1. Honors Physics Chapter 12
Thermal Energy

2. 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).

3. What makes an object hot?
High temperature, but what causes a high temperature?

4. 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.
Temperature is the measure of the average speed of the random motion.
Fast  more energy  high temperature
Slow  less energy  low temperature
Collisions between molecules are elastic.

5. KMT
Gas: gas molecules
Wood: organic molecules
Metal: electrons

6. 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.

7. Practice
The measure of the average kinetic energy of individual molecules is referred to as
temperature.
thermal energy.
heat.
internal energy.

8. 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

9. 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.

10. 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.

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

12. Example
The process whereby heat flows by means of molecular collisions is referred to as
radiation.
conduction.
convection.
inversion.

13. Practice:
By what primary heat transfer mechanism does one end of an iron bar become hot when the other end is placed in a flame?
forced convection
onduction
radiation
natural convection

14. Practice
The process whereby heat flows by the mass movement of molecules from one place to another is referred to as
conduction.
inversion.
radiation.
convection.

15. Temperature Scales: Fahrenheit, Celsius, and Kelvin

16. Example Convert 0 oC to Kelvins 0 K to degrees Celsius

17. Solution: a) TC = 0, TK = TC + 273.15 =0+273.15=273.15= 273
Convert
0 oC to Kelvins
0 K to degrees Celsius
273 oC to Kelvins
273 K to degrees Celsius
Solution:
a) TC = 0, TK = TC = =273.15= 273
 0 oC = 273 K
b) TK = 0, TC = TK = -273
 0 K = -273 oC
c) TC = 273, TK = TC = = 546
 273 oC = 546 K
d) TK = 273, TC = TK = 273 – = 0
 273 K = 0 oC

18. Practice: Convert these Celsius temperatures to Kelvin temperatures.
27 oC = _______
560 oC = _______
-184 oC = _______
-300 oC = _______
300 K
833 K
89 K
-27 K
Impossible temperature
0 K is the lowest possible temperature.
Kelvin scale = absolute scale

19. 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.

20. Practice Which temperature scale never gives negative temperatures?
Celsius
Kelvin
Fahrenheit
all of the above

21. Temperature and temperature change
1 oC ___1 K 
(because 1 oC = 274 K)
Temperature Change:
1 oC ___ 1 K =
Ti = 100 oC =373 K 
Tf = 101 oC = 374 K

22. Practice Which two temperature changes are equivalent? 1 eF = 1 eC
1 eC = 1 K
1 K = 1 eF
none of the above

23. Thermal Energy (Heat) Transfer and Temperature Change
Q: heat gain (+) or loss (-)
m: mass of object
T = Tf – Ti: temperature change
T is the result of Q
C: specific heat of object
Unit of C is
Table 12-1 on Pg318
Water: C =
Temperature change

24. Example: A 38-kg block of lead is heated from –26 oC to 180 oC
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 = ?

25. 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 = 80.0oC
Q = ?

26. Another Unit of Heat: calorie
1 cal = 4.18 J
1 Cal = 1000 cal = 1 k cal

27. Calorimetry Conservation of Energy:
Only when there is no heat loss or gain.
The system of A and B is isolated.

28. 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

29. Example: Pg321pp6 A 2. 00  102-g sample of water at 80
Example: Pg321pp6 A 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?

30. Practice: Pg321pp7 A 4. 00  102-g sample of methanol at 16
Practice: Pg321pp7 A 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?

31. Practice: Pg321pp9 A 1. 00  102-g aluminum block at 100
Practice: Pg321pp9 A 1.00  102-g aluminum block at 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?

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

33. State (Phase) of Matter
Solid
Particles (molecules or electrons) pretty much cannot move. They move slowly.
Liquid
Particles free to move within the liquid.
But they don’t move fast enough (or have enough energy) to jump out of the liquid.
Gas
Particles move fast enough (or have enough energy) to escape out of the liquid into the air.
Plasma
Atoms have so much energy that electrons are taken away from nuclei.

34. 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.)

35. 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.

36. 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.

37. Change of States Gas Liquid Solid reverse reverse Vaporize
_______ heat
Qv ___ 0
reverse
Condensate
_______ heat
Qc ___ 0
Absorb
Release
>
<
Liquid
reverse
Freeze (solidify)
_______ heat
Qs ___ 0
Melt (fusion)
_______ heat
Qf ___ 0
Release
Absorb
<
>
Solid

38. Temperature of Water vapor water ice/water Heat ice Temperature
Boiling Point (100 oC)
water
ice/water
Melting Point (0 oC)
Heat
ice

39. Heat Transfer Associated with Change of States
Table 12-2 on Pg324
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
Heat needed to melt ice of mass m:
Hf: heat of fusion
Melting (fusion) is the reverse of freezing (solidification):
Hf = – Hs
Notice: Hv  7 Hf

40. Example Phase changes occur as the temperature remains the same.
as the temperature increases.
as the temperature decreases.
all of the above

41. Practice
The heat required to change a substance from the solid to the liquid state is referred to as the
heat of melting.
heat of vaporization.
heat of fusion.
heat of freezing.

42. Practice When a solid melts heat energy leaves the substance.
the temperature of the substance decreases.
heat energy enters the substance.
the temperature of the substance increases.

43. Example: Pg325pp19 How much heat is absorbed by 1
Example: Pg325pp19 How much heat is absorbed by 1.00  102 g of ice at –20.0 oC to become water at 0.0 oC?
-20oC ice to 0oC ice:
0oC ice to 0oC water:
So total heat is:

44. Practice: Pg325pp20 A 2. 00  102-g sample of water at 60
Practice: Pg325pp20 A 2.00  102-g sample of water at 60.0 oC is heated to steam at oC. How much heat is absorbed?
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!

45. Continue… Q2: 100.0oC water  100.0oC steam
Q3: 100.0oC steam  140.0oC steam
So total heat is

46. First Law of Thermodynamics
Conservation of Energy
U: change in internal energy of a system
Q: heat flow into/out of system
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.

47. Sample The internal energy of an ideal gas depends on its temperature.
its pressure.
its volume.
all of the above

48. So which one is right? But we also have So which formula is correct?
General formula, valid for any W
Valid only when W = 0.

49. 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?

50. 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.
The change in the internal energy of the aluminum can be found from
But this is the work done by aluminum (on drill), and the work done by drill on aluminum is the opposite:

51. Practice: Pg328pp25 When you stir a cup of tea, you do about 0
Practice: Pg328pp25 When you stir a cup of tea, you do about 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?
But this is the work done by cup of tea (on you), and the work done by you is the negative of it.

52. Second Law of Thermodynamics
Entropy: level of messiness, disorder
S: Change in entropy
Q: Heat transfer
T: Temperature in Kelvin scale
Second Law: Natural processes go in a direction that increases the total entropy of the universe.

53. Lab: Calorimetry
See Slide 28