Thermodynamics

1. Thermodynamics Chapter 15

2. Thermodynamics Thermodynamics is the set of ideas and rules that explain how to convert heat energy into useful work. Heat engines provide power for transportation. Heat engines powering electric generators provide our electricity.

3. Life with and without heat engines

4. Zeroth law of thermodynamics Objects that have the same temperature are in thermal equilibrium with each other. Thermometers can be used to detect equilibrium.

5. input heat Work output unused heat Heat engine cycle Goal: Use heat energy to produce mechanical work. Input: Hot heat energy Output: Mechanical work Unused heat energy Heat engine always starts each cycle with the same amount of internal energy. All of the energy that goes into the engine during the cycle must go out of the engine by the end of the cycle. Conservation of energy requires (energy in) = (energy out) | QH | = | QC | + | W |

6. Thermodynamic system Heat energy and mechanical work can add or subtract energy to a part of the universe we call the system. The system is a gas (usually air). The state of the system is specified by the pressure, volume, temperature, number of moles, and internal energy of the gas.

7. internal energy U internal energy U HeatQ system system HeatQ Heat energy added to a system Positive heat input Heat flow into a system increases the system’s internal energy. Negative heat flowExhaust heat Heat flow out of a system decreases the system’s internal energy. Heat is positive when the system gainsheat. Heat is negative when the system losesheat.

8. internal energy U internal energy U WorkW system system WorkW Negative work Work done on the system Work done by a system Positive work output Work done by the system Work done by a system decreases the system’s internal energy. Work done on a system increases the system’s internal energy. Work is positive when work is done by the system. Work is negative when work is done on the system.

9. internal energy U WorkW HeatQ system - Work output + - Heat input + WorkW HeatQ First law of thermodynamics First law is based on the conservation of energy principle. Internal energy of a system changes due to heat and work. Positive work is done by a gas during expansion.Negative work is done by a gas during compression. (During compression work is done on the gas.) The state of the gas is specified by n, T, P, and V.

10. internal energy U WorkW HeatQ system - Work output + - Heat input + WorkW HeatQ Example 1 Positive and Negative Work a) A system gains 1500 J of heat and does 2200 J of work. Find the change in the internal energy. 2200 J 0 J 0 J 1500 J

11. internal energy U WorkW HeatQ system - Work output + - Heat input + WorkW HeatQ Example 1 Positive and Negative Work b) A system gains 1500 J of heat and 2200 J of work is on the system. Find the change in internal energy. 0 J 2200 J 0 J 1500 J

12. The amount of work done each cycle is equal to the area inside the P vs. V cycle diagram. Heat engine pressure vs. volume cycles To continually produce mechanical work, heat engines use a repeating cycle of expansion and compression of a gas.

13. Four important thermodynamic processes isobaric:constant pressure isochoric:constant volume isothermal:constant temperature adiabatic:no heat transfer

14. Displacement s Area A Pressure P Isobaric process: Isobaric process: constant pressure

15. internal energy U WorkW HeatQ system Positive heat input Isobaric process example How much work is done by a gas that expands from 0.2 m3 to 0.6 m3 at a constant 250,000 Pa pressure. Isobaric process: +100,000 J Since PV=nRT, when V increases, heat makes T increase.

16. internal energy U HeatQ Positive heat input system Isochoric process: Isochoric process: constant volume No displacement  no work is done. Since PV=nRT, when P increases, heat makes T increase.

17. Isothermal process: Isothermal process: Isothermal process: constant temperature Since PV=nRT, when T stays the same, PV also stays the same, so P decreases when V increases.

18. Isothermal process: example 5 Two moles of 298K argon gas expand isothermally from 0.025m3 to 0.050m3. a) Find the work done by the gas. b) Find the change in internal energy of the gas. c) Find the heat supplied to the gas. Internal energy doesn't change for an isothermal process. First law of thermodynamics Heat added equals work done for an isothermal process.

19. internal energy U WorkW HeatQ system 3435 J Work output + Heat input + 3435 J Isothermal process: example 5 (continued) Internal energy stays the same for an isothermal process. First law of thermodynamics For an isothermal process, heat added equals work done .

20. Adiabatic process: Adiabatic gas law Adiabatic process: no heat transfer An adiabatic process has no heat transfer because a) heat transfer blocked by insulation, or b) process happens so quickly that there is not enough time for heat to transfer. Temperature drops during adiabatic expansion.Temperature rises during adiabatic compression.

21. internal energy U system -114120 J WorkW Negative work by system (Work done on the system) Adiabatic process: example 6 mol of air initially at 101,000 Pa, 0.15 m3, and 300 K was compressed in a diesel engine to a final pressure of 9,123,000 Pa and volume of 0.01 m3. How much work was done by the gas system during this compression? Adiabatic process: What was the final temperature? Hot enough to ignite the fuel.

22. Isochoric process: Adiabatic process: Isobaric process: Summary internal energy change = net heat input - net work output isobaric: constant pressure isochoric: constant volume no motion  no work isothermal: constant temperatureΔU = 0 Isothermal process: adiabatic: no heat transfer Q = 0

23. Second law of thermodynamics Heat energy transfer has a natural direction.Hot Cold Heat flow statement of the second law Heat energy flows spontaneously from a substance at a higher temperature to a substance at a lower temperature. Heat energy does not flow spontaneously in the opposite direction.

24. input heat Work output unused heat Heat engine Heat engines take advantage of the natural tendency of heat to flow from hot to cold and converts part of the heat flow into mechanical energy. It is impossible to convert 100% of the heat into mechanical energy. Efficiency measures how much of the heat flow gets converted into mechanical energy. Conservation of energy requires (energy in) = (energy out) | QH | = | QC | + | W |

25. input heat Goal: produce work output We pay for the input energy Work unused heat Efficiency Efficiencyof a heat engine is defined as the ratio of the work output to the heat input. Efficiency tells us the fraction of the input heat energy that produces work. QH = 1 QH W = e QH QC = (1-e) QH It is usually easier to use the efficiency as a percent rather than as a decimal fraction.

26. input heat Work unused heat Efficiency example What is the efficiency of an engine that uses 1000 J of heat input to produce 150 J of work? 100% How much heat was rejected to the environment? 15% 85%

27. input heat Work output unused heat Example 6 An Automobile Engine An automobile engine has an efficiency of 22% and produces 2510 J of work. How much input heat is supplied? How much unused heat is exhausted? 100% solve using proportions 2510 J 22% 78% Check: QH = QC + W = 8899 J + 2510 J = 11, 409 J

28. Second law of thermodynamics A reversible process is one in which both the system and the environment can be returned to exactly the same states they were in before the process occurred. Reversible processes describe idealized, frictionless systems. All natural, spontaneous processes are irreversible. Carnot statement of the second law All reversible engines operating between the same two temperatures have the same efficiency. No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures.

29. Carnot heat engine: maximum possible efficiency The Carnot heat engine is an ideal, reversible, heat engine. All of the hot heat originates from a single temperature, and all the cold heat is rejected to a single temperature. Temperatures must be in kelvins.

30. Carnot heat engine example What is the Carnot efficiency of a heat engine that operates between the temperatures of 1000 K and 400 K?

31. Refrigeration process Spontaneous heat flow is from hot to cold. Refrigerators, air conditioners, and heat pumps move heat energy from cold to hot by running a heat engine backwards.

32. Air conditioner Refrigerator Input work used to move cold inside heat to the hot outside.

33. Conceptual Example 9: Is it possible to cool your kitchen by leaving the refrigerator door open or to cool your room by putting a window air conditioner on the floor of your room?

34. Coefficient of performance Refrigerator or air conditioner

35. Coefficient of performance example An air conditioner has a coefficient of performance of 4. How much input work will it take to remove 1000 J from the room? How much heat will be delivered to the environment? 5 1 4 1000 J so W = 250 J

36. Seasonal Energy Efficiency Ratio (SEER) Efficiency rating for air conditioners in the USA. SEER = (coefficient of performance) x 3.792 Starting 2006 the minimum SEER was 13 for new units.

37. air conditioner Heat pump A heat pump uses input work to move cold outside heat to the hotter inside of a building. A heat pump is an air conditioner that is cooling the outdoors instead of the building.

38. Entropy The universe has more order when there are hot and cold objects compared to when all objects are at the same temperature. As heat spontaneously flows from hot to cold, the universe goes from a more ordered state to a less ordered state. The less ordered state is called a more random state. The universe is going in the direction of increasing randomness. Entropy is how we measure the amount of randomness in the universe. For any process, entropy either stays the same or increases.

39. entropy change entropy reversible absolute temperature Entropy Entropy changes when heat energy Q flows from one object to another. Q is positive when heat is added. Q is negative when heat is lost.

40. Example 11 Entropy Change of the Universe 1200 J of heat spontaneously flowing through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount by which this process changes the entropy of the universe. Heat is lost from the hot object. Heat is added to the cold object. A positive change means that the entropy of the universe increased.

41. The Entropy Change of the Universe 1200 J of heat spontaneously flowing through a copper rod from a hot reservoir at 350.1 K to a cold reservoir at 350 K. Determine the amount by which this process changes the entropy of the universe. A reversible process doesn't alter the entropy of the universe. TH = 350.1 K Changing TH to 349.9 K would reverse the heat flow so this process is nearly reversible. The small change shows the entropy of the universe almost stayed the same.

42. Any irreversible process increases the entropy of the universe. Irreversible process: A reversible process doesn't change the entropy of the universe. Reversible process: Entropy Entropy statement of the second law The total entropy of the universe stays the same when a reversible process occurs, but always increases when an irreversible process occurs. Entropy is on the rise !!!!

43. The End