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Honors Physics Chapter 12

Thermal Energy

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NGSS

HS-PS3-4.Plan and conduct an investigation to provide evidence that the transfer of thermal energy when two components of different temperature are combined within a closed system results in a more uniform energy distribution among the components in the system (second law of thermodynamics).

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What makes an object hot?

High temperature, but what causes a high temperature?

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Kinetic Molecular Theory (KMT)

Every thing is made from molecules (or atoms). Molecules are constantly at movement. The movement is completely random (vibration).

All direction.

All different speeds.

Fast more energy high temperature

Slow less energy low temperature

Temperature is the measure of the average speed of the random motion.

Collisions between molecules are elastic.

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KMT

Gas: gas molecules Wood: organic molecules Metal: electrons

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Thermal Energy, (Heat*)

Random motion speed Kinetic Energy Thermal Energy of a sample of gas is the total kinetic

energy of all the vibrating gas molecules. Temperature and heat are closely related to each

other, but they are not exactly the same.– Heat is the total kinetic energy collective property– Temperature is the average speed.– Two bottles of gas at same temperature are added together:

Temperature stays the same. Thermal energy is doubled.

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Practice

)A )B )C )D

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Thermal Energy (Heat) Transfer

Naturally, heat transfers from place/object of higher temperature to place/object of lower temperature until both at the same temperature (thermal equilibrium).

Reverse: forced heat transfer. Three mechanisms of heat transfer:

– Conduction– Convection– Radiation

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Conduction

Two objects must be in contact. Collision transfers kinetic energy from faster

moving molecules to slower moving ones. No macroscopic material is moving from one

object to another. Imagine fast-running person hitting a slower-

moving person.

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Convection

Actual moving of higher temperature material to place of lower temperature (due to difference in density).

Imagine you are taking hot water from the hot water bucket and pour it into the cold water bucket.

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Radiation

In form of visible or invisible light. Can take place in empty space. Imagine the sun sending light in all direction.

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Example

)A )B )C )D

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Practice:

)A )B )C )D

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Practice

)A )B )C )D

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Temperature Scales: Fahrenheit, Celsius, and Kelvin

1.8 32F CT T

59 15 288 o oF C K

F C Ko oF C KT T T

273.15K CT T

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1.8F

C

TT

273.15C KT T

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Example

Convert

a. 0 oC to Kelvins

b. 0 K to degrees Celsius

c. 273 oC to Kelvins

d. 273 K to degrees Celsius

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Solution:

a) TC = 0, TK = TC + 273.15 =0+273.15=273.15= 273

0 oC = 273 K

b) TK = 0, TC = TK - 273.15 = -273

0 K = -273 oC

c) TC = 273, TK = TC + 273.15 = 273 + 273.15 = 546

273 oC = 546 K

d) TK = 273, TC = TK - 273.15 = 273 – 273.15 = 0

273 K = 0 oC

Convert

a. 0 oC to Kelvins

b. 0 K to degrees Celsius

c. 273 oC to Kelvins

d. 273 K to degrees Celsius

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Practice:

Convert these Celsius temperatures to Kelvin temperatures.

a. 27 oC = _______

b. 560 oC = _______

c. -184 oC = _______

d. -300 oC = _______

300 K

833 K89 K

-27 K

0 K is the lowest possible temperature.

Kelvin scale = absolute scale

Impossible temperature

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Absolute Zero Temperature: 0K

No temperature can be lower than 0 K. Temperature is a measure of the average

speed of the random motion of molecules. – Kelvin scale is defined such that at 0 K, the average

speed is zero.– No speed slower than zero, so no temperature lower

than 0 K.

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Practice

)A )B )C )D

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Temperature and temperature change

Temperature:

1 oC ___1 K

Ti = 100 oC =373 K

Temperature Change:

1 oC ___ 1 K

(because 1 oC = 274 K)

=

KT

CT 1 1oK C

Tf = 101 oC = 374 K

374 373 1K K K

101 100 1o o oC C C

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Practice

)A F)B )C )D

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Thermal Energy (Heat) Transfer and Temperature Change

Q: heat gain (+) or loss (-) m: mass of object

Tm

QC

Q mC T

Unit of C is

T = Tf – Ti: temperature change

T is the result of Q

Temperature change

Table 12-1 on Pg318

Water: C =

or o

J J

kg C kg K

4180 or 4180o

J J

kg C Kkg

C: specific heat of object

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Example: A 38-kg block of lead is heated from –26 oC to 180 oC. How much heat does it absorb during the heating?

m = 38kg, Ti = -26oC, Tf =180 oC, C = 130 J/(kg oC)

Q = ?

Q mC T f imC T T

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13038 180 26

13038 206 1.0 10

o oo

oo

Jkg C C

kg C

Jkg C J

kg C

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Practice: Pg319pp3

When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe with a mass of 2.3 kg when its temperature is raised from 20.oC to 80.0oC?

m = 2.3 kg, C = 385 J / (kg · oC), Ti = 20.0oC, Tf = 80.0oCQ = ?

43852.3 80.0 20.0 5.31 10

f i

o oo

Q mC T mc T T

Jkg C C J

kg C

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Another Unit of Heat: calorie

1 cal = 4.18 J

1 Cal = 1000 cal = 1 k cal

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Calorimetry

Conservation of Energy:

Only when there is no heat loss or gain. The system of A and B is isolated.

constantA BE E

0A BE E

0A BQ Q

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Mixing liquids at different temperature.

When liquid 1 of mass m1 of specific heat C1, initially at temperature T1i, is mixed with liquid 2 of mass m2 of specific heat C2, initially at

temperature T2i, together, the final equilibrium temperature Tf is

1 2 0Q Q

1 1 2 2 2 2 21

1 1

f ii

T mC m C m C TT

mC

1 1 1 2 2 2

1 1 2 2

i if

mCT m C TT

mC m C

2 2 2

1

1 1

f i

i f

m C T TC

m T T

1 1 1 2 2 2 1 1 2 2i i f fmCT m C T mCT m C T

1 1 1 2 2 2 0f i f imC T T m C T T

1 1 2 2 fmC m C T

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2 21 1 2

2 1 2

2.00 10 0.200 , 80.0 , 2.00 10 0.200 ,

10.0 , 4180

?

oi

oi o

f

m g kg T C m g kg

JT C C C

kg C

T

1 1 1 2 2 2

1 1 2 2

i if

mCT m C TT

mC m C

Example: Pg321pp6A 2.00 102-g sample of water at 80.0 oC is mixed with 2.00 102 g of water at 10.0 oC. Assume no heat loss to the surroundings. What is the final temperature of the mixture?

0.200 4180 80.0 0.200 4180 10.045.0

0.200 4180 0.200 4180

o oo o

o

o o

J Jkg C kg C

kg C kg CC

J Jkg kg

kg C kg C

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1 1 1 2 2 2

1 1 2 2

i if

mCT m C TT

mC m C

21 1 1

22 2 2

4.00 10 0.400 , 2450 , 16.0 ,

4.00 10 0.400 , 4180 , 85.0 ,

?

oio

oio

f

Jm g kg C T C

kg C

Jm g kg C T C

kg C

T

Practice: Pg321pp7A 4.00 102-g sample of methanol at 16.0oC is mixed with 4.00 102-g of water at 85.0oC. Assume that there is no heat loss to the surroundings. What is the final temperature of the mixture?

0.400 2450 16.0 0.400 4180 85.0

0.400 2450 0.400 4180

59.5

o oo o

o o

o

J Jkg C kg C

kg C kg C

J Jkg kg

kg C kg C

C

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21 1

22 2 2

1

1.00 10 0.100 , 100.0 ,

1.00 10 0.100 , 4180 , 10.0 , 25.0

?

oi

o oi fo

m g kg T C

Jm g kg C T C T C

kg C

C

1 2 0Q Q

Practice: Pg321pp9A 1.00 102-g aluminum block at 100.0 oC is placed in a 1.00 102 g of water at 10.0 oC. The final temperature of the mixture is 25.0oC. What is the specific heat of the aluminum?

1 1 1 2 2 2 0f i f imC T T m C T T

2 2 2

1

1 1

f i

f i

m C T TC

m T T

0.100 4180 25.0 10.0836

0.100 100.0 25.0

o oo

oo o

Jkg C C

Jkg Ckg Ckg C C

2 2 2

1 1

f i

i f

m C T T

m T T

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Thermodynamics

Thermodynamics: the study of the properties of thermal energy and its changes

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State (Phase) of Matter

Solid

Liquid

Gas

Plasma

Particles free to move within the liquid. But they don’t move fast enough (or have enough energy) to jump out

of the liquid.

Particles move fast enough (or have enough energy) to escape out of the liquid into the air.

Atoms have so much energy that electrons are taken away from nuclei.

Particles (molecules or electrons) pretty much cannot move. They move slowly.

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Vaporization

Normally when heat is added into water, its temperature increases. But once up to 100 oC, any heat added to water will be absorbed by

a few water molecules. Then these few water molecules will move fast enough to jump out

of the water. But the remaining water stays at the same temperature (100 oC). So at 100 oC, adding heat does not increase the water’s

temperature, it only turns liquid water into water vapor. Water vapor temperature stays at 100 oC until all the water is

vaporized. Water vapor at 100 oC has more energy than liquid water at 100 oC.

(Water molecules need energy to escape away from each other when they become gas.)

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Condensation

Reverse process of vaporization. Normally when water vapor loses heat, the molecules of water

vapor slow down, and the temperature decreases. But once down to 100 oC, any heat loss will slow down a few water

molecules so that they move slow enough to stick to each other and become liquid water. (They cannot escape the attraction of each other.)

But the remaining water vapor stays at the same temperature (100 oC).

So at 100 oC, removing heat does not decrease the temperature of water vapor, it only turns water vapor into liquid water.

Water temperature stays at 100 oC until all the vapor condenses. When 1 gram of water vapor condenses, it releases the same

amount of heat that is needed to vaporize 1 gram of liquid water.

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Melting and Freezing

Similar to vaporization, ice melts at 0 oC into liquid water without temperature change. – Each water molecule has enough energy to move away from any specific

water molecule,– But it doesn’t have enough energy to completely escape from all other

water molecules.– Temperature stays at 0 oC.

Similar to condensation, liquid water freezes at 0 oC into ice without temperature change.– Each water molecule moves slow enough to be captured by some

specific water molecules.– Each water molecule can still move a little bit.

Freezing is the opposite of melting.– When 1 gram of water freezes, it releases the same amount of heat that

is needed to melt 1 gram of ice.

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Change of States

Gas

Melt (fusion) _______ heat Qf ___ 0

Vaporize _______ heat Qv ___ 0

Freeze (solidify) _______ heat Qs ___ 0

Condensate _______ heat Qc ___ 0

reverse

reverse

Liquid

Solid

Absorb

Absorb

Release

Release

>

>

<

<

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Temperature of Water

Heat

Temperature

ice

ice/waterwater

Water/vapor vapor

Melting Point (0 oC)

Boiling Point (100 oC)

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Heat Transfer Associated with Change of States

Heat needed to vaporize liquid of mass m: HV: heat of vaporization

Condensation is the reverse process of vaporization: HC = – HV

Melting (fusion) is the reverse of freezing (solidification): Hf = – Hs

vQ mH

Table 12-2 on Pg324

Heat needed to melt ice of mass m:fQ mH

Hf: heat of fusion

Notice: Hv 7 Hf

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Example

)A )B )C )D

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Practice

)A )B )C )D

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Practice

)A )B )C )D

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2 1

5

1.00 10 1.00 10 , 20.0 , 0 ,

3.34 10 , 2060

?

o oi f

f o

m g kg T C T C

J JH C

kg kg C

Q

0 20 20o o of iT T T C C C

11 1.00 10 2060 20 4120o

o

JQ mC T kg C J

kg C

1 5 42 1.00 10 3.34 10 3.34 10f

JQ mH kg J

kg

4 41 2 4120 3.34 10 3.75 10Q Q Q J J J

-20oC ice to 0oC ice:

0oC ice to 0oC water:

So total heat is:

Example: Pg325pp19How much heat is absorbed by 1.00 102 g of ice at –20.0 oC to become water at 0.0 oC?

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10.200 , 60.0 , 100 , 4180 , ?o oi f o

Jm kg T C T C C Q

kg C

Q1: 60.0oC water 100.0oC water

Q2: 100.0oC water 100.0oC steam

Q3: 100.0oC steam 140.0oC steam

Q1: 60.0oC water 100.0oC water

Next Page!

Practice: Pg325pp20A 2.00 102-g sample of water at 60.0 oC is heated to steam at 140.0 oC. How much heat is absorbed?

1 0.200 4180 100 60.0 33440o of i o

JQ mc T mc T T kg C C J

kg C

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Continue…

Q2: 100.0oC water 100.0oC steam6

20.200 , 2.26 10 , ?v

Jm kg H Q

kg

Q3: 100.0oC steam 140.0oC steam

0.200 , 100.0 , 140 , 2020 , ?o oi f o

Jm kg T C T C C Q

kg C

So total heat is 5 5

1 2 3 33440 4.52 10 16160 5.0 10Q Q Q Q J J J J

6 52 0.200 2.26 10 4.52 10v

JQ mH kg J

kg

3 0.200 2020 140 100.0 16160o of i o

JQ mc T mc T T kg C C J

kg C

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First Law of Thermodynamics

U: change in internal energy of a system Q: heat flow into/out of system

U Q W Conservation of Energy

• Into system: Q > 0 • Out of system: Q < 0

W: work done by/on system• System does work to surrounding: W > 0• Surrounding does work to system: W < 0

ΔU = mC T• Heat and work together cause the temperature change.

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Sample

)A )B )C )D

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So which one is right?

U mC T U Q W

Q W mC T

Q mC T But we also have

So which formula is correct?

Q mC T W General formula, valid for any W

Q mC T Valid only when W = 0.

Q mC T W

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Example: Pg328pp22

A gas balloon absorbs 75 J of heat. The balloon expands but stays at the same temperature. How much work did the balloon do in expanding?

75 , constant, ?Q J T W

U mc T

U Q W

W

0,

Q U 75J

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Example: Pg328pp23:A drill bores a small hole in a 0.40-kg block of

aluminum and heats the aluminum by 5.0oC. How much work did the drill do in boring the hole?

The change in temperature of the aluminum is a result of work done by drill on aluminum, not because of heat added to it. Actually Q = 0 since we are not heating the aluminum with a heater.

0.40 , 897 , 5.0 , ?odrillo

Jm kg C T C W

kg C

The change in the internal energy of the aluminum can be found from

0.40 897 5.0 1794oo

JU mC T kg C J

kg C

U Q W

But this is the work done by aluminum (on drill), and the work done by drill on aluminum is the opposite:

31794 1.8 10 1.8drillW J J kJ

1794W Q U J

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Practice: Pg328pp25When you stir a cup of tea, you do about 0.050 J of work each time you circle the spoon in the cup. How many times would you have to stir the spoon to heat a 0.15-kg cup of tea by 2.0 oC?

1 0.050 , 0.15 , 4180 , 2.0 , 0, ?oo

Jstir J m kg C T C Q n

kg C

0.15 4180 2.0 1254oo

JU mC T kg C J

kg C

1254W U J

But this is the work done by cup of tea (on you), and the work done by you is the negative of it.

1254W J

U Q W

11254 25,080

0.050

stirJ stirs

J

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Second Law of Thermodynamics

Entropy: level of messiness, disorder

Second Law: Natural processes go in a direction that increases the total entropy of the universe.

QS

T S: Change in entropy

Q: Heat transfer

T: Temperature in Kelvin scale

0S

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Lab: Calorimetry

1 2 0Q Q

Q mC T f imC T T

1 1 1 2 3 3 0f i f imC T T m C T T

See Slide 28