physics 7a -- lecture 2 winter 2009 prof. robin d. erbacher 343 phy/geo bldg...
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Physics 7A -- Lecture 2
Winter 2009
Physics 7A -- Lecture 2
Winter 2009
Prof. Robin D. Erbacher343 Phy/Geo Bldg
Prof. Robin D. Erbacher343 Phy/Geo Bldg
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This course has two instructors: Prof. Robin Erbacher (me) rderbacher @ucdavis Ruggero Tacchi (Lead DL Instructor) rtacchi @ucdavis
You are enrolled in one of two 7A classes. 7A-C/D has lectures on Tuesdays. We are independent courses but cover the same material, so you
can attend any review session. We have a common final exam.
The final exam is Friday March 20th, 6:00 pm-8:00 pm If you know you cannot make the final, you should take 7A in
a different quarter. There are no make-up exams.
This course has two instructors: Prof. Robin Erbacher (me) rderbacher @ucdavis Ruggero Tacchi (Lead DL Instructor) rtacchi @ucdavis
You are enrolled in one of two 7A classes. 7A-C/D has lectures on Tuesdays. We are independent courses but cover the same material, so you
can attend any review session. We have a common final exam.
The final exam is Friday March 20th, 6:00 pm-8:00 pm If you know you cannot make the final, you should take 7A in
a different quarter. There are no make-up exams.
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• Join this Class Session with your PRS clicker! (Practice run today, credit begins next time.)
• Quiz today! Lecture 1, DLM 1 + FNTs. Must take it in correct lecture time slot.
• Check Physics 7 website frequently for calendar &announcements. DL Instructors have PTA numbersfor adding this class. No Lecture next week!
• Turn off cell phones and pagers during lecture.
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• Three-phase model of matter
• Energy-interaction model
• Mass-spring oscillator
• Particle model of matter Particle model of bond energy Particle model of thermal energy
• Thermodynamics
• Ideal gas model
• Statistical model of thermodynamics
We started withthese two…
We introduce this one next(chapter 2)
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3-Phase ModelRevisited
3-Phase ModelRevisited
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• Solid: Keeps its shape without a container.• Liquid: Takes the shape of the (bottom of) the container. Keeps its volume the same.• Gas: Takes the shape and volume of the container.
Example H2O
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• Tbp: Temperature at which a pure substance changes phase from liquid to gas (boiling point).• Tmp: Temperature at which a pure substance changes phase from solid to liquid (melting point).
Tbp
Tmp
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You take ice out of the freezer at -300C and place it in a sealed container and slowly heat it on the stove. You would find:
• the temperature of the ice rises,
• remains fixed at 00C for an extended time while it is a mixture of ice and water,
• the temperature rises again after it all melted,
• remains fixed at 1000C for an extended time while it is a mixture of liquid and gas,
• the temperature rises again after it is all gas (steam).
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Q How do we change the phase of matter?
How do we change the temperature of matter?
A By adding or removing energy. In some cases this energy is transferred from, or to, the substance as heat, “Q”.
Example H2O
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Thermal Equilibriumand Heat
Thermal Equilibriumand Heat
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An ice-cube sits in a bath of water. Water and ice can exchange heat with each other
but not with the environment.What is the direction of heat transfer?
A. From ice-cube to waterB. From water to ice-cubeC. Neither of aboveD. Impossible to tell
00 CWater
Ice-cube00 C
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Starting definition of heat (to be revised much later):
Heat (Q) is the transfer of energy from a hot object to a cold object because the objects are at different temps.
Corollary: If the two objects are at the same temperature, no Q (heat) flows between them.
Energy leaves hot objects in the form of heat.Energy enters cold objects in the form of heat.
Low temp High tempQ
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The Zeroth law of thermodynamics says:
Since they are in thermal equilibrium with each other, there is no net energy exchanged among them.
If objects A and B are separately in thermal
equilibrium with a third object C, then A and B
are in thermal equilibrium with each other
If objects A and B are separately in thermal
equilibrium with a third object C, then A and B
are in thermal equilibrium with each other
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If the two objects are at the same temperature, no heat flows between them.
A system in thermal equilibriumin thermal equilibriumis a system whose temperature is not changing in time.
Tfinal
Energy leaves hot objects in the form of heat Energy enters cold objects in the form of heat
Low temp High temp
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The Zeroth law of thermodynamics example:
• Let the third object C be the thermometer.• If the two readings are the same, then A and B are also in thermal equilibrium.
• Energy (heat) will not flow between A and B if put together.
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A cup of hot coffee left in a room…
A thermometer
Cold beer
It can take some time for things to reach thermal equilibrium with its environment. ~ what is happening at microscopic level? => more to
come when we cover Particle models of thermal energy
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C =
[C] = J/K
Coffee cup: ceramic material
A thermometerTip: metalBody: glass, plastic
Beer glass:glass
Heat capacity [C] of substances:A measure of the amount of energy required to
increase the temperature of the substance a certain amount
€
QΔT
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Heat capacity C is an extensive property:2kg of water will have twice the heat capacity of 1kg water
Heat capacity of substances:A measure of the amount of energy required to
increase the temperature of the substance a certain amount
C =
[C] = J/K
€
QΔT
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Porcelain 1.1kJ/kgK
Tip:metal
(Silver: 0.24kJ/kgK) Body: plastic~ 1.2kJ/kgK
Glass 0.84kJ/kgK
Specific heat capacity Cp is an intensive property:Specific heat capacity only depends on the substance
Specific heat capacity Cp of substances:the amount of energy per unit mass/unit mole required to
increase the temperature of the substance by one degree Kelvin
[Cp] = kJ/kgK = kJ/moleK
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The scientific "calorie" is spelled with a lower-case "c".
One "calorie" = 4.184 Joules
The "dieter's" calorie is spelled with an upper-case "C".
One "Calorie" = 1000 calories
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You heat 1 L of water and raise its temperature by 100 C. (Water~1g/ml)
Question: If you add the same quantity of heat to 2 L of water, how much will the temperature rise?
a) Not enough information is given.
b) Twice as much.
c) Half as much.
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You heat 1 L of water and raise its temperature by 100 C. (Water~1g/ml)
Question: If you add the same quantity of heat to 5 L of water, how much will the temperature rise?
• Not enough information is given.
• 20 C.
• 500 C.
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Heat capacity C – sort of the slope here of A, C, E
Heat of fusion Heat of vaporization
ΔE
€
C = Q
ΔT
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Temperature (K)
Energy added (J)
solid
liquid
gas
∆T
∆E
C =
[C] = J/K
€
QΔT
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Temperature (K)
Energy added (J)
solid
liquid
gas
Tb
∆E
C =
[C] = J/K
€
QΔT
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Temperature (K)
Energy added (J)
solid
liquid
gas
Tb
∆E
“Heat” of vaporization : ∆Hthe amount of energy per unit mass/unit mole required for a substance to change its phase
from liquid to gas or vice versa
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Temperature (K)
Energy added (J)
solid
liquid
gas
Tm
∆E
“Heat” of fusion (melting) : ∆Hthe amount of energy per unit mass/unit mole required for a substance to change its phase
from solid to liquid or vice versa
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Temperature (K)
Energy added (J)
solid
liquid
gas
Tm
∆E
Typically, ∆Hv >> ∆Hm
e.g. It takes 6 times more energy to vaporize 1kg of water
than to melt the same amount of ice
Tb
∆E
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In our notation, we always have ΔE = Efinal - Einitial .
ΔE negative: Energy is released from the system. (“Neg. energy added.”)
ΔE positive: Energy is put into the system. Be sure to select the correct sign for all energy transfers!
=> Note also: ΔT is always Tf - Ti .
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Heat capacity - Extensive-How much energy it takes to change the temperature of this amount of pure substance (see parts A, C and E in graph).
Specific heat capacity (or specific heat) - Intrinsic-How much energy it takes to change the temperature per unit of pure substance (mass/mole) (parts A, C and E).
Heat of fusion - Intrinsic-How much energy it takes to melt all of the ice to water (see isothermal part B of graph).
Heat of vaporization - Intrinsic- How much energy it takes to boil all the water to steam (see isothermal part D of graph).
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You put a red hot iron 1.0 kg mass into 1.0 L of cool water.
1) The increase in the water temperature is equal to the decrease in the iron’s temperature. True or False?
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You put a red hot iron 1.0 kg mass into 1.0 L of cool water.
1) The increase in the water temperature is equal to the decrease in the iron’s temperature. True or False?
2) The iron and the water will both reach the same temperature.
True or False?
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Conservation of Energy and the
Energy Interaction Model
Conservation of Energy and the
Energy Interaction Model
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• Energy is a thing (quantity). You & I contain energy, as do the chairs you sit on and the air we breathe. • We cannot see it, but we can measure the transformation of energy (or change, ΔE) through measuring a process.
Conservation of EnergyEnergy cannot be created nor destroyed, simply
converted from one form to another.
Conservation of EnergyEnergy cannot be created nor destroyed, simply
converted from one form to another.
• If the energy of an object increases, something else must have given that object its energy.
• If it decreases, it has given its energy to something else.
• A transfer of energy is when one object gives energy to another.There are 2 types of energy transfers ΔE -- Heat and Work.
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Protons + anti-protons New particles!
Fermilab
Conservation of EnergyEnergy cannot be created nor destroyed, simply
converted from one form to another.
E=Mc2
!
x
x
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Etherma
l
EbondEmovement
(KE)
Egravit
y
Eelectri
c
Esprin
g
There are many different types of energies called energy systems:
........
For each energy system, there is an indicator that tells us how that energy system can change:
Ethermal: indicator is temperatureEbond: indicator is the mass of the initial and final phases
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•Ethermal = C ΔT, Temperature is the indicator.
• Between phase changes, only thermalenergy changes.
• Ebond = |Δm ΔH|, Δm is the indicator.
• At a physical phase change, only the bond-energy system changes. ΔH is the heat of the particular phase change. Δm is the amount that changed phase.
• In a chemical reaction, there are several bond energy changes corresponding to diff. molecular species (reactants or products). Here ΔH is the heat of formation for a particular species.
Etherma
l
Ebond
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Ea Eb Ec
Conservation of EnergyThe total energy of a closed physical system must remain constant. So, the change of the energies of all energy systems associated with the physical system must sum to zero.
Change in closed system energy = ∆Ea + ∆ Eb + ∆ Ec = 0
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Ea Eb Ec
Conservation of EnergyThe change of the energies of all systems associated with an open physical system must sum to the net energy added or removed. Energy is added or removed as Heat or Work.
Change in open system energy = ∆Ea + ∆ Eb + ∆ Ec
= (Energy added) - (Energy removed) = Q + W.
Energy added Energy removed
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Ea
Energy added = + 100 J
Suppose we have a system where 100J of heat comes in from the outside. Joe claims that the only energy system that changes is Ea and that ΔEa is negative (Ea decreases).
Can Joe be correct?1) Yes, its possible that he is correct.2) Yes, Joe is definitely correct.3) No way is Joe’s description correct.
Clicker!
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Example: Melting IceTi= 0°C Tf = room temperature
Tem
pera
ture
Energy of substance
solid
liquid
gas
l-g coexist
s-l coexist
Initial
TMP
TBP
Final
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Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Process 2: Water at T=0ºC Water at room temperatureTem
pera
ture
Energy of substance
solid
liquid
gas
l-g coexist
s-l coexist
Process 1Initial
TMP
TBP
Process 1Final /
Process 2Initial
Process 2Final
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Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Ice
∆T = 0
∆Eth = mCpΔT = 0
Initial phase Solid, Final phase Liquid
Etherm
al
Ebond
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Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Ice
∆T=0
∆Eth = mCpΔT =0
Initial phase Solid, Final phase Liquid
Etherm
al
EbondHeat
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Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Ice
Initial phase Solid, Final phase Liquid
∆Eth + ∆Ebond= Q+W
∆Ebond= ±|Δm||ΔH| = Q
Etherm
alEbond
HeatMw
∆T=0
∆Eth = mCpΔT =0
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Example: Melting IceProcess 2: Water at T=0ºC Water at room temperature
Ice
Initial phase Liquid, Final phase Liquid
Etherm
al
Ebond
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Example: Melting IceProcess 2: Water at T=0ºC Water at room temperature
Ice
Initial phase Liquid, Final phase Liquid
∆Ebond= ±|Δm||ΔH| = 0
Etherm
al
Ebond
T
Heat
∆Eth + ∆Ebond= Q+W
∆Eth= mCpΔT = Q
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Example: Melting Ice
Ice
Initial phase Liquid, Final phase Solid
Etherm
al
Ebond
Freezing (Water at T=0°C Ice at T=0°C)
∆T=0
∆Eth= mCpΔT= 0
Heat
NOTE: Heat is released when bonds are formed! (In general ΔE is negative)
Mw
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• For a closed system:
(Is it clear why there’s no Q or W for a closed system?)
• For an open system:
(Q and W can be positive or negative, as can ΔEs.)
€
ΔE total = ΔE1 + ΔE 2 + ΔE 3 + ... = 0
€
ΔE total = ΔE1 + ΔE 2 + ΔE 3 + ... = Q + W
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Next Time:Two New Energy
Systems
Next Time:Two New Energy
Systems
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Backup Information:
Backup Information:
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• Kelvin: the standard for scientific use.Increasing the temperature by 1 K =Increasing the temperature by 10C
• Celsius/CentigradeSame as Kelvin except 0 in a different place
• Fahrenheit Smaller unit of temperature
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The heat capacity, C, of a particular substance is defined as the amount of energy needed to raise the temperature of that sample by 1° C.
If energy (heat, Q) produces a change of temperature, ΔT, then:
Heat capacity depends on the amount of a substance we have, since it will take more energy to change the temperature of a larger quantity of something.
It is thus called an extensive quantity, or dependent upon the quantity/mass of a substance (kg or mole).
Q = C ΔTQ = C ΔT
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The specific heat capacity, often simply called specific heat, is a particular number for a given substance and does not depend on quantity.
Specific heat is thus an intensive property.
The specific heat of water
is one calorie per gram
per degree Celsius.
The specific heat of water
is one calorie per gram
per degree Celsius.
SI units for heat capacity and specific heat:• heat capacity J/K• specific heat J/kg•K, or J/mol•K (molar specific heat)