Water
NaCl(s) + H2O Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Specific Gravity
Jaffe New World of Chemistry 1955 page 66
09025
water 10
ice
cork
aluminum
27
>
>
>
Archimedes Principle
Vfinal = 985 cm3
- Vinitial = 445 cm3
Vfishing sinker = 540 cm3
Before immersion
Water
445 cm3
After immersion
Fishing sinker
985 cm3
Thread
>
Dissolving of Salt in Water
NaCl(s) + H2O Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Archimedes Principle
Vfinal = 985 cm3
- Vinitial = 445 cm3
Vfishing sinker = 540 cm3
Before immersion
Water
445 cm3
After immersion
Fishing sinker
985 cm3
Thread
>
Dissolving of Salt in Water
NaCl(s) + H2O Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dissolving of Salt in Water
NaCl(s) + H2O Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Liquids
The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Some Properties of Solids Liquids and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of the container
Arrangement of Fixed very close Random close Random far apartParticles
Interactions between Very strong Strong Essentially noneparticles
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
bull To evaporate molecules must have sufficient energy to break IM forces
bull Molecules at the surface break away and become gas
bull Only those with enough KE escapebull Breaking IM forces requires energy The
process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat
Evaporation
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Change from gas to liquid
Achieves a dynamic equilibrium with vaporization in a closed system
What is a closed system
A closed system means matter canrsquot go in or out (put a cork in it)
What the heck is a ldquodynamic equilibriumrdquo
Condensation
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
When first sealed the molecules gradually escape the surface of the liquid
As the molecules build up above the liquid - some condense back to a liquid
The rate at which the molecules evaporate and condense are equal
Dynamic Equilibrium
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase ldquodynamicrdquo
The total amount of liquid and vapor remains constant ldquoequilibriumrdquo
Dynamic Equilibrium
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
bull Vaporization is an endothermic process - it requires heat
bull Energy is required to overcome intermolecular forces
bull Responsible for cool earthbull Why we sweat
Vaporization
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Ene
rgy
of s
yste
m
Brown LeMay Bursten Chemistry 2000 page 405
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
solid
liquid
gas
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Tem
pera
ture
(oC
)
A
B
C
DE
Heating Curve for Water
0
100
LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Latent Heat
bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat
water
steam(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 calg Lv = 540 calg
Lf is the latent heat of fusionLv is the latent heat of vaporization
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Elements
only one kindof atom atomsare bonded ifthe element
is diatomic orpolyatomic
Compounds
two ormore kindsof atomsthat arebonded
substancewith
definitemakeup
andproperties
Mixtures
two or moresubstances
that arephysically
mixed
two ormore
kinds ofand
Both elements and compounds have a definite makeup and definite properties
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Pure Substances
Elementndash composed of identical atomsndash examples copper wire aluminum foil
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Pure Substances
Compoundndash composed of 2 or more elements
in a fixed ratio
ndash properties differ from those of individual elements
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Pure Substances
Law of Definite Compositionndash A given compound always contains the same
fixed ratio of elements
Law of Multiple Proportionsndash Elements can combine in different ratios to
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Mixtures
Variable combination of two or more pure substances
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Mixtures
Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect
ndash particles donrsquot settlendash EX rubbing alcohol
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Mixtures
Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Classification of Matter
Materials
HomogeneousHeterogeneous
Heterogeneousmixture
Homogeneousmixture
Substance
Element Compound Solution Mixture
Speci
fic
Gen
eral
Order Disorder
Smoot Smith Price Chemistry A Modern Course 1990 page 43
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Elements Compounds and Mixtures
(a)an element(hydrogen)
(b)a compound(water)
(c)a mixture(hydrogen and oxygen)
(d)a mixture(hydrogenand oxygen)
Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68
hydrogenatoms hydrogen
atoms
oxygen atoms
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Mixture vs Compound
Mixture
FixedComposition
Bonds between
components
Can ONLY beseparated by
chemical means
VariableComposition
No bondsbetween
components
Can beseparated by
physical means
Alike Different
Contain two or more
elements
Can beseparated
intoelements
Involvesubstances
Compound
Different
Topic Topic
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Compounds vs Mixtures
bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2
bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Diatomic Elements 1 and 7H2
N2 O2 F2
Cl2
Br2
I2
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
MatterMatter
SubstanceDefinite composition
(homogeneous)
SubstanceDefinite composition
(homogeneous)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Element(Examples iron sulfur
carbon hydrogenoxygen silver)
Mixture ofSubstances
Variable composition
Mixture ofSubstances
Variable composition
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Compound(Examples water
iron (II) sulfide methaneAluminum silicate)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Homogeneous mixtureUniform throughoutalso called a solution
(Examples air tap watergold alloy)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Heterogeneous mixtureNonuniform
distinct phases(Examples soup concrete granite)
Chemicallyseparable
Physicallyseparable
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
The Organization of Matter
MATTER
PURESUBSTANCES
HETEROGENEOUSMIXTURE
HOMOGENEOUSMIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical methods
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorousspontaneously ignites
Red phosphorousused for matches
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Allotropes of Carbon
Graphite BuckminsterfullereneDiamond
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
1824 atoms Au2424 atoms Au 1424 atoms Au
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Solid Brass
An alloy is a mixture of metals
bull Brass = Copper + Zincbull Solid brass
bull homogeneous mixturebull a substitutional alloy
Copper
Zinc
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Brass Plated
bull Brass = Copper + Zincbull Brass plated
bull heterogeneous mixturebull Only brass on outside
Copper
Zinc
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Galvanized Nails and Screws
bull Zinc coating prevents rustndash Use deck screws for any outdoor project
bull Iron will rust if untreated ndash Weaken and break
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Chromatography
bull Tie-dye t-shirt
bull Black pen ink
bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Ion chromatogram of orange juice
time (minutes)
de
tec
tor
res
po
ns
e
0 5 10 15 20 25
Na+
K+
Mg2+ Fe3+
Ca2+
A Distillation Apparatus
liquid with a soliddissolved in it
thermometer
condenser
tube
distillingflask
pure liquid
receiving flaskhose connected to
cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282
Centrifugation
bull Spin sample very rapidly denser materials go to bottom (outside)
bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood
cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
A Distillation Apparatus
liquid with a soliddissolved in it
thermometer
condenser
tube
distillingflask
pure liquid
receiving flaskhose connected to
cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282
Centrifugation
bull Spin sample very rapidly denser materials go to bottom (outside)
bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood
cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Centrifugation
bull Spin sample very rapidly denser materials go to bottom (outside)
bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood
cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Electrolysis
Must add acid catalyst to conduct electricity
H1+
water oxygen hydrogen
ldquoelectrordquo = electricity ldquolysisrdquo = to split
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32
Water
Hydrogengas forms
Oxygengas forms
ElectrodeSource ofdirect current
H2O(l) O2 (g) + 2 H2 (g)
Reviewing ConceptsPhysical Properties
bull List seven examples of physical properties
bull Describe three uses of physical propertiesbull Name two processes that are used to
separate mixturesbull When you describe a liquid as thick are
you saying that it has a high or low viscosity
Reviewing ConceptsPhysical Properties
bull Explain why sharpening a pencil is an example of a physical change
bull What allows a mixture to be separated by distillation
Reviewing ConceptsChemical Properties
bull Under what conditions can chemical properties be observed
bull List three common types of evidence for a chemical change
bull How do chemical changes differ from physical changes
Reviewing ConceptsChemical Properties
bull Explain why the rusting of an iron bar decreases the strength of the bar
bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical
ELEMENT
hydrogen molecule H2
ELEMENT
oxygen molecule O2
MIXTURE
a mixture ofhydrogen and oxygen molecules
CHEMICAL REACTION
if molecules collide with enoughforce to break them into atoms a can take place
COMPOUND
water H2O
>
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Reviewing ConceptsPhysical Properties
bull List seven examples of physical properties
bull Describe three uses of physical propertiesbull Name two processes that are used to
separate mixturesbull When you describe a liquid as thick are
you saying that it has a high or low viscosity
Reviewing ConceptsPhysical Properties
bull Explain why sharpening a pencil is an example of a physical change
bull What allows a mixture to be separated by distillation
Reviewing ConceptsChemical Properties
bull Under what conditions can chemical properties be observed
bull List three common types of evidence for a chemical change
bull How do chemical changes differ from physical changes
Reviewing ConceptsChemical Properties
bull Explain why the rusting of an iron bar decreases the strength of the bar
bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical
ELEMENT
hydrogen molecule H2
ELEMENT
oxygen molecule O2
MIXTURE
a mixture ofhydrogen and oxygen molecules
CHEMICAL REACTION
if molecules collide with enoughforce to break them into atoms a can take place
COMPOUND
water H2O
>
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Reviewing ConceptsPhysical Properties
bull Explain why sharpening a pencil is an example of a physical change
bull What allows a mixture to be separated by distillation
Reviewing ConceptsChemical Properties
bull Under what conditions can chemical properties be observed
bull List three common types of evidence for a chemical change
bull How do chemical changes differ from physical changes
Reviewing ConceptsChemical Properties
bull Explain why the rusting of an iron bar decreases the strength of the bar
bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical
ELEMENT
hydrogen molecule H2
ELEMENT
oxygen molecule O2
MIXTURE
a mixture ofhydrogen and oxygen molecules
CHEMICAL REACTION
if molecules collide with enoughforce to break them into atoms a can take place
COMPOUND
water H2O
>
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Reviewing ConceptsChemical Properties
bull Under what conditions can chemical properties be observed
bull List three common types of evidence for a chemical change
bull How do chemical changes differ from physical changes
Reviewing ConceptsChemical Properties
bull Explain why the rusting of an iron bar decreases the strength of the bar
bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical
ELEMENT
hydrogen molecule H2
ELEMENT
oxygen molecule O2
MIXTURE
a mixture ofhydrogen and oxygen molecules
CHEMICAL REACTION
if molecules collide with enoughforce to break them into atoms a can take place
COMPOUND
water H2O
>
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Reviewing ConceptsChemical Properties
bull Explain why the rusting of an iron bar decreases the strength of the bar
bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical
ELEMENT
hydrogen molecule H2
ELEMENT
oxygen molecule O2
MIXTURE
a mixture ofhydrogen and oxygen molecules
CHEMICAL REACTION
if molecules collide with enoughforce to break them into atoms a can take place
COMPOUND
water H2O
>
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
ELEMENT
hydrogen molecule H2
ELEMENT
oxygen molecule O2
MIXTURE
a mixture ofhydrogen and oxygen molecules
CHEMICAL REACTION
if molecules collide with enoughforce to break them into atoms a can take place
COMPOUND
water H2O
>
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
2 H22 H2 O2
O2 2 H2O2 H2O++ ++ EE
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Copyright copy 2007 Pearson Benjamin Cummings All rights reserved
(a) Radiant energy (b) Thermal energy
(c) Chemical energy (d) Nuclear energy (e) Electrical energy
The energy something possesses due to its motion depending on mass and velocity
Potential energy
Energy in Energy out
kinetic energy kinetic energy
Energy
Kinetic Energy ndash energy of motion
KE = frac12 m v 2
Potential Energy ndash stored energy
Batteries (chemical potential energy)
Spring in a watch (mechanical potential energy)
Water trapped above a dam (gravitational potential energy)
mass velocity (speed)
B
AC
>
School Bus or Bullet
Which has more kinetic energy a slow moving school bus or a fast moving bullet
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
The energy something possesses due to its motion depending on mass and velocity
Potential energy
Energy in Energy out
kinetic energy kinetic energy
Energy
Kinetic Energy ndash energy of motion
KE = frac12 m v 2
Potential Energy ndash stored energy
Batteries (chemical potential energy)
Spring in a watch (mechanical potential energy)
Water trapped above a dam (gravitational potential energy)
mass velocity (speed)
B
AC
>
School Bus or Bullet
Which has more kinetic energy a slow moving school bus or a fast moving bullet
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Energy
Kinetic Energy ndash energy of motion
KE = frac12 m v 2
Potential Energy ndash stored energy
Batteries (chemical potential energy)
Spring in a watch (mechanical potential energy)
Water trapped above a dam (gravitational potential energy)
mass velocity (speed)
B
AC
>
School Bus or Bullet
Which has more kinetic energy a slow moving school bus or a fast moving bullet
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
School Bus or Bullet
Which has more kinetic energy a slow moving school bus or a fast moving bullet
Either may have more KE it depends on the mass of the bus and the velocity of the bullet
Which is a more important factor mass or velocity Why (Velocity)2
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Kinetic Energy and Reaction Rate
Kinetic energy
Fra
ctio
ns o
f pa
rtic
les
lower temperature
higher temperature
minimum energyfor reaction
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Hot vs Cold Tea
Kinetic energy
Many molecules have anintermediate kinetic energy
Few molecules have avery high kinetic energy
Low temperature(iced tea)
High temperature(hot tea)
Perc
ent o
f mol
ecul
es
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Decomposition of Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Ene
rgy
Energy of reactants
Energy of products
Reaction Progress
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Effect of Catalyst on Reaction Rate
reactants
products
Ene
rgy
activation energy for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Burning of a Match
Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293
Energy released to the surrounding as heat
SurroundingsSystem
(Reactants)
D(PE)
Pot
entia
l ene
rgy
(Products)
>
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Conservation of Energy in a Chemical Reaction
Surroundings
System
Surroundings
SystemEn
erg
y
Beforereaction
Afterreaction
In this example the energy of the reactants and products increases while the energy of the surroundings decreases
In every case however the total energy does not change
Myers Oldham Tocci Chemistry 2004 page 41
Endothermic Reaction
Reactant + Energy Product
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Direction of Heat Flow
Surroundings
ENDOthermicqsys gt 0
EXOthermicqsys lt 0
System
Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207
System
H2O(s) + heat H2O(l)
melting
H2O(l) H2O(s) + heat
freezing
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Caloric Values
Food joulesgrams caloriesgram Caloriesgram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
Smoot Smith Price Chemistry A Modern Course 1990 page 51
1000 calories = 1 Calorie
science food
1calories = 4184 joules
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Units of energy
Most common units of energy
1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also
expressed in kilojoules (1 kJ = 103J)
2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1degC
One cal = 4184 J or 1J = 02390 cal
Units of energy are the same regardless of the form of energy
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance
Experimental Determination of Specific Heat of a Metal
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
A Bomb Calorimeter
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Heating CurvesTe
mp
erat
ure
(oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Calculating Energy Changes - Heating Curve for Water
Tem
per
atu
re (
oC
)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x Cfus
DH = mol x Cvap
DH = mass x DT x Cp liquid
DH = mass x DT x Cp gas
DH = mass x DT x Cp solid
Cp gas = 187 JgoC
Cp liquid = 4184 JgoC
Cp solid = 2077 JgoC
Cf water = 333 Jg
Cv water = 2256 Jg
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C333
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system
a) 60oC b) 30oC c) 20oC d)
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
Assume NO heat energy is ldquolostrdquo to the surroundings from the system
20 g (40oC) 10 g (20oC) 333oC
C726
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 267oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 300oC
Block ldquoArdquo Block ldquoBrdquoFinal
Temperature
20 g (40oC) 10 g (20oC) 333oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 267oC
AgH2O
Real Final Temperature = 266oC
Why
Wersquove been assuming ALL materialstransfer heat equally well
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Specific Heat
bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC
bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC
bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy
bull Lets look at the math
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
ldquolosesrdquo heat
Calorimetry
C266 x
3208x 8550
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 266oC
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Calorimetry
C266 x
8550 3208x
7845 3138x x 057 705
algebra the solve and units Drop
C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350
equation into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
1 Calorie = 1000 calories
ldquofoodrdquo = ldquosciencerdquo
Candy bar300 Calories = 300000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4184 Joules
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Cp(ice) = 2077 Jg oC
It takes 2077 Joules to raise 1 gram ice 1oC
X Joules to raise 10 gram ice 1oC
(10 g)(2077 Jg oC) = 2077 Joules
X Joules to raise 10 gram ice 10oC
(10oC)(10 g)(2077 Jg oC) = 2077 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DTTe
mpe
ratu
re (
o C)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Heat = (specific heat) (mass) (change in temperature)
q = Cp m DT
T m C q p(ice)
initialfinalp(ice) TT m C q
C)30(C20- g 10 C g
J 2077 q oo
o
Given Ti = -30oC
Tf = -20oC
q = 2077 Joules
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp liquid
Heat = mass x Dt x Cp gas
Heat = mass x Dt x Cp solid
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron
Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)
2059 X = 22091
X = 1073 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter
Calorimetry Problems 2 question 8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units
- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)
-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104
3 x 104 = 136 x 103 Tf
Tf = 221oC
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system
Calorimetry Problems 2 question 9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units
- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)
-2468 Tf + 3208 = 364 Tf - 26208
29416 = 6108 Tf
Tf = 482oC
T = 72oC
mass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead
Calorimetry Problems 2 question 12
PbT = oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)
- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units
- [(4444) (46oC - Ti)] = (11046) (21oC)
- 2044 + 4444 Ti = 23197
4444 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice
Calorimetry Problems 2 question 13
H2OT = -12oCmass = g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]
4582 m = - 17339
m = 378 g
iceTf = 24oC
qA = [(CpH2O) (mass) (DT)]
qC = [(CpH2O) (mass) (DT)]
qB = (CfH2O) (mass)
qA = [(2077 JgoC) (mass) (12oC)]
qB = (333 Jg) (mass)
qC = [(4184 JgoC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]
249 m
333 m
1003 m
4582 mqTotal = qA + qB + qC
4582 4582
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Endothermic Reaction
Energy + Reactants Products
+DH Endothermic
Reaction progress
Ene
rgy
Reactants
ProductsActivation Energy
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
O
Catalytic Converter
C O
N O
CO
OCO
NN
One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas
CO
N
NN
OO
OC
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Enthalpy Diagram
H2O(g)H2O(g)
H2O(l)H2O(l)
H2(g) + frac12 O2(g)
- 44 kJExothermic
+44 kJEndothermic
DH = +242 kJEndothermic
-242 kJExothermic
-286 kJEndothermic
DH = -286 kJExothermic
Ene
rgy
H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements
C(g) + 2O(g) CO2(g)
Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ
Smith Smoot Himes pg 141
2O(g) O2(g) H = - 250 kJ
C(g) + 2O(g) CO2(g) H = -1360 kJ
C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Fission vs Fusion
Fuse small atoms2H2 He
NO Radioactive
waste
Very HighTemperatures~5000000 oC
(SUN)
Split large atoms
U-235
Radioactive waste
(long half-life)
NuclearPowerPlants
Alike Different
Create Large Amounts
of EnergyE = mc2
Transmutationof Elements
Occurs
Change Nucleus
of Atoms
Fusion
Different
Topic Topic
Fission
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
bull Use fear and selective facts
to promote an agenda
bull Eating animalsbull Radiation = Bad
Look who is funding research it may bias the results
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Shielding Radiation
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Nuclear Fission
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Nuclear Fission
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Nuclear Power Plants
map Nuclear Energy Institute
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Conservation of Masshellipmass is converted into energy
Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu
FUSION
2 H2 1 He + ENERGY
1008 amux 440032 amu = 4004 amu + 0028 amu
This relationship was discovered by Albert EinsteinE = mc2
Energy= (mass) (speed of light)2
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Tokamak Reactor
bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial
(doughnut shaped) ring
bull Magnetic field contains plasma
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Cold Fusion
bull Fraudbull Experiments must
be repeatable to
be valid
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
0 1 2 3 4Number of half-lives
Rad
iois
otop
e re
mai
ning
(
)
100
50
25
125
Half-life of Radiation
Initial amountof radioisotope
t12
t12
t12
After 1 half-life
After 2 half-lives
After 3 half-lives
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Objectives - Matter
bull Explain why mass is used as a measure of the quantity of matter
bull Describe the characteristics of elements compounds and mixtures
bull Solve density problems by applying an understanding of the concepts of density
bull Distinguish between physical and chemical properties and physical and chemical changes
bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Objectives - Energy
bull Identify various forms of energybull Describe changes in energy that take place
during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between
electrostatic charges
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Law of Conservation of EnergyEafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Law of Conservation of Energy
ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2 H2 + O2 2 H2O + energy
+ + WOOF
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
Law of Conservation of Energy
ENERGY
C2H2 + O2C2H2 + O2
PEreactants
PEproducts
KEstopper
heat light sound
Eafter = Ebefore
2C2H2 + 5O2 4 CO2 + 2H2O + energy
Energy Changes
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures
First experimental image showing internal atomic structures
copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs
Energy and Matter
Slide 2
Slide 3
Physical and Chemical Properties
Three Possible Types of Bonds
Shattering an Ionic Crystal Bending a Metal
Chemical Bonds
Covalent vs Ionic
Temperature Scales
Heat versus Temperature
Molecular Velocities
Temperature vs Heat
Conservation of Matter
Density
Properties of Matter
Volume and Density
Density (2)
Two ways of viewing density
Specific Gravity
Archimedes Principle
Dissolving of Salt in Water
Liquids
States of Matter
Some Properties of Solids Liquids and Gases
Slide 25
Condensation
Dynamic Equilibrium
Dynamic Equilibrium (2)
Vaporization
Energy Changes Accompanying Phase Changes
Heating Curve for Water
Heating Curve for Water (2)
Latent Heat
Slide 34
Slide 35
Matter Flowchart
Pure Substances
Pure Substances (2)
Pure Substances (3)
Mixtures
Mixtures (2)
Mixtures (3)
Mixtures (4)
Classification of Matter
Classification of Matter (2)
Classification of Matter (3)
Elements Compounds and Mixtures
Mixture vs Compound
Compounds vs Mixtures
Diatomic Elements 1 and 7
Slide 51
The Organization of Matter
Phosphorous (P4)
Allotropes of Carbon
Gold
Solid Brass
Slide 57
Galvanized Nails and Screws
Methods of Separating Mixtures
Chromatography
Paper Chromatography of Water-Soluble Dyes
Separation by Chromatography
Ion chromatogram of orange juice
A Distillation Apparatus
Centrifugation
The decomposition of two water molecules
Electrolysis
Reviewing Concepts Physical Properties
Reviewing Concepts Physical Properties (2)
Reviewing Concepts Chemical Properties
Reviewing Concepts Chemical Properties (2)
Slide 72
Slide 73
Specific Heats of Some Substances
Slide 75
Slide 76
Energy
School Bus or Bullet
Kinetic Energy and Reaction Rate
Hot vs Cold Tea
Decomposition of Nitrogen Triiodide
Exothermic Reaction
Endothermic Reaction
Effect of Catalyst on Reaction Rate
Burning of a Match
Conservation of Energy in a Chemical Reaction
Direction of Heat Flow
Slide 88
Units of energy
Slide 90
A Bomb Calorimeter
Heating Curves
Calculating Energy Changes - Heating Curve for Water
Heat Transfer
Heat Transfer (2)
Heat Transfer (3)
Heat Transfer (4)
Specific Heat
Calorimetry
Calorimetry (2)
Slide 101
Slide 102
Slide 103
Slide 104
Slide 105
Slide 106
Slide 107
Slide 108
Slide 109
Slide 110
Endothermic Reaction (2)
Catalytic Converter
Enthalpy Diagram
Hessrsquos Law
Fission vs Fusion
Irradiated Spam
Shielding Radiation
Nuclear Fission
Nuclear Fission (2)
Nuclear Power Plants
Nuclear Fusion
Conservation of Mass
Tokamak Reactor
Cold Fusion
Half-life of Radiation
Objectives - Matter
Objectives - Energy
Law of Conservation of Energy
Law of Conservation of Energy (2)
Law of Conservation of Energy (3)
First experimental image showing internal atomic structures