3. Plan for Today (AP Physics 2)
Lecture/Notes on Heat Engines, Carnot Engines, and Entropy

4. Energy Transfer by Conduction
Kinetic energy exchange between microscopic particles
Collisions take place between particles, energy transfer, and gradually it works through a material

5. Conduction Rate depends on properties of the substance
Metals are good thermal conductors (free electrons to transfer energy)

6. Conduction Only occurs if there is a difference in temperature
Temperature difference drives the flow in energy

7. Conduction
Rate of energy transfer is proportional to the cross sectional area of the slab and the temperature difference
H = Q/t
And is inversely proportional to the thickness of the slab

9. Conduction Equation
k is proportionality constant, depending on the material

10. Heat engine
A device that takes in heat energy and converts some of it to other forms of energy (like mechanical)
Work done by a heat engine through a cyclic process is
W = |Qh| - |Qc|
Qh is energy absorbed from hot reservoir, Qc is energy expelled to cold

11. HEAT ENGINES Qhot Wout Qcold
A heat engine is any device which through a cyclic process:
Cold Res. TC
Engine
Hot Res. TH
Qhot
Wout
Qcold
Absorbs heat Qhot
Performs work Wout
Rejects heat Qcold

12. THE SECOND LAW OF THERMODYNAMICS
Wout
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.
Not only can you not win (1st law); you can’t even break even (2nd law)!

13. THE SECOND LAW OF THERMODYNAMICS
Cold Res. TC
Engine
Hot Res. TH
400 J
300 J
100 J
A possible engine.
An IMPOSSIBLE engine.
Cold Res. TC
Engine
Hot Res. TH
400 J

14. EFFICIENCY OF AN ENGINE
The efficiency of a heat engine is the ratio of the net work done W to the heat input QH.
Cold Res. TC
Engine
Hot Res. TH QH W QC
e = = W QH
QH- QC
e = 1 - QC QH

15. EFFICIENCY EXAMPLE 800 J W 600 J
An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency?
Cold Res. TC
Engine
Hot Res. TH
800 J W
600 J
e = 1 - QC QH
e = 1 -
600 J
800 J
e = 25%
Question: How many joules of work is done?

16. Carnot Engine
No real engine operating between two reservoirs can be more efficient than a Carnot engine

17. Carnot Engine Substance whose temperature varies between Tc and Th
Ideal gas in a cylinder with a movable piston
Cycle consists of 2 adiabatic and 2 isothermal processes

19. EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine)
Cold Res. TC
Engine
Hot Res. TH QH W QC
For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.
e =
TH- TC TH
e = 1 - TC TH

20. Example 3: A steam engine absorbs 600 J of heat at 500 K and the exhaust temperature is 300 K. If the actual efficiency is only half of the ideal efficiency, how much work is done during each cycle?
Actual e = 0.5ei = 20%
e = 1 - TC TH
e = W QH
e = 1 -
300 K
500 K
W = eQH = 0.20 (600 J)
e = 40%
Work = 120 J

21. REFRIGERATORS Qhot Win Win + Qcold = Qhot Qcold WIN = Qhot - Qcold
A refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat into hot reservoir.
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
Win
Win + Qcold = Qhot
WIN = Qhot - Qcold

22. THE SECOND LAW FOR REFRIGERATORS
It is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0.
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
If this were possible, we could establish perpetual motion!

23. Second Law of Thermodynamics
Energy will not flow spontaneously by heat from a cold object to a hot object
No heat engine operating in a cycle can absorb energy from a reservoir and perform an equal amount of work

24. Entropy Systems tend toward disorder
A disorderly arrangement is more probable than an orderly one
Entropy is a measure of the disorder
The entropy of the Universe increases in all natural processes (alternate 2nd law)

25. Entropy
The change in entropy is equal to the energy flowing into a system (while the system changes from one state to another) divided by the absolute temperature
Change in S = Qr/T

26. COEFFICIENT OF PERFORMANCE
Cold Res. TC
Engine
Hot Res. TH QH W QC
The COP (K) of a heat engine is the ratio of the HEAT Qc extracted to the net WORK done W. QC W
K = = QH
QH- QC
K = TH
TH- TC
For an IDEAL refrigerator:

27. COP EXAMPLE
A Carnot refrigerator operates between 500 K and 400 K. It extracts 800 J from a cold reservoir during each cycle. What is C.O.P., W and QH ?
Cold Res. TC
Eng ine
Hot Res. TH
800 J W QH
500 K
400 K
K =
400 K
500 K K TC
TH- TC =
C.O.P. (K) = 4.0

28. COP EXAMPLE (Cont.) QH W 800 J 500 K 400 K
Cold Res. TC
Eng ine
Hot Res. TH
800 J W QH
500 K
400 K
Next we will find QH by assuming same K for actual refrigerator (Carnot).
K = QC
QH- QC
800 J
QH J =
4.0
QH = 1000 J

29. COP EXAMPLE (Cont.) 1000 J W 800 J 500 K
Cold Res. TC
Engine
Hot Res. TH
800 J W
1000 J
500 K
400 K
Now, can you say how much work is done in each cycle?
Work = 1000 J J
Work = 200 J

30. Q = U + W final - initial)
Summary
The First Law of Thermodynamics: The net heat taken in by a system is equal to the sum of the change in internal energy and the work done by the system.
Q = U + W final - initial)
Isochoric Process: V = 0, W = 0
Isobaric Process: P = 0
Isothermal Process: T = 0, U = 0
Adiabatic Process: Q = 0

31. The following are true for ANY process:
Summary (Cont.)
The Molar Specific Heat capacity, C:
Units are:Joules per mole per Kelvin degree
c = Q
n T
The following are true for ANY process:
Q = U + W
U = nCv T
PV = nRT

32. Summary (Cont.) Qhot Wout Qcold
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
Wout
The Second Law of Thermo: It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.
Not only can you not win (1st law); you can’t even break even (2nd law)!

33. Summary (Cont.) The efficiency of a heat engine: e = 1 - QC QH e = 1 -TC TH
The coefficient of performance of a refrigerator: