1. 11/12/2013PHY 113 C Fall 2013 -- Lecture 211 PHY 113 C General Physics I 11 AM - 12:15 pM MWF Olin 101 Plan for Lecture 21: Chapter 20: Thermodynamics 1. Heat and internal energy 2.Specific heat; latent heat 3.Thermodynamic work 4.First “law” of thermodynamics

2. 11/12/2013 PHY 113 C Fall 2013 -- Lecture 212

3. 11/12/2013PHY 113 C Fall 2013 -- Lecture 213 In this chapter T ≡ temperature in Kelvin or Celsius units

4. 11/12/2013PHY 113 C Fall 2013 -- Lecture 214 Webassign question from Assignment #18 A rigid tank contains 1.50 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 28.5 atm to 5.10 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.

5. 11/12/2013PHY 113 C Fall 2013 -- Lecture 215 T1T1 T2T2 Q Heat – Q -- the energy that is transferred between a “system” and its “environment” because of a temperature difference that exists between them. Sign convention: Q<0 if T 1 < T 2 Q>0 if T 1 > T 2

6. 11/12/2013PHY 113 C Fall 2013 -- Lecture 216 Heat withdrawn from system Thermal equilibrium – no heat transfer Heat added to system

7. 11/12/2013PHY 113 C Fall 2013 -- Lecture 217 Units of heat: Joule Other units: calorie = 4.186 J Kilocalorie = 4186 J = Calorie Note: in popular usage, 1 Calorie is a measure of the heat produced in food when oxidized in the body

8. 11/12/2013PHY 113 C Fall 2013 -- Lecture 218 iclicker question: According to the first law of thermodynamics, heat and work are related through the “internal energy” of a system and generally cannot be interconverted. However, we can ask the question: How many times does a person need to lift a 500 N barbell a height of 2 m to correspond to 2000 Calories (1 Calorie = 4186 J) of work? A.4186 B.8372 C.41860 D.83720 E.None of these

9. 11/12/2013PHY 113 C Fall 2013 -- Lecture 219 Heat capacity: C = amount of heat which must be added to the “system” to raise its temperature by 1K (or 1 o C). Q = C  T Heat capacity per mass: C=mc Heat capacity per mole (for ideal gas): C=nC v C=nC p

10. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2110 Some typical specific heats MaterialJ/(kg· o C)cal/(g· o C) Water (15 o C)41861.00 Ice (-10 o C)22200.53 Steam (100 o C)20100.48 Wood17000.41 Aluminum 9000.22 Iron 4480.11 Gold 1290.03

11. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2111 iclicker question: Suppose you have 0.3 kg of hot coffee (at 100 o C). How much heat do you need to remove so that the temperature is reduced to 40 o C? (Note: for water, c=1000 kilocalories/(kg o C)) A.300 kilocalories (1.256 x 10 6 J) B.12000 kilocalories (5.023 x 10 7 J) C.18000 kilocalories (7.535 x 10 7 J) D.30000 kilocalories (1.256 x 10 8 J) E.None of these Q=m c  T = (0.3)(1000)(100-40) = 18000 kilocalories

12. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2112 Q=333J Heat and changes in phase of materials Example: A plot of temperature versus Q added to 1g = 0.001 kg of ice (initially at T=-30 o C) Q=2260J

13. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2113 TiTi TfTf C/LQ A-30 o C0 o C2.09 J/ o C62.7 J B0 o C 333 J C0 o C100 o C4.186 J/ o C418.6 J D100 o C 2260 J E100 o C120 o C2.01 J/ o C40.2 J m=0.001 kg

14. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2114 Some typical latent heats MaterialJ/kg Ice  Water (0 o C) 333000 Water  Steam (100 o C) 2260000 Solid N  Liquid N (63 K) 25500 Liquid N  Gaseous N 2 (77 K) 201000 Solid Al  Liquid Al (660 o C) 397000 Liquid Al  Gaseous Al (2450 o C) 11400000

15. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2115 iclicker question: Suppose you have 0.3 kg of hot coffee (at 100 o C) which you would like to convert to ice coffee (at 0 o C). What is the minimum amount of ice (at 0 o C) you must add? (Note: for water, c=4186 J/(kg o C) and for ice, the latent heat of melting is 333000 J/kg. ) A.0.2 kg B.0.4 kg C.0.6 kg D.1 kg E.more

16. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2116 Review Suppose you have a well-insulated cup of hot coffee (m=0.3kg, T=100 o C) to which you add 0.3 kg of ice (at 0 o C). When your cup comes to equilibrium, what will be the temperature of the coffee?

17. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2117 Work defined for thermodynamic process the “system” In most text books, thermodynamic work is work done on the “system”

18. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2118

19. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2119 Thermodynamic work Sign convention: W > 0 for compression W < 0 for expansion

20. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2120 Work done on system -- “Isobaric” (constant pressure process) P (1.013 x 10 5 ) Pa PoPo ViVi VfVf W= -P o (V f - V i )

21. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2121 Work done on system -- “Isovolumetric” (constant volume process) P (1.013 x 10 5 ) Pa ViVi PiPi PfPf W=0

22. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2122 Work done on system which is an ideal gas “Isothermal” (constant temperature process) P (1.013 x 10 5 ) Pa PiPi ViVi VfVf For ideal gas: PV = nRT

23. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2123 Work done on a system which is an ideal gas: “Adiabatic” (no heat flow in the process process) P (1.013 x 10 5 ) Pa PiPi ViVi VfVf Q i  f = 0 For ideal gas: PV  = P i V i 

24. 11/12/2013PHY 113 C Fall 2013 -- Lecture 21 24

25. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2125 Thermodynamic processes: Q: heat added to system (Q>0 if T E >T S ) W: work done on system (W<0 if system expands

26. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2126 iclicker question: What happens to the “system” when Q and W are applied? A.Its energy increases B.Its energy decrease C.Its energy remains the same D.Insufficient information to answer question

27. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2127 Internal energy of a system E int (T,V,P….) The internal energy is a “state” property of the system, depending on the instantaneous parameters (such as T, P, V, etc.). By contrast, Q and W describe path-dependent processes.

28. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2128 First law of thermodynamics: EiEi EfEf Q W

29. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2129 Applications of first law of thermodyamics System  ideal gas pressure in Pascals volume in m 3 # of moles temperature in K 8.314 J/(mol K)

30. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2130 Ideal gas -- continued

31. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2131 iclicker question: We are about to see several P-V diagrams. Why is this helpful? A.It will help us analyze the thermodynamic work B.Physicists like nice graphs C.It is not actually helpful

32. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2132 Constant volume process on an ideal gas P V P f V f =V i P i V i

33. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2133 Constant pressure process on an ideal gas P V P f = P i V f P i V i

34. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2134 Constant temperature process on an ideal gas

35. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2135 Consider the process described by A  B  C  A iclicker exercise: What is the net change in total energy? A.0 B.12000 J C.Who knows?

36. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2136 Consider the process described by A  B  C  A iclicker exercise: What is the net heat added to the system? A.0 B.12000 J C.Who knows?

37. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2137 Consider the process described by A  B  C  A iclicker exercise: What is the net work done on the system? A.0 B.-12000 J C.Who knows?

38. 11/12/2013PHY 113 C Fall 2013 -- Lecture 2138 Webassign problem from Assignment #19