Ch 15 Thermodynamics

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Ch 15 Thermodynamics

  1. 1. Chapter 15 Thermodynamics
  2. 2. AP Learning Objectives Kinetic theory and thermodynamics  Ideal gases  Students should know how to apply the ideal gas law and thermodynamic principles, so they can:  Relate the pressure and volume of a gas during an isothermal expansion or compression.  Relate the pressure and temperature of a gas during constant-volume heating or cooling, or the volume and temperature during constant-pressure heating or cooling.  Calculate the work performed on or by a gas during an expansion or compression at constant pressure.  Understand the process of adiabatic expansion or compression of a gas.  Identify or sketch on a PV diagram the curves that represent each of the above processes.
  3. 3. AP Learning Objectives Laws of Thermodynamics  Students should know how to apply the first law of thermodynamics, so they can:  Relate the heat absorbed by a gas, the work performed by the gas, and the internal energy change of the gas for any of the processes above.  Relate the work performed by a gas in a cyclic process to the area enclosed by a curve on a PV diagram.  Students should understand the second law of thermodynamics, the concept of entropy, and heat engines and the Carnot cycle, so they can:  Determine whether entropy will increase, decrease, or remain the same during a particular situation.  Compute the maximum possible efficiency of a heat engine operating between two given temperatures.  Compute the actual efficiency of a heat engine.  Relate the heats exchanged at each thermal reservoir in a Carnot cycle to the temperatures of the reservoirs.
  4. 4. Table of Contents 1. Thermodynamic Systems & Surroundings 2. The Zeroth Law of Thermodynamics 3. The First Law of Thermodynamics 4. Thermal Processes 5. Thermal Processes Using an Ideal Gas 6. Specific Heat Capacities 7. The Second Law of Thermodynamics 8. Heat Engines 9. Carnot’s Principle and the Carnot Engine 10. Refrigerators, Air Conditioners, and Heat Pumps 11. Entropy 12. The Third Law of Thermodynamics
  5. 5. Chapter 15 Thermodynamics Section 1: Thermodynamic Systems & Surroundings
  6. 6. Thermodynamic Systems and Their Surroundings  Thermodynamics is the branch of physics that is built upon the fundamental laws that heat and work obey.  The collection of objects on which attention is being focused is called the system, while everything else in the environment is called the surroundings.  Walls that permit heat flow are called diathermal walls, while walls that do not permit heat flow are called adiabatic walls.  To understand thermodynamics, it is necessary to describe the state of a system.
  7. 7. Chapter 15 Thermodynamics Section 2: The Zeroth Law of Thermodynamics
  8. 8. Zeroth Law of Thermodynamics  The forgotten Law of Science  Two systems are said to be in thermal equilibrium if there is no heat flow between them when they are brought into contact.  Temperature is the indicator of thermal equilibrium in the sense that there is no net flow of heat between two systems in thermal contact that have the same temperature.
  9. 9. Two systems individually in thermal equilibrium with a third system are in thermal equilibrium with each other. The Zeroth Law of Thermodynamics
  10. 10. 15.2.1. Which one of the following situations is described by the zeroth law of thermodynamics? a) An air conditioner transfers heat from the inside of a house to the outside of the house. b) A monatomic gas is held within a container that has a moveable piston. The gas absorbs heat from the surroundings and expands at constant pressure and temperature. c) A container with adiabatic walls holds boiling water. A thermometer is calibrated by inserting it into the boiling water and allowing it to reach thermal equilibrium with the water. d) A pot contains oil at 175 °C. When frozen sliced potatoes are dropped into the oil, heat is transferred from the oil to the potatoes. e) A physicist removes energy from a system in her laboratory until it reaches a temperature of 3 × 10−10 K, a temperature very close to (but still greater than) absolute zero.
  11. 11. Chapter 15 Thermodynamics Section 3: The First Law of Thermodynamics
  12. 12.  Suppose that a system gains heat Q and that is the only effect occurring.  Consistent with the law of conservation of energy, the internal energy of the system changes:  Heat is positive when the system gains heat and negative when the system loses heat. QUUU if =−=∆ Effect of Heat on a System
  13. 13.  If a system does work W on its surroundings and there is no heat flow, conservation of energy indicates that the internal energy of the system will decrease:  Work is positive when it is done on the system and negative when it is done by the system. WUUU if =−=∆ Effect of Work on a System
  14. 14. The First Law of Thermodynamics  The internal energy of a system changes due to heat and work:  Heat is positive when the system gains heat and negative when the system loses heat.  Work is positive when it is done on the system and negative when it is done by the system. if UUU −=∆ WQU +=∆
  15. 15. Example 1 Positive and Negative Work In part a of figure, the system gains 1500J of heat and 2200J of work is done by the system on its surroundings. In part b, the system also gains 1500J of heat, but 2200J of work is done on the system. In each case, determine the change in internal energy of the system. (a) (b) WQU +=∆ WQU +=∆ ( ) ( )J2200J1500 −++=∆U J700−= ( ) ( )J2200J1500 +++=∆U J3700+=
  16. 16. Example 2 An Ideal Gas The temperature of three moles of a monatomic ideal gas is reduced from 540K to 350K as 5500J of heat flows into the gas. Find (a) the change in internal energy and (b) the work done by the gas. TnRU ∆=∆ 2 3 WQUUU if +=−=∆ if nRTnRTU 2 3 2 3 −=∆ QUW −∆= (a) (b) ( ) ( )( )( )K540K350KmolJ31.8mol0.32 3 −⋅=∆U J7100−= J5500J7100 −−= J12600−=
  17. 17. 15.3.1. An insulated container is filled with a mixture of water and ice at zero °C. An electric heating element inside the container is used to add 1680 J of heat to the system while a paddle does 450 J of work by stirring. What is the increase in the internal energy of the ice-water system? a) 450 J b) 1230 J c) 1680 J d) 2130 J e) zero J
  18. 18. 15.3.2. The internal energy of a system increases during some time interval. Which one of the following statements concerning this situation must be true? a) The increase in internal energy indicates that work was done on the system. b) The increase in internal energy indicates that heat was added to the system. c) The increase in internal energy indicates that work was done by the system. d) The increase in internal energy indicates that heat was removed from the system. e) The information given is insufficient to indicate the reason for the increase.
  19. 19. 15.3.3. A gas is enclosed in a cylinder by a piston. The volume of the gas is then reduced to one half its original value by applying a force to the piston. Which one of the following statements concerning the internal energy of the gas is true? a) The internal energy of the gas will decrease. b) The internal energy of the gas will increase. c) The internal energy of the gas will neither increase nor decrease. d) The internal energy of the gas will equal the work done in moving the piston. e) The internal energy of the gas may increase, decrease, or remain the same depending on the amount of heat that is gained or lost by the gas.
  20. 20. 15.3.4. Which one of the following statements is not consistent with the first law of thermodynamics? a) The internal energy of a finite system must be finite. b) An engine may be constructed such that the work done by the machine exceeds the energy input to the engine. c) An isolated system that is thermally insulated cannot do work on its surroundings nor can work be done on the system. d) The internal energy of a system decreases when it does work on its surroundings and there is no flow of heat. e) An engine may be constructed that gains energy while heat is transferred to it and work is done on it.
  21. 21. Chapter 15 Thermodynamics Section 4: Thermal Processes
  22. 22. A quasi-static process is one that occurs slowly enough that a uniform temperature and pressure exist throughout all regions of the system at all times. isobaric: constant pressure isochoric: constant volume isothermal: constant temperature adiabatic: no transfer of heat Thermal Processes
  23. 23. An isobaric process is one that occurs at constant pressure. FsW −= Isobaric process: ( )if VVPVPW −−=∆−= Isobaric Process ( )AsP−= VP∆−= Sign Convention of AP is opposite that of our book!
  24. 24. ( )if VVPVPW −−=∆−= Isobaric Process Work by the system is the area under a PV graph. Work is path dependent
  25. 25. Example 3 Isobaric Expansion of Water One gram of water is placed in the cylinder and the pressure is maintained at 2.0x105 Pa. The temperature of the water is raised by 31o C. The water is in the liquid phase and expands by the small amount of 1.0x10-8 m3 . Find the work done and the change in internal energy. VPW ∆−= WQU +=∆ TmcQ ∆= J0020.0−=( )( )385 m100.1Pa100.2 − ××−=W J130=J0020.0J130 −= ( ) ( )[ ]( ) C31CkgJ4186kg0010.0 ⋅= J130=
  26. 26. isochoric: constant volume Q= 0= Isochoric Process VPW ∆−= WQU +=∆
  27. 27. Example 4 Work and the Area Under a Pressure-Volume Graph Determine the work for the process in which the pressure, volume, and temperature of a gas are changed along the straight line in the figure. Since the volume increases, the work is positive. ( )( )345 m100.1Pa100.29.8 − ××−=W Estimate that there are 8.9 colored squares in the drawing. J180−=W
  28. 28. 15.4.1. Consider the pressure-versus-volume plot shown. There are eight points labeled and the choices below indicate possible multi- step processes. In which one of the processes does the work done have the largest value? a) G−H−B−D b) G−F−B−D c) H−A−B−D d) E−D−F−H e) C−B−F−G
  29. 29. 15.4.2. Consider the pressure-versus-volume plot shown. There are eight points labeled and the choices below indicate possible multi-step processes. If the initial state of the system is at A and the final state is at E, which of the following paths between these two states results in the largest increase in internal energy of the system? a) A−H−D−E b) A−B−F−E c) A−G−E d) A−C−E e) All paths between A and E are equivalent for internal energy.
  30. 30. 15.4.3. An isobaric process is represented by which one of the following graphs? a) A b) B c) C d) D e) E
  31. 31. 15.4.4. An insulated container with rigid walls has two compartments within. One compartment contains n moles of an ideal gas and the other compartment has been evacuated. A valve connecting the two chambers is opened at time t = 0 s. Which one of the following statements concerning this situation is true? a) There is no change in the internal energy of the gas. b) There is no change in the pressure of the gas. c) The temperature of the gas decreases with time. d) Work is done by the gas as it fills the previously evacuated compartment. e) The gas will remain in the first compartment unless heat is added to the system.
  32. 32. 15.4.5. In which of the following cases is a system undergoing an isobaric process? a) The system is placed within a thermal bath held at constant temperature. b) The system is an ideal gas enclosed in a container with a piston that may move up or down. A heavy object is placed on top of the piston. c) The system is an ideal gas enclosed in a container that is in contact with an object that is continually kept warmer or cooler than the gas within the system. d) The system is an ideal gas enclosed in a container has a constant volume. e) The system is an ideal gas enclosed in a container that is connected to a source of the gas from which gas may be added or removed to maintain a constant pressure.
  33. 33. 15.4.6. In which of the following cases is a system undergoing an adiabatic process? a) The system is placed within a thermal bath held at constant temperature. b) The system is an ideal gas enclosed in a container with a piston that may move up or down. A heavy object is placed on top of the piston. c) The system is an ideal gas enclosed in a container that is in contact with an object that is continually kept warmer or cooler than the gas within the system. d) The system is an ideal gas enclosed in a container has a constant volume. e) The system volume is changed rapidly.
  34. 34. Chapter 15 Thermodynamics Section 5: Thermal Processes Using an Ideal Gas
  35. 35. Isothermal expansion or compression of an ideal gas         = f i V V nRTW ln Isothermal Expansion or Compression of an Ideal Gas Calculus Alert! ∫−= f i V V PdVW ∫−= f i V V dV V nRT W ∫−= f i V V V dV nRTW if x x xx x dxf i lnln −=∫       = i f V V ln Work is the area under a PV graph  integral
  36. 36. Example 5 Isothermal Expansion of an Ideal Gas Two moles of the monatomic gas argon expand isothermally at 298K from and initial volume of 0.025m3 to a final volume of 0.050m3 . Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the change in internal energy of the gas, and (c) the heat supplied to the gas. (a)         = f i V V nRTW ln if nRTnRTU 2 3 2 3 −=∆(b) WQU +=∆(c) J3400−= ( ) ( )( )( )       ⋅= 3 3 m050.0 m250.0 lnK298KmolJ31.8mol0.2W J3400−= 0= WQ =
  37. 37. ( )if TTnRW −= 2 3 Adiabatic Expansion/Compression of a Monatomic Ideal Gas Adiabatic: no heat transfer WQU +=∆ TnRU ∆=∆ 2 3 γγ ooii VPVP = If you want to see the proof, go here
  38. 38. 15.5.1. A cylinder with a moveable piston contains an ideal gas. The gas is subsequently compressed adiabatically. Which of the following choices correctly identifies the signs of (1) the heat exchanged with the environment, (2) the work done, and (3) the change in the internal energy? a) (1) is zero, (2) is negative, and (3) is negative b) (1) is negative, (2) is positive, and (3) is negative c) (1) is zero, (2) is negative, and (3) is positive d) (1) is zero, (2) is positive, and (3) is positive e) (1) is positive, (2) is negative, and (3) is zero
  39. 39. 15.5.2. Two moles of an ideal gas have an initial Kelvin temperature T0 and absolute pressure P0. The gas undergoes a reversible isothermal compression from an initial volume V0 to a final volume 0.5 V0. How much heat is exchanged with the environment, specifying whether it is absorbed or released? a) Heat is released to the environment and its value is Q = 0.5P0V0. b) Heat is absorbed from the environment; and its value is Q = 0.5P0V0. c) No heat is exchanged with the environment. d) Heat is released to the environment; and its value is Q = P0V0 ln 2. e) Heat is absorbed from the environment; and its value is Q = P0V0 ln 2.
  40. 40. 15.5.3. Consider the pressure-volume graph shown for an ideal gas that may be taken along one of two paths from state A to state B. Path “1” is directly from A to B via a constant volume path. Path “2” follows the path A−C−B. How does the amount of work done along each path compare? a) W1 = W2; and the value is not equal to zero b) W1 = W2 = 0 c) W1 > W2 d) W1 < W2 e) It is not possible to compare the work done along each path without knowing the values of the temperature, pressure, and volume for each state.
  41. 41. 15.5.4. Consider the following pressure-volume graphs. Which of these graphs represents the behavior of a gas undergoing free expansion? a) A b) B c) C d) D e) None of the graphs represent a gas undergoing free expansion.
  42. 42. 15.5.5. A gas is enclosed in a cylinder by a piston. The volume of the gas is then reduced to one half its original value by applying a force to the piston. Which one of the following statements concerning the internal energy of the gas is true? a) The internal energy of the gas will decrease. b) The internal energy of the gas will increase. c) The internal energy of the gas will neither increase nor decrease. d) The internal energy of the gas will equal the work done in moving the piston. e) The internal energy of the gas may increase, decrease, or remain the same depending on the amount of heat that is gained or lost by the gas.
  43. 43. Chapter 15 Thermodynamics Section 6: Specific Heat Capacities
  44. 44. To relate heat and temperature change in solids and liquids, we used: TmcQ ∆= specific heat capacity The amount of a gas is conveniently expressed in moles, so we write the following analogous expression: TnCQ ∆= molar specific heat capacity Specific Heat Capacities
  45. 45. For gases it is necessary to distinguish between the molar specific heat capacities which apply to the conditions of constant pressure and constant volume: PV CC , WUQ +∆=pressureconstant constant pressure for a monatomic ideal gas RCP 2 5 = Specific Heat Capacities – Constant P ( ) ( )ifif TTnRTTnR −+−= 2 3 TnR∆= 2 5 first law of thermodynamics TnRU ∆=∆ 2 3 VPW ∆= Tn Q CP ∆ = Tn TnR ∆ ∆ = 2 5
  46. 46. constant volume for a monatomic ideal gas RCV 2 3 = monatomic ideal gas 3 5 2 3 2 5 === R R C C V P γ any ideal gas RCC VP =− Specific Heat Capacities – Constant V WUQ +∆=pressureconstant ( ) 02 3 +−= if TTnR TnR∆= 2 3 first law of thermodynamics TnRU ∆=∆ 2 3 γ is the ratio of the molar specific heat capacities at constant pressure and volume
  47. 47. Chapter 15 Thermodynamics Section 7: The Second Law of Thermodynamics
  48. 48. THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction. The second law is a statement about the natural tendency of heat to flow from hot to cold, whereas the first law deals with energy conservation and focuses on both heat and work. Second Law of Thermodynamics
  49. 49. Chapter 15 Thermodynamics Section 8: Heat Engines
  50. 50. heatinputofmagnitude=HQ heatrejectedofmagnitude=CQ doneworktheofmagnitude=W Heat Engine  A heat engine is any device that uses heat to perform work. It has three essential features.  Heat is supplied to the engine at a relatively high temperature from a place called the hot reservoir.  Part of the input heat is used to perform work by the working substance of the engine.  The remainder of the input heat is rejected to a place called the cold reservoir
  51. 51. The efficiency of a heat engine is defined as the ratio of the work done to the input heat: HQ W e = If there are no other losses, then CH QWQ += H CH Q QQ e − = Efficiency CH QQW −= H C Q Q −=1
  52. 52. Example 6 An Automobile Engine An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine? HQ W e = CH QWQ += e W QH = J8900= WQQ HC −= W e W QC −= ( )       −= 1 220.0 1 J2510CQ       −= 1 1 e W
  53. 53. 15.8.1. An automobile engine that burns gasoline has been engineered to have a relatively high efficiency of 22 %. While a car is being driven along a road on a long trip, 14 gallons of gasoline are consumed by the engine. Of the 14 gallons, how much gasoline was used in doing the work of propelling the car? a) 14 gallons b) about 11 gallons c) about 8 gallons d) about 3 gallons e) about 1 gallon
  54. 54. Chapter 15 Thermodynamics Section 9: Carnot’s Principle & the Carnot Engine
  55. 55. A reversible process is one in which both the system and the environment can be returned to exactly the states they were in before the process occurred. CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICS No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency. Carnot Principle
  56. 56. H C Q Q e −=1 H C H C T T Q Q = Carnot Engine  The Carnot engine is useful as an idealized model.  All of the heat input originates from a single temperature, and all the rejected heat goes into a cold reservoir at a single temperature.  Since the efficiency can only depend on the reservoir temperatures, the ratio of heats can only depend on those temperatures H C T T −=1 H C c T T e −=1
  57. 57. Example 7 A Tropical Ocean as a Heat Engine Water near the surface of a tropical ocean has a temperature of 298.2 K, whereas the water 700 meters beneath the surface has a temperature of 280.2 K. 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 and engine. H C T T e −=1c K298.2 K2.280 1−= 060.0=
  58. 58. Conceptual Example 8 Natural Limits on the Efficiency of a Heat Engine Consider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and delivers 1000J of work, rejecting no heat to a cold reservoir whose temperature is above 0 K. Decide whether this engine violates the first or second law of thermodynamics.
  59. 59.  Reasoning The first law of thermodynamics is an expression of energy conservation. The second law states that no irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. The efficiency of such a reversible engine is eC the efficiency of a Carnot engine.  From the point of view of energy conservation, nothing is wrong with an engine that converts 1000 J of heat into 1000 J of work. Energy has been neither created nor destroyed; it has only been transformed from one form (heat) into another form (work). Therefore, this engine does not violate the first law of thermodynamics.  Since all of the input heat is converted into work, the efficiency of the engine is 1, or 100%. But Equation 15.15, which is based on the second law of thermodynamics, indicates that the maximum possible efficiency is eC = 1 - TC/TH where TC and TH are the temperatures of the cold and hot reservoirs, respectively. Since we are told that TC is above 0 K, it is clear that the ratio TC/TH is greater than zero, so the maximum possible efficiency is less than 1 (or less than 100%). The engine, therefore, violates the second law of thermodynamics, which limits the efficiencies of heat engines to values less than 100%
  60. 60. 15.9.1. Which one of the following statements is not consistent with the first law of thermodynamics? a) The internal energy of a finite system must be finite. b) An engine may be constructed such that the work done by the machine exceeds the energy input to the engine. c) An isolated system that is thermally insulated cannot do work on its surroundings nor can work be done on the system. d) The internal energy of a system decreases when it does work on its surroundings and there is no flow of heat. e) An engine may be constructed that gains energy while heat is transferred to it and work is done on it.
  61. 61. Chapter 15 Thermodynamics Section 10: Refrigerators, Air Conditioners, and Heat Pumps
  62. 62. Refrigerators, air conditioners, and heat pumps are devices that make heat flow from cold to hot. This is called the refrigeration process. Refrigeration
  63. 63. Refrigeration
  64. 64. Conceptual Example 9 You Can’t Beat the Second Law of Thermodynamics 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 by the bed? Neither, the work done by the motor would be added to the hot reservoir, the room, and make it hotter.
  65. 65. W QC =eperformancoftCoefficienRefrigerator or air conditioner “Efficiency”
  66. 66. The heat pump uses work to make heat from the wintry outdoors flow into the house or to hot Summer weather outside. Heat Pumps heat pump W QH =Coefficient of performance
  67. 67. Example 10 A Heat Pump An ideal, or Carnot, heat pump is used to heat a house at 294 K. How much work must the pump do to deliver 3350 J of heat into the house on a day when the outdoor temperature is 273 K? H C H C T T Q Q = CH QQW −= H C HC T T QQ = H C HH T T QQW −=       −= H C H T T QW 1 ( )       −= K294 K273 1J3350 J240=
  68. 68. 15.10.1. A house that is heated using a heat pump with an ideal coefficient of performance loses heat to its surroundings at a rate of Z1(Thouse − Tsurr.), where Z1 is a constant, Thouse is the temperature inside the house; and Tsurr. is the temperature of its surroundings. In this process, heat is taken from the surroundings and heats the house at a rate of Z2(Tout − Thouse) where Tout is the temperature of the air output from the heat pump, which has a constant value. Which one of the following expressions is equal to the efficiency of the heat pump? a) d) b) e) c) surr.out out TT T − outsurr. surr. TT T − surr.out house TT T − housesurr. house TT T − housesurr. out TT T −
  69. 69. 15.10.2. An air conditioner pumps heat from a cold room to the hot outdoors in a three step cyclic process: (1) Room temperature, low pressure refrigerant gas passes through a compressor and comes out with increased temperature and increased pressure. The hot gas passes through piping on the outside, where heat is rejected to the surroundings. (2) The gas then passes through a narrower pipe before entering a compressor. Work is done by the compressor to increase the pressure enough for the gas to turn into a liquid. (3) The liquid then undergoes free expansion into a gas and cools. The cool gas passes through pipes that are inside the house. The inside air is cooled by coming into contact with these pipes. The refrigerant gas exits these pipes as a room temperature, low pressure gas. The cycle is then repeated. Why doesn’t this system violate the second law of thermodynamics? a) The internal energy of the gas is constant. b) Heat is normally taken from a warm place and transported to a warmer place. c) The system involves a closed cycle. d) Work is continually done on the system. e) Since the compressor adds entropy, the total entropy increases.
  70. 70. Chapter 15 Thermodynamics Section 11: Entropy
  71. 71. In general, irreversible processes cause us to lose some, but not necessarily all, of the ability to do work. This partial loss can be expressed in terms of a concept called entropy. H C H C T T Q Q = H H C C T Q T Q = R       =∆ T Q Sentropy change reversible Carnot engine Entropy
  72. 72. Entropy, like internal energy, is a function of the state of the system. Consider the entropy change of a Carnot engine. The entropy of the hot reservoir decreases and the entropy of the cold reservoir increases. R       =∆ T Q S 0=−+=∆ H H C C T Q T Q S Reversible processes do not alter the entropy of the universe. Entropy
  73. 73. Entropy  What happens to the entropy change of the universe in an irreversible process is more complex.  Since ΔS is independent of the path, it does not matter what path is used.
  74. 74. Example 11 The Entropy of the Universe Increases The figure shows 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. H H C C T Q T Q S −+=∆ universe K650 J1200 K350 J1200 −+= KJ6.1+=
  75. 75. Any irreversible process increases the entropy of the universe. 0universe >∆S THE SECOND LAW OF THERMODYNAMICS STATED IN TERMS OF ENTROPY The total entropy of the universe does not change when a reversible process occurs and increases when an irreversible process occurs. Restating the Second Law
  76. 76. Example 12 Energy Unavailable for Doing Work Suppose that 1200 J of heat is used as input for an engine under two different conditions (as shown on the right). Determine the maximum amount of work that can be obtained for each case. HQ W e = The maximum amount of work will be achieved when the engine is a Carnot Engine, where (a) H C T T e −=1carnot ( ) HQeW carnot= (b) H C T T e −=1carnot ( ) HQeW carnot= The irreversible process of heat through the copper rod causes some energy to become unavailable. 77.0= K650 K150 1−= ( )( )J120077.0= J920= K350 K150 1−= 57.0= ( )( )J120057.0= J680=
  77. 77. universeeunavailabl STW o∆= More on Entropy
  78. 78. 15.11.1. A box with five adiabatic sides contains an ideal gas with an initial temperature T0. The sixth side is diathermal and is placed in contact with a reservoir with a constant temperature T2 > T0. Assuming the specific heat capacity of the system does not change with temperature, why must the entropy change of the universe always be increasing as the box warms? a) Entropy will always be increasing since the work done on the gas in the box is negative. b) Entropy will always be increasing since the temperature of the box is always less than or equal to T2. c) Entropy will always be increasing since this process is reversible. d) Entropy will always be increasing since the temperature of the box is always greater than absolute zero. e) Entropy will always be increasing since in any process entropy increases.
  79. 79. 15.11.2. A leaf is growing on a tree. Does this growth process violate the second law of thermodynamics when it is stated in terms of entropy? a) Yes, but the law does not apply to living things. It only applies to inanimate objects. b) Yes, because this law is not applicable in situations involving radiant energy from the Sun. c) No, because the entropy of the Sun has decreased while the entropy of the leaf increases as it grows. d) No, because while the entropy of the leaf is decreasing as it grows, there is a net increase in entropy because of the light emitted from the leaf. e) No, because there is no net increase in the energy of the leaf.
  80. 80. 15.11.3. While watching a fantasy film, you observe a wizard wave his arms and six potion vials that had fallen to the floor suddenly piece themselves back together with the potions inside and rise up with a table. In the end, the table is upright and the six vials with their potions are sitting on the table as if nothing had happened. Which of the following principles or laws of physics is disobeyed by this scene from the movie? a) conservation of energy b) second law of thermodynamics c) Newton’s laws of motion d) time dilation e) the work-energy theorem
  81. 81. Chapter 15 Thermodynamics Section 12: The Third Law of Thermodynamics
  82. 82. It is not possible to lower the temperature of any system to absolute zero in a finite number of steps. The 3rd Law of Thermodynamics

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