Chapter 4

Heat Energy

What is heat energy?

• What is the difference between a hot piece of chalk and a cold one?

The difference between Heat and Temperature

• Heat Energy is the kinetic energy of the molecules.

• Temperature is how “hot” or “cold” something is.

• Temperature is related to the average velocity of the molecules.

• The amount of heat energy in a substance depends on more than just the temperature.– It depends on the type of material

• It requires more energy to change the temperature of water than most other substances.

Temperature Changes

• An increase in the temperature causes the motion of the molecules to increase– This causes expansion

• A physically more correct way of looking at temperature is to say that we perceive (feel) things as being hotter when their molecules are moving more rapidly.

• The rate at which heat (temperature change) moves though substances varies– Insulators – heat does not move readily through them– Conductors – heat moves faster through them

Temperature Scales• Fahrenheit

– Water freezes at 32 °F

– Water boils at 212 °F

• Celsius – Water freezes at 0 °C

– Water boils at 100 °C

• Kelvin or Absolute Scale– Used in many scientific calculations

– More on this scale later

– Add 273 to the Celsius temperature

– TK=TC+273

Converting between Scales

• TF = (9/5)TC + 32°

• TC= (5/9)(TF - 32°)

• TK = TC + 273

Change 30° C to Fahrenheit

Data TC = 30°

TF = ?

TF = (9/5)TC + 32°

TF = (9/5)(30°) + 32°

TF = (54°) + 32°

TF = 86° F

Thirty degrees Celsius corresponds to 86° Fahrenheit

Change 30° C to Absolute (Kelvin)

DataTC = 30°

TK = ?

TK = 30° + 273

TK = 303 K

Note that in the absolute temperature scale, no degree symbol is needed, as this is not relative to anything, but is “absolute”.

Units of heat energy

• Joules may be used.

• Calories also may be used– One calorie is the amount of energy needed to

change the temperature of one gram of water one degree Celsius.

– One calorie = 4.185 J (may round to 4.2 J in problems)

– One Calorie = 1 kilocalorie = 1000 calories• Used in diets (one diet coke contains 1 Calorie =

1000 calories)

metabolic energy

• resting basal– Basal energy is the energy necessary for vital

functions such as heart, breathing, etc.; (the things necessary to keep you alive).

• 5.2 W/kg pigeon

• 1.2 W/kg person

• .67 W/kg cow

Metabolic Energy

• 70 kg person– 80 W resting– 125 W sitting– 300 W standing, walking– 1200 W hard work, running

Metabolic Energy

Human brain uses 20% of basal energy

• 9% monkeys

• 5% dogs and cats

• One glass of beer has 80 Calories. How many times would you need to lift barbells with 20 kg (~50 lbs) 60 cm to use up this energy? (note that Calories = kilocalories)

• Write the principle• How many joules in 80 Calories?• 334,880j• How many joules are used each time you lift

the barbells?• 117.6j• How many times do you need to lift?• 2848 times

Temperature and Color

• 500°C - Begins to glow

• 600°C - Red hot

• 1000°C - Orange

• 1400°C - White hot

Density

• Density is the mass per unit volume.

• Density = mass/volume

• D=m/v

• Put on formula sheet!

Calculate the weight of a block of gold 11 cm x 17 cm x 25 cm.

• Volume = length x width x height

• Volume = 11 cm x 17 cm x 25 cm = 4675 cm3

• Density of gold = 19 g/cm3

D = m/v v = m/D

m = D·v (all of these may be on your formula sheet)

m = 19 g/cm3·4675 cm3 = 88,825 g = 88.8 kg

88.8 kg x 2.21 lbs/kg = 196 lbs

Pressure

• Pressure = force/area• P = F/A• What is the pressure when a 130 pound woman

puts her weight on a shoe heel that is ½ inch square?

Data: F = 130 lbs, A = .5 in x .5 in = .25 in2, P=?P=F/A = 130 lbs/.25 in2

P = 520 lbs/in2

A wind creates a pressure of 2 lbs/in2 on a window that is 6ft x 15 ft. What is the force on the window?

Data

P = 2 lbs/in2

A = (12 in/ft•6ft) (12 in/ft•15ft) = 12960 in2

F = ?

P = F/A

F= P•A=12960 in2 • 2 lbs/in2 = 25,920 lbs

Metric units of pressure

• 1 Pascal (Pa) = 1 Newton/meter2

• Atmospheric pressure = 14.7 lbs/in2

• Atmospheric pressure = 101,300 Pa = 1.013 bar

• 1 bar = 100,000 Pa = 105 Pa

If a person’s lungs are considered to be 9 in square, with what force does atmospheric pressure push on

your chest?

• Data

• A = (9 in)2 = 81 in2

• P = 14.7 lbs/in2

• F = ?

• F = P • A = 14.7 lbs/in2 • 81 in2 = 1191 lbs

So how can we breathe?

Archimedes's Principle

• When an object is submersed in a fluid, there is a buoyant force on the object that is equal to the weight of the fluid it displaces– The weight of the fluid that would occupy the

space of the object.

How deep will a 1 m2 box sink when it is floating on water and a 80 kg person steps

in it?

• From Archimedes’s principle, we know that the box will sink until 80 kg of water has been displaced.

• The question then becomes, “How deep will water in a 1 m2 box be to weigh 80 kg.

• Each cm of water in the box weighs 100 cm x100 cm x 1 g/cm3 = 10,000 g = 10 kg

• Therefore the box will float 8 cm below the surface of the water.

How big must a balloon filled with hydrogen be to lift a 70 kg person?

• Density of air = 1.3 kg/m3

• Density of hydrogen = 0.09 kg/m3

• Each m3 of balloon will lift 1.21 kg• To lift 70 kg we will need

of hydrogen

• Volume of a sphere = V = 4/3 π r3

33

9.57/21.1

70m

mkg

kg

mmV

r 4.214.34

9.573

4

33

3

3

The kinetic theory of matter

Solids• The molecules in a solid vibrate more

rapidly as the solid gets hotter but they stay in the same location.– This more rapid movement means they take up

more space (expand).– The molecules themselves do not change size

they just move around more and take up more space.

– This makes the substance expand.

liquids

As heat is added to a solid, the vibrations become sufficiently large to break the bonds that hold the molecules in place and they become free to change places and the substance melts to a liquid.

The energy needed to break the bonds of the solid is called the heat of fusion.

In a liquid the molecules can change places but there are still large forces between molecules

These forces can be seen when water is on a waxed surface, etc.

packets

Heat of fusion

• The heat energy necessary to change one gram of a solid to a liquid.

• For water at zero degrees Celsius, this is 335 j/gram

• Melting – freezing

Heat of Vaporization

• Is the amount of heat energy necessary to change one gram of liquid to a gas.– This is the energy necessary to break the bonds

of the liquid.• Breaking the bonds of the liquid completely, so that

the substance becomes a gas, requires more than six times as much energy as going from a solid to a liquid (for water).

• For water this is 2260 j/gram• Vaporization – Condensation• 1 calorie = 4.2 joules

Gases

• As the liquid is heated the molecules move about more and more until they break completely away from the liquid.– The energy necessary to break these bonds is called

the heat of vaporization– In a gas there is no attraction between molecules– The heat of vaporization is much larger than the heat

of fusion. (for water 2260 j versus 335 j)

Absolute Zero

• Is the temperature at which the volume and pressure of an ideal gas is zero

• Is the temperature at which there is zero kinetic energy in the molecules.

• Is the lowest possible temperature

• Absolute zero = - 273° C

• Absolute zero = - 459° F

Energy Equations

fhmE

vhmE

Specific Heat

• The amount of heat energy necessary to change the temperature of one gram of a material, one degree Celsius.

• water = 1 cal/gr – Learn this!– Necessary for your pre-lab

• aluminum = .22 cal/ gram• iron = .11 cal/ gram

Heat Energy

• The amount of heat energy needed to change the temperature of m grams of a substance with specific heat c by an amount ΔT is given by the equation:

Put on your formula sheet

TmcEnergyheat

Calculate the energy necessary to heat 30 g of aluminum from 20ºC to 50ºC

• E=?• m = 30 g• Change in temp = ΔT = 30ºC• c(for aluminum) = .22 cal/g·ºC• Energy to change the temp =

= mass●specific heat●temp change• E=m·c· ΔT (Put on formula sheet)• E = 30 g· .22 cal/g·ºC·30ºC• E = 198 cal• E = 198 cal x 4.2 j/cal =832 joules

Absolute zeroAbsolute temperature (Kelvin)

Temperature

Pre

ssur

e

Absolute zeroAbsolute temperature (Kelvin)

Temperature

Vol

ume

Four Postulates of the kinetic theory of ideal gases

• There is no attraction between the molecules in an ideal gas.

• The molecules lose no energy when colliding with the container or other molecules

• The molecules occupy no space (They are very small)

• The kinetic energy of the molecules is proportional to the absolute temperature of the gas

Ideal Gas Law

T in this equation must be absolute temperature

Put on your formula sheet!

T(K) = T(ºC) + 273

2

22

1

11

T

VP

T

VP

The pressure of air in tire is 32 lbs/in2 when the temperature is 5°C, what is the pressure when the temperature is 30°C? Assume the volume of the tire does not change.Data

• P1= 32 lbs/in2 P2= ?

• V1=V2= V

• T1= 5°C; T(K) = T(°C) + 273 K = 5 + 273 = 278 K

• T2= 30°C T(K) = T(°C) + 273 K = 30 + 273 =303 K

Solution

• P1V1/T1 = P2V2/T2

• P2= P1V1 T2 /V2T1

• P2= (32 lbs/in2 • V • 303 K)

(V • 278K)

• P2= 34.9 lbs/in2

First Law of Thermodynamics

• Also known as the conservation of energy

• FIRST LAW = The total amount of energy in any closed system remains unchanged, but the energy may change from one type to another

• Energy is never lost, but can be changed from one type of energy to another.

Second Law of Thermodynamics

• Also applies to all systems, living and non-living

• Tells the direction energy moves through systems

• Energy flows from hot to cold

• Systems tend toward disorder (Entropy increases)

• SECOND LAW = All energy can not be used as work, some is changed to heat

The second law of thermodynamics

Thermodynamic Efficiency of Heat Engines

• Gives the maximum possible efficiency, if there is no friction, no heat lost except through the exhaust.

• Tc= temperature of the exhaust

• Th= temperature in the hottest place in the system.– For example inside the cylinder where the gasoline burns.

• Note that both of these temperatures must be Kelvin!

h

c

T

Teff 1

What is the maximum possible efficiency of a car whose exhaust

temperature is 150ºC and cylinder temperature is 1200ºC

• Tc=150ºC = 423 K

• Th=1200ºC=1473 K

• Eff =1-Tc/Th

• Eff =1-423K/1473K=1-0.287=.713• Maximum efficiency = 71%

h

c

T

Teff 1

What is the maximum possible efficiency of a car whose exhaust temperature is 120ºC

and cylinder temperature is 1000ºC

• Tc=120ºC = 393 K

• Th=1000ºC=1273 K

• Eff =1-Tc/Th

• Eff =1-393K/1273K=1-0.309=.691

• Maximum efficiency = 69.1%

Refrigerator

Freon boils using energy from the food.

Freon condenses here and gives off

energy to the room.

The net effect is to move the heat from the freezing

compartment to the outside of the refrigerator.

Calculate the amount of energy needed to change 300 grams of ice at 0° C to steam at 100 ° C

In this problem we need to do three separate calculations:1) Melt the ice2) Heat the water from 0° C to 100 ° C3) Boil the water

To answer the question, we then need to add the answers to all three.

1) Melt the ice

Data

m = 300 g

Heat of fusion of water: hf= 335 j/g

Equation: For melting

Heat energy = mass x heat of fusion

E = m • hf (formula sheet)

E = 300g • 335 j/g = 100500 j

2) Heat the water

Data:

m = 300 g

ΔT = 100° C (Start at zero, end at 100 degrees)

Specific heat of water = 4.2 j/g °C

Formula: for changing temperature

Energy = mass • specific heat • Δ T = m•hs• Δ T

E = 300 g • 4.2 j/g °C • 100° C

E = 126000 j

3) Boiling the water

Data:

m = 300 g

Heat of vaporization of water: he= 2260 j/g

Equation: For boiling

Heat energy = mass x heat of vaporization

E = m • he

E = 300g • 2260 j/g = 678000 j

The final answer

Add the three energies to get the total energy

Total Energy = Energy to melt + energy to heat the water + energy to change the water to steam

E = 100,500 j + 126,000 j + 678000 j

E = 904,500 j

Calculate the work necessary to raise a 10 kg rock 30 m high.

• Data m = 10 kg, h = 30 m, W=?W=F·d=mgh=10kg·9.8m/s2·30m = 2940j

• What is the potential energy at this height (25m)?

• No calculation is necessary, the P.E. is 2940J.• What is the kinetic energy at this height?• No calculation is necessary, as the rock is

sitting still, the K.E. is zero!• Total energy = K.E. + P.E. = 2940j at all

heights.

• What is the total energy at this height?• No calculation is necessary, T.E. is 2940j.• Calculate the P.E., the K.E. and the T.E when

the rock is 10 m from the ground and when it just begins to touch the ground (the height is zero and the velocity is maximum).

• The total energy remains the same in all cases, 2940j,

• At ground level, h=0 so P.E. is zero, thus the K.E. is 2940j,

• To calculate the P.E. at 10 m height,

Data: m=10 kg, h=10m, P.E.=?

P.E. = mgh = 10kg·9.8m/s2·10m = 980j

The K.E. at this point is:

K.E. = T.E. – P.E. = 2940j – 980j = 1960j

What is the heat energy produced when the rock hits the ground?

No calculation is necessary, all the energy is changed to heat, so the answer is 2940j.

If the rock falls into a bucket containing 4 kg of water, what is the

temperature change of the water?

• Data E = 2940 j

• M = Mass of water = 4 kg = 4000 grams

• ΔT=?

• c (water ) = 4.2 j/gºC

• Equation E=M c ΔT

• ΔT = E/(cΔT)= 2940j/(4.2 j/gºC4000 g) = 0.175ºC

A candy bar has 250 Calories. You exercise by lifting 20 kg (about 50 lbs)

moving your arms .5 m up and down. How many times do you need to lift the weight to

use up the 200 Calories?

• Steps:

• Change the Calories to calories.

• 250 Cal x 1000 cal/Cal = 250,000 cal

• Change the calories to joules

• 250,000 cal x 4.2 j/cal = 1,050,000 j

How much work is done in lifting 20 kg, 0.5 meters high?

• Data

• m=20kg, h=0.5 m

• Formula W=mgh=20kg·9.8m/s2·0.5m = 98 j

• How many times must the weight be lifted to use the 1,050,000 j?

• No. of times = 1,050,000/98 = 10,715 times

The average velocity of the molecules of air at 20°C is 330 m/s. What will the

velocity be at 60°C

Data

T1 = 20°C = 273 + 20 = 293 K

T2 = 60°C = 273 + 60 = 333 K

V1= 330 m/s

V2= ?

The volume of air in a balloon is 300 cm3 when the pressure is 5 Pa and the temperature is 15°C. What is the volume when the pressure is 10 Pa and the temperature is 30°C? Data

• P1= 5 Pa P2= 10 Pa

• V1= 300 cm3 V2= ?

• T1= 15°C T2= 30°C

• T1= 273 + 15 = 288 K

• T2= 273 + 30 = 303 K

Work and solution

• P1V1/T1 = P2V2/T2

• V2 = P1V1 T2 /T1P2

• V2 = 5 Pa • 300 cm3 • 303 K /288 K • 10 Pa

• V2 = 158 cm3

Principle, Work and Answer• Principle

– The kinetic energy of the molecules of an ideal gas is proportional to its absolute temperature.

• Equation– KE2/ KE1= T2/T1

– ½mv22/½mv12 = T2/T1

– v22/v1

2 = T2/T1

– v1/v2 =√ T2/T1 = √333/293

– v1 = v2 •√333/293 = 330 m/s •√333/293 = 351m/s