2. Chapter 20 - Thermodynamics A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

3. A THERMODYNAMIC SYSTEM A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.) Work done on gas or work done by gas

4. INTERNAL ENERGY OF SYSTEM The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system. Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.

5. TWO WAYS TO INCREASE THE INTERNAL ENERGY,  U. HEAT PUT INTO A SYSTEM (Positive) +U+U+U+U WORK DONE ON A GAS (Positive)

6. WORK DONE BY EXPANDING GAS:  W is positive -  U Decrease Decrease TWO WAYS TO DECREASE THE INTERNAL ENERGY,  U. HEAT LEAVES A SYSTEM  Q is negative  Q is negative Q out hot W out hot

7. THERMODYNAMIC STATE The STATE of a thermodynamic system is determined by four factors: Absolute Pressure P in PascalsAbsolute Pressure P in Pascals Temperature T in KelvinsTemperature T in Kelvins Volume V in cubic metersVolume V in cubic meters Number of moles, n, of working gas Number of moles, n, of working gas

8. THERMODYNAMIC PROCESS Increase in Internal Energy,  U. Initial State: P 1 V 1 T 1 n 1 Final State: P 2 V 2 T 2 n 2 Heat input Q in W out Work by gas

9. The Reverse Process Decrease in Internal Energy,  U. Initial State: P 1 V 1 T 1 n 1 Final State: P 2 V 2 T 2 n 2 Work on gas Loss of heat Q out W in

10. THE FIRST LAW OF THERMODYAMICS: The net heat put into a system is equal to the change in internal energy of the system minus any work done BY the system.The net heat put into a system is equal to the change in internal energy of the system minus any work done BY the system.  U =  Q +  W  final - initial) Conversely, the work done ON a system is equal to the change in internal energy minus any the heat gained in the process.Conversely, the work done ON a system is equal to the change in internal energy minus any the heat gained in the process.

11. SIGN CONVENTIONS FOR FIRST LAW Heat Q input is positive Heat Q input is positive  U=  Q +  W  final - initial) Heat OUT is negative Heat OUT is negative Work ON a gas is positive Work BY a gas is negative +Q in -W out  U +W in -Q out  U

12. APPLICATION OF FIRST LAW OF THERMODYNAMICS Example 1:In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of expansion on the piston. What is the change in internal energy of the system? Example 1: In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of expansion on the piston. What is the change in internal energy of the system?  U =  Q +  W Apply First Law: Q in 400 J W out = -120 J

13. Example 1 (Cont.): Apply First Law    U = +280 J Q in 400 J W out = -120 J  U =  Q +  W = (+400 J) + (-120 J) = (+400 J) + (-120 J) = +280 J = +280 J  W is negative: -120 J (Work OUT)  U =  Q +  W  Q is positive: +400 J (Heat IN)

14. Example 1 (Cont.): Apply First Law    U = +280 J The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J Q in 400 J W out =120 J The increase in internal energy is: Energy is conserved:

15. FOUR THERMODYNAMIC PROCESSES: Isochoric Process:  V = 0,  W = 0 Isobaric Process:  P = 0 Isothermal Process:  T = 0,  U = 0 Adiabatic Process:  Q = 0 Isochoric Process:  V = 0,  W = 0 Isobaric Process:  P = 0 Isothermal Process:  T = 0,  U = 0 Adiabatic Process:  Q = 0  Q =  U +  W

16. Absorbs heat Q hot Performs work W out Rejects heat Q cold A heat engine is any device which through a cyclic process: Cold Res. T C Engine Hot Res. T H Q hot W out Q cold HEAT ENGINES

17. THE SECOND LAW OF THERMODYNAMICS 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)! W out Cold Res. T C Engine Hot Res. T H Q hot Q cold

18. THE SECOND LAW OF THERMODYNAMICS Cold Res. T C Engine Hot Res. T H 400 J 300 J 100 J A possible engine. An IMPOSSIBLE engine. Cold Res. T C Engine Hot Res. T H 400 J

19. EFFICIENCY OF AN ENGINE Cold Res. T C Engine Hot Res. T H QHQH W QCQC The efficiency of a heat engine is the ratio of the net work done W to the heat input Q H. e = 1 - QCQHQCQH e = = WQHWQH Q H - Q C Q H

20. EFFICIENCY EXAMPLE Cold Res. T C Engine Hot Res. T H 800 J W 600 J An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency? e = 1 - 600 J 800 J e = 1 - QCQHQCQH e = 25% Question: How many joules of work is done?

21. EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine) For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T. e = 1 - TCTHTCTH e = T H - T C T H Cold Res. T C Engine Hot Res. T H QHQH W QCQC

22. 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? e = 1 - TCTHTCTH 300 K 500 K e = 40% Actual e = 0.5e i = 20% e = WQHWQH W = eQ H = 0.20 (600 J) Work = 120 J

23. REFRIGERATORS A refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat into hot reservoir. W in + Q cold = Q hot W IN = Q hot - Q cold Cold Res. T C Engine Hot Res. T H Q hot Q cold W in

24. 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. If this were possible, we could establish perpetual motion! Cold Res. T C Engine Hot Res. T H Q hot Q cold

25. COEFFICIENT OF PERFORMANCE Cold Res. T C Engine Hot Res. T H QHQH W QCQC The COP (K) of a heat engine is the ratio of the HEAT Q c extracted to the net WORK done W. K = T H T H - T C For an IDEAL refrigerator: QCWQCW K = = Q H Q H - Q C

26. 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 Q H ? Cold Res. T C Eng ine Hot Res. T H 800 J W QHQH 500 K 400 K K = 400 K 400 K 500 K - 400 K T C T H - T C = C.O.P. (K) = 4.0

27. COP EXAMPLE (Cont.) Next we will find Q H by assuming same K for actual refrigerator (Carnot). Cold Res. T C Eng ine Hot Res. T H 800 J W QHQH 500 K 400 K K = Q C Q H - Q C Q H = 1000 J 800 J 800 J Q H - 800 J =4.0

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

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

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

31. Summary (Cont.) 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. Cold Res. T C Engine Hot Res. T H Q hot Q cold W out Not only can you not win (1st law); you can’t even break even (2nd law)!

32. Summary (Cont.) The efficiency of a heat engine: e = 1 - QCQHQCQH TCTHTCTH The coefficient of performance of a refrigerator: