1. Chapter 18 The Second Law of Thermodynamics

2. Irreversible Processes Irreversible Processes: always found to proceed in one direction Examples: free expansion of ideal gas heat flow (for finite  T): from hot to cold friction: converts mechanical energy to heat

3. Reversible Processes Reverisble processes: just an idealization, but we can come close put system in equilibrium (with itself and its surroundings) make slow, infinitesimal changes in order to reverse a process

4. Heat Engine Heat engine: transforms heat into mechanical energy, W Reservoirs: hot reservoir at T H cold reservoir at T C

5. Heat Engine engine absorbs heat Q H from hot reservoir engine discards heat Q C to cold reservoir engine does work W

6. Heat Engine Working substance: material in engine undergoing the heat transfer Cyclic process: reuse same substance, return it to initial state at end of each cycle

7. Q = net heat input For each cycle: heat Q H enters engine heat Q C leaves engine net heat in per cycle: Q = Q H + Q C = Q H – |Q C |

8. Heat Engine cyclic process:  U = 0 for engine during one cycle 1st Law: Q = W +  U Thus for each cycle: Q = W

9. Heat Engine For each cycle: Q = Q H – |Q C | So by 1st Law: W = Q = Q H – |Q C |

10. Engine Efficiency W= Q H – |Q C | Perfect engine: |Q C | = 0 All experiments: |Q C | > 0 Efficiency: e measures how close an engine gets to |Q C | = 0

11. Engine Efficiency

12. Ideal Engine = limiting behavior of real engines engine: converts heat to mechanical energy recall: converting mechanical energy to heat is ‘irreversible’

13. Ideal Engine avoid irreversibility carry out engine cycle slowly and with reversible heat flow the reversibility of heat flow motivates the allowed processes

14. Ideal Engine For reversible heat flow, there can be no finite  T between engine and reservoir so during heat transfer we need an isothermal process

15. Ideal Engine If there is a finite  T between engine and reservoir, any heat flow would be irreversible So when there is a finite  T we need an adiabatic process

16. Ideal Engine so we are motivated to consider a cycle with only two processes: isothermal adiabatic called a ‘Carnot cycle’

17. Ideal Engine efficiency: e = ? working substance = ? We’ll calculate e for substance = ideal gas Actually, e is independent of the working substance

18. Carnot Cycle

19. Example engine: internal combustion engine

20. T H is provided by combustion of air-fuel mixture T C is provided by exhaust gases venting to outside air

21. fuel = gasoline working substance = mixture of air and burned fuel this engine doesn’t actually recycle the same mixture in each cycle, but we can idealize with simple models

22. Problem 18-44

23. Refrigerator Refrigerator: like a heat engine, but operating in reverse Reservoirs: hot reservoir at T H cold reservoir at T C

24. Refrigerator refrigerator absorbs heat Q C refrigerator discards heat Q H work W done on refrigerator

25. Refrigerator cyclic process:  U = 0 during one cycle 1st Law: Q = W +  U Thus for each cycle: Q = W

26. Refrigerator So by 1st Law: W = Q = Q H + Q C – |W| = – |Q H | + Q C |Q H | = Q C + |W|

27. Announcements Today: finish Chapter 18 Wednesday: review Thursday: no class (office hours 12:30-2:30) HW 6 (Ch. 15):returned at front Midterms: returned at front (midterm scores: classweb, solutions: E-Res)

28. The Second Law of Thermodynamics

29. 2nd Law of Thermodynamics Four equivalent approaches, in terms of: Engines Refrigerators Carnot Cycle Entropy

30. Heat Engine engine absorbs heat Q H engine discards heat Q C engine does work W engine efficiency :

31. 2nd Law: Engines There are no perfect engines (real engines have e < 1) It is impossible for a system to change heat completely into work, with no other change to the system taking place

32. 2nd Law: Engine Version If 2nd Law weren’t true: ships could move by cooling the ocean cars could move by cooling surrounding air

33. Refrigerator refrigerator absorbs heat Q C refrigerator discards heat Q H work W done on refrigerator

34. 2nd Law: Refrigerator Version There are no perfect refrigerators (for real refrigerators |W|> 0). It is impossible for heat to flow from hot to cold with no other change to the system taking place

35. ‘Engine’ and ‘Refrigerator’ versions of 2nd Law are equivalent

36. (a)If you can build a perfect refrigerator, then you can build a perfect engine (perfect refrigerator) + (real engine) = perfect engine See notes

37. (b)If you can build a perfect engine, then you can build a perfect refrigerator (perfect engine) + (real refrigerator) = perfect refrigerator See notes

38. Engine Efficiency Revisited 2nd Law: Real engines have efficiency e < 1. How close can we get to e = 1? Is there a maximum possible efficiency? Yes!

39. Carnot Cycle

40. 2nd Law: Carnot Cycle Version e < e Carnot The efficiency e of a real engine operating between temperatures T H and T C can never exceed that of a Carnot engine operating between the same T H and T C : Proof

41. Problem 18-44, Revisited

42. Defining Entropy For Carnot cycle: See notes

43. Any reversible cycle = sum of many Carnot cycles For cycle, in limit of infinitesimally close isotherms:

44. Entropy (S) Result for a reversible cycle: This defines a new state variable, entropy (S):

45. Entropy measures the disorder of a system The more heat dQ you add, the higher the entropy increase dS that results The smaller the temperature T, the larger the entropy increase dS that results

46. Entropy (S) For any reversible path from state 1 to 2: This must also equal  S for an irreversible path from 1 to 2, since S is a state variable Exercises 18-20, 18-26 Example 18-8, Problem 18-50

47. Example 18-8: Free Expansion of Ideal Gas

48. Entropy (S) For any reversible path from state 1 to 2: This must also equal  S for an irreversible path from 1 to 2, since S is a state variable Exercises 18-20, 18-26 Example 18-8, Problem 18-50

49. 2nd Law: Entropy Version If we consider all systems taking part in a process, For a reversible process:  S total = 0 For an irreversible process:  S total > 0

50. Laws of Thermodynamics Law state variable Zeroth Law temperature: T 1st Law  U = Q – W internal energy: U 2nd Law entropy: S

51. Announcements Today: finish Chapter 18 Wednesday: review Thursday: no class (office hours 12:30-2:30) HW 6 (Ch. 15):returned at front Midterms: returned at front (midterm scores: classweb, solutions: E-Res)