Chapter 14 The Second Law of Thermodynamics

1. Chapter14 The Second Law of Thermodynamics

2. Main Points of Chapter 14 • Engines and refrigerators • Second law of thermodynamics • Carnot cycle • Other types of engines • Entropy

3. 14-1 Engines and Refrigerators An engine is a device that cyclically transforms thermal energy into mechanical energy. 1. An engine must work in cycles if it is to be useful – otherwise, when one cycle does work the engine stops. 2. A cyclic engine must include more than one thermal reservoir.

4. In a cyclic process, it is impossible to avoid dumping heat. The following situation is impossible: Hot reservoir Qh Engine W = Qh Why is this impossible? We shall see that, for a cyclic process, the Second Law of Thermodynamics prevents perfect heat-to-work conversion.

5. Qh Ql P a A W B b V Every engine has an efficiency, which is a measure of what fraction of the heat flow from the hotter thermal reservoir is converted into work.

6. Qh Ql A Refrigerator is an engine run in reverse P a A W B b V Thermal energy will be taken from a cold reservoir and released to a hotter reservoir Work will be done on the system by surroundings.

7. Qh P Reservoir Th a Qabs A B b Reservoir Tc V W The coefficient of performance

8. 14-2 The Second Law of Thermodynamics The First Law of Thermodynamics tells us that energy is always conserved for a thermal system. When ice cubes are put into hot tea, the ice cubes melt and the tea cools. But there are processes that do not occur spontaneously even though energy would be conserved, such as a lukewarm drink forming ice cubes and becoming hot.

9. 1. Spontaneous processes A crate sliding over an ordinary surface eventually stops. When ice cubes are put into hot tea ,the ice cubes melt and the tea cools. If you puncture a helium-filled balloon in a closed room, the helium gas spreads throughout the room. These processes occur spontaneously.

10. These phenomena can be summarized as follows: • Not all the thermal energy in a thermal system • is available to do work. b. Thermal systems spontaneously change only in certain ways, and in particular, spontaneous heat flow always goes from a body at higher temperature to a body at lower temperature.

11. 2. Statistical Interpretation of spontaneous processes Spontaneous processes: compressed gas in a bottle will expand when bottle is opened, but won’t spontaneously go back in – why not? Suppose a box contains four molecules of gas There are five possible configurations. There are different arrangements of the molecules microstate for one configuration All microstates are equally probable.

12. number of microstates Probability configuration 1 L4，R0 L3，R1 4 L2，R2 6 L1，R3 4 L0，R4 1 L2,R2 is the most probable configuration

13. For large values of N there are extremely large numbers of microstates, but nearly all the microstates belong to the configuration in which the molecules are divided equally between the two halves of the box. Free expansion of an ideal gas is a spontaneous process. Its directionality is determined by what takes you to the state with the highest probability. It is a irreversible processes. The reverse process could happen, but the probability is impossibly small

14. P P1 P2 V o V1 V2 3 The Second Law of Thermodynamics There are many ways of expressing the second law of thermodynamics. The Kelvin form. It is impossible to construct a cyclic engine that converts thermal energy from a body into equivalent amount of mechanical work without a further change in its surroundings. Thermal energy can be converted into equivalent amount of mechanical work , but the engine is not a cyclic.

15. W Q T air or ocean If the Kelvin formulation were not true It is the second law that limits the efficiencies of heat engines to values less than 100%.

16. The Clausius form. It is impossible to construct a cyclic engine whose only effect is to transfer thermal energy from a colder body to a hotter body. Refrigerator is a device that transfer thermal energy from a colder body to a hotter body, but work must be done on it.

17. Hot reservoir Q W+ Ql Ql W E F Ql Ql Cold reservoir Two formulations are equivalent If the Kelvin formulation were not true It shows that the Clausius formulation would be not true

18. Hot reservoir Q Q W W F E Q Ql Q-Ql Cold reservoir If the Clausius formulation were not true It shows that the Kelvin formulation would be not true

19. 14-3 The Carnot cycle • A minimal version of an engine has two reservoirs at different temperatures Th and Tc, and follows a reversible cycle known as the Carnot cycle. • The Carnot cycle has four steps: • Isothermal expansion • Adiabatic expansion • Isothermal compression • Adiabatic compression

20. How the cycle might be realized

21. p A Qh 1 4 B Th 2 D 3 C TC QC V Va Vd Vb Vc Step 1: Step 3: Step 2: Step 4:

22. p A Qh 1 4 B Th 2 D 3 C TC QC V Va Vd Vb Vc A Carnot cycle is more efficient if the two temperatures are far apart

23. ACT Three Carnot engines operate between temperature limits of (a) 400 and 500 K, (b) 500 and 600 K, and (c) 400 and 600 K. Each engine extracts the same amount of energy per cycle from the high-temperature reservoir. Rank the magnitudes of the work done by the engines per cycle, greatest first. (c), (a), (b),

24. The importance of Carnot engine Two results: • All Carnot cycles that operate between the same • two temperatures have the same efficiency. The • efficiency of a Carnot cycle does not depend on • the use of an ideal gas. 2. The Carnot engine is the most efficient engine possible that operates between any two given temperatures.

25. To demonstrate the first result We show if it were not true we would violate the second law of dynamics Suppose the efficiency of Carnot engine A is higher than Carnot engine B. let B run in reverse. Such arrangement violates the Kelvin form the second law of thermodynamics

26. Th QB QA DW W Carnot engine irreversible engine Qc Qc Tc To demonstrate the second result We show if an irreversible engine is more efficient than a Carnot engine we would violate the second law of dynamics. let Carnot engine run in reverse. The net effect is to have a net heat flow from Thwith work done, which violates the Kelvin statement. The argument can not be reversed, because irreversible engine is not reversible.

27. ACT The heat engine below is: 1) a reversible (Carnot) heat engine. 2) an irreversible heat engine. 3) a hoax. 4) none of the above.

28. Example Water near the surface of a tropical ocean has a temperature of 298.2 K (25.0 °C), while water 700 m beneath the surface has a temperature of 280.2 K (7.0 °C). It has been proposed that the warm water be used as the hot reservoir and the cool water as the cold reservoir of a heat engine. Find the maximum possible efficiency for such an engine. Solution The maximum possible efficiency is the efficiency of a Carnot engine Using TH = 298.2 K and TC = 280.2 K

29. P dT=0 C B A V o • Example Prove that two reversible adiabatic paths cannot intersect Proof Assume that two reversible adiabatic paths intersect at point A Draw an isotherm which intersect with adiabats at B and C respectively. This cycle would violate the second law of thermodynamics. So two reversible adiabatic paths cannot intersect.

30. ACT Is it possible to cool a house by leaving a refrigerator door open? What would be the net effect if you were to leave the door open? Solution No A refrigerator is essentially a heat engine operated in reverse. It takes in a certain amount of work (W), in the form of electrical energy from the power outlet, extracts a certain amount of heat (Qc) from the low-temperature source (the interior of the refrigerator), and released some heat (Qh) into the high temperature environment (the house). According to the first law of thermodynamics Qh= Qc+ W. So, the net effect is that the refrigerator consumes some electrical energy and turns it into heat that is released into the house. Keeping the refrigerator door open only makes the house even warmer over all.

31. 14-4 Other Types of Engines Any closed curve in the P-V diagram represents a reversible cycle The Stirling Engine 1mol ideal gas

32. Otto cycle (Gasoline Engine) P combustion 3 1mol ideal gas adiabatic exhaust 2 4 adiabatic 1 V o V2 V1

33. P 3 adiabatic 2 4 adiabatic 1 V o V2 V1 compression ratio

34. There are other possible cycles; here are a few:

35. A • Example a reversible cycle P T1 Isotherms :AB,CD,EF B C T2 D F E T3 Adiabats: BC,DE,FA G V Find o Solution To extrapolate Adiabat BC we have two Carnot cycles

36. Heat Pumps Heat pumps and refrigerators are engines run in reverse: Refrigerator removes heat from cold reservoir, puts it into surroundings, keeping food in reservoir cold. Heat pump takes energy from cold reservoir and puts it into a room or house, thereby warming it. In either case, energy must be added!

37. W W In winter In summer Air conditioner Heat pump It pumps heat “uphill” from a lower temperature to a higher temperature, just as a water pump forces water uphill from a lower elevation to a higher elevation. The coefficient of performance It is easier to transfer thermal energy from the cold ground to a warm house if the temperature difference is small

38. Example There is a 70 W heat leak from a room at temperature 22 °C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10 °C? Solution For the fridge, Qc must exactly compensate the heat leak. So Qc = 70 J per second For Carnot cycle Power required = 8.5 W

39. ACT An ideal or Carnot heat pump is used to heat a house to a temperature of TH = 294 K (21 °C). How much work must be done by the pump to deliver QH = 3350 J of heat into the house when the outdoor temperature TC is (a) 273 K (0 °C) and (b) 252 K (-21 °C)? Solution For Carnot cycle It is more difficult to transfer thermal energy from the cold ground to a warm house if the temperature difference is larger.

40. Example Work from a hot brick Heat a brick to 400 K. Connect it to a Carnot Engine. How much work can we extract if the cold reservoir is 300 K? The brick has a constant heat capacity of C = 1 J/K. Solution Did you use the relation: W = Qh(1 -Tc/Th) ? If so, you missed that the brick was cooling during the process. Suppose the temperature of the brick was T dW = dQh (1 -Tc/T) =-CdT (1 -Tc/T)

41. 14-5 Entropy and the Second Law 1. Entropy as a Thermodynamic Variable In a Carnot cycle To generalize to any reversible cycle Clausius’ equation

42. The integral depends only on the initial and final states, and not on the path rev rev There must be a function S that depends only on the state of the gas and not on how it got that way; this is called entropy. Definition of Entropy:

43. rev • Example Calculate the entropy change of 1 mol of ideal gas that undergoes an isothermal transformation from an initial state to a final state Solution • Act Calculate the entropy change of 1 mol of ideal gas that undergoes an reversible adiabatic transformation

44. A • Example a reversible cycle P T1 Isotherms :AB,CD,EF B C T2 D F E T3 Adiabats: BC,DE,FA G V Find o Solution Clausius’ equation

45. 2. Entropy of an Ideal Gas To calculate the entropy change of 1mol ideal gas from initial state (pi.Vi,Ti) to final state (pf,Vf,Tf) We design a reversible process that takes the system from the initial state to the final state.

46. isochore isotherm (pf,Vf,Tf) (pi,Vi,Ti) (p,Vf,Ti) The general form of the entropy for ideal gas

48. P T S V

49. ACT Point i in Figure below represents the initial state of an ideal gas at temperature T. Taking algebraic signs into account, rank the entropy changes that the gas undergoes as it moves, successively and reversibly, from point i to points a, b, c, and d, greatest first. b, a, c, d

50. T S TH TC S2 S1 T-S diagram For a Carnot cycle Q=W