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Termodinamika

1. 1. TERMODINAMIKATERMODINAMIKA
2. 2. Thermodynamic Systems andThermodynamic Systems and Their SurroundingsTheir Surroundings ThermodynamicsThermodynamics is the branch of physicsis the branch of physics that is built upon the fundamental laws thatthat is built upon the fundamental laws that heat and work obey.heat and work obey.
3. 3. ThermodynamicsThermodynamics  In thermodynamics the collection of objectsIn thermodynamics the collection of objects upon which attention is being focused isupon which attention is being focused is called thecalled the system,system, while everything else in thewhile everything else in the environment is called theenvironment is called the surroundings.surroundings.  The system and its surroundings are separatedThe system and its surroundings are separated by walls of some kind. Walls that permit heatby walls of some kind. Walls that permit heat to flow through them, such as those of theto flow through them, such as those of the engine block, are calledengine block, are called diathermal walls.diathermal walls. Perfectly insulating walls that do not permitPerfectly insulating walls that do not permit heat to flow between the system and itsheat to flow between the system and its surroundings are calledsurroundings are called adiabatic walls.adiabatic walls.
4. 4. State of a SystemState of a System  To understand what the laws ofTo understand what the laws of thermodynamics have to say about thethermodynamics have to say about the relationship between heat and work, it isrelationship between heat and work, it is necessary to describe the physical conditionnecessary to describe the physical condition oror state of a system.state of a system.  The state of the system would be specifiedThe state of the system would be specified by giving values for the pressure, volume,by giving values for the pressure, volume, temperature, and mass of the hot air.temperature, and mass of the hot air.
5. 5. The Zeroth Law ofThe Zeroth Law of ThermodynamicsThermodynamics  (a) Systems A and B are(a) Systems A and B are surrounded by adiabaticsurrounded by adiabatic walls and register the samewalls and register the same temperature on thetemperature on the thermometer. (b) When A isthermometer. (b) When A is put into thermal contact withput into thermal contact with B through diathermal walls,B through diathermal walls, no net flow of heat occursno net flow of heat occurs between the systems.between the systems.
6. 6. Two systems individually in thermalTwo systems individually in thermal equilibrium with a third systemequilibrium with a third system are in thermalare in thermal equilibrium with each other.equilibrium with each other. The Zeroth Law ofThe Zeroth Law of ThermodynamicsThermodynamics
7. 7. Temperature is the indicator of thermalTemperature is the indicator of thermal equilibrium in the sense that there is noequilibrium in the sense that there is no net flow of heat between two systems innet flow of heat between two systems in thermal contact that have the samethermal contact that have the same temperature.temperature. There can be no flow of heat within aThere can be no flow of heat within a system in thermal equilibrium.system in thermal equilibrium. The Zeroth Law ofThe Zeroth Law of ThermodynamicsThermodynamics
8. 8. The First Law ofThe First Law of ThermodynamicsThermodynamics
9. 9.  The internal energy of a system changesThe internal energy of a system changes from an initial valuefrom an initial value UUii to a final value ofto a final value of UUff due to heatdue to heat QQ and workand work WW:: ∆∆UU == UUff -- UUii == Q – WQ – W  Q is positive when the system gains heatQ is positive when the system gains heat and negative when it loses heat. W isand negative when it loses heat. W is positive when work is done by the systempositive when work is done by the system and negative when work is done on theand negative when work is done on the system.system. The First Law ofThe First Law of ThermodynamicsThermodynamics
10. 10. The Internal EnergyThe Internal Energy  The internal energy depends only on theThe internal energy depends only on the state of a system, not on the method bystate of a system, not on the method by which the system arrives at a given state.which the system arrives at a given state.
11. 11. EXAMPLE : An Ideal GasEXAMPLE : An Ideal Gas  The temperature of three moles of aThe temperature of three moles of a monatomic ideal gas is reduced frommonatomic ideal gas is reduced from TTii == 540 K to540 K to TTff = 350 K by two different= 350 K by two different methods. In the first method 5500 J of heatmethods. In the first method 5500 J of heat flows into the gas, while in the second,flows into the gas, while in the second, 1500 J of heat flows into it. In each case1500 J of heat flows into it. In each case find (a) the change in the internal energyfind (a) the change in the internal energy and (b) the work done by the gas.and (b) the work done by the gas.
12. 12. SolutionSolution  Since the internal energy of a monatomic ideal gasSince the internal energy of a monatomic ideal gas isis UU = (= (33 //22)) nRTnRT and since the number of molesand since the number of moles nn is fixed, only a change in temperatureis fixed, only a change in temperature TT can altercan alter the internal energy.the internal energy.  Since the change inSince the change in TT is the same in both methods,is the same in both methods, the change inthe change in UU is also the same. From the givenis also the same. From the given temperatures, the changetemperatures, the change ∆∆UU in internal energyin internal energy can be determined. Then, the first law ofcan be determined. Then, the first law of thermodynamics can be used withthermodynamics can be used with ∆∆UU and theand the given heat values to calculate the work.given heat values to calculate the work.
13. 13. (a)(a) Using EquationUsing Equation UU = (= (33 //22)) nRTnRT for thefor the internal energy of a monatomic ideal gas,internal energy of a monatomic ideal gas, we find for each method of adding heat thatwe find for each method of adding heat that (b)(b) SinceSince ∆∆UU is now known and the heat isis now known and the heat is given in each method,given in each method, ∆∆UU == Q – WQ – W can becan be used to determine the work:used to determine the work: SolutionSolution
14. 14. Thermal ProcessesThermal Processes  A system can interact with its surroundingsA system can interact with its surroundings in many ways, and the heat and work thatin many ways, and the heat and work that come into play always obey the first law ofcome into play always obey the first law of thermodynamics.thermodynamics.  thermal process is assumed to bethermal process is assumed to be quasi-quasi- static,static, which means that it occurs slowlywhich means that it occurs slowly enough that a uniform pressure andenough that a uniform pressure and temperature exist throughout all regions oftemperature exist throughout all regions of the system at all times.the system at all times.
15. 15. An Isobaric ProcessAn Isobaric Process The substance in the chamberThe substance in the chamber is expanding isobaricallyis expanding isobarically because the pressure is heldbecause the pressure is held constant by the externalconstant by the external atmosphere and the weight ofatmosphere and the weight of the piston and the block.the piston and the block. An isobaric process is oneAn isobaric process is one that occurs at constantthat occurs at constant pressure.pressure.
16. 16.  The Figure shows a substance (solid, liquid, orThe Figure shows a substance (solid, liquid, or gas) contained in a chamber fitted with agas) contained in a chamber fitted with a frictionless piston.frictionless piston.  The pressureThe pressure PP experienced by the substance isexperienced by the substance is always the same and is determined by the externalalways the same and is determined by the external atmosphere and the weight of the piston and theatmosphere and the weight of the piston and the block resting on it.block resting on it.  Heating the substance makes it expand and doHeating the substance makes it expand and do workwork WW in lifting the piston and block through thein lifting the piston and block through the displacementdisplacement ss.. An Isobaric ProcessAn Isobaric Process
17. 17.  The work can be calculated fromThe work can be calculated from W = FsW = Fs, where, where FF is the magnitude of the force andis the magnitude of the force and ss is theis the magnitude of the displacement.magnitude of the displacement.  The force is generated by the pressureThe force is generated by the pressure PP acting onacting on the bottom surface of the piston (area =the bottom surface of the piston (area = AA),), according toaccording to F = PAF = PA..  With this substitution forWith this substitution for FF, the work becomes, the work becomes WW = (= (PAPA))ss. But the product. But the product AA ·· ss is the change inis the change in volume of the material,volume of the material, ∆∆V = VV = Vff -- VVii, where, where VVff andand VVii are the final and initial volumes, respectively.are the final and initial volumes, respectively. An Isobaric ProcessAn Isobaric Process
18. 18.  Thus, the expression for the work isThus, the expression for the work is Isobaric processIsobaric process W = PW = P∆∆V = P(VV = P(Vff – V– Vii ))  A positive value for the work done by aA positive value for the work done by a system when it expands isobarically (Vsystem when it expands isobarically (Vff exceeds Vexceeds Vii).).  That Equation also applies to an isobaricThat Equation also applies to an isobaric compression (Vcompression (Vff less than Vless than Vii). Then, the). Then, the work is negative, since work must be donework is negative, since work must be done on the system to compress it.on the system to compress it. An Isobaric ProcessAn Isobaric Process
19. 19. An Isobaric ProcessAn Isobaric Process For an isobaric process, aFor an isobaric process, a pressure-versus-volumepressure-versus-volume plot is a horizontal straightplot is a horizontal straight line, and the work doneline, and the work done [[WW == PP((VVff 22 –– VVii 22 )] is the)] is the colored rectangular areacolored rectangular area under the graph.under the graph.
20. 20. An Isochoric ProcessAn Isochoric Process a) The substance in the chambera) The substance in the chamber is being heated isochoricallyis being heated isochorically because the rigid chamber keepsbecause the rigid chamber keeps the volume constant. (b) Thethe volume constant. (b) The pressure–volume plot for anpressure–volume plot for an isochoric process is a verticalisochoric process is a vertical straight line. The area under thestraight line. The area under the graph is zero, indicating that nograph is zero, indicating that no work is done.work is done. an isochoric process, one thatan isochoric process, one that occurs at constant volume.occurs at constant volume.
21. 21.  FigureFigure aa illustrates an isochoric process inillustrates an isochoric process in which a substance (solid, liquid, or gas) iswhich a substance (solid, liquid, or gas) is heated. The substance would expand if itheated. The substance would expand if it could, but the rigid container keeps thecould, but the rigid container keeps the volume constant, so the pressure–volumevolume constant, so the pressure–volume plot shown in Figureplot shown in Figure bb is a vertical straightis a vertical straight lineline  Because the volume is constant, theBecause the volume is constant, the pressure inside rises, and the substancepressure inside rises, and the substance exerts more and more force on the walls.exerts more and more force on the walls. An Isochoric ProcessAn Isochoric Process
22. 22.  While enormous forces can be generated inWhile enormous forces can be generated in the closed container, no work is done, sincethe closed container, no work is done, since the walls do not move. Consistent with zerothe walls do not move. Consistent with zero work being done, the area under the verticalwork being done, the area under the vertical straight line in Figurestraight line in Figure bb is zero.is zero.  Since no work is done, the first law ofSince no work is done, the first law of thermodynamics indicates that the heat inthermodynamics indicates that the heat in an isochoric process serves only to changean isochoric process serves only to change the internal energy:the internal energy: ∆∆U = Q - W = QU = Q - W = Q.. An Isochoric ProcessAn Isochoric Process
23. 23. An Isothermal ProcessAn Isothermal Process The colored area gives theThe colored area gives the work done by the gas for thework done by the gas for the process fromprocess from XX toto YY..  A third important thermalA third important thermal process is anprocess is an isothermalisothermal process, one that takes place atprocess, one that takes place at constant temperature.constant temperature.
24. 24. The Adiabatic ProcessThe Adiabatic Process  Last, there is theLast, there is the adiabatic process, oneadiabatic process, one that occurs without the transfer of heat.that occurs without the transfer of heat.  Since there is no heat transfer,Since there is no heat transfer, QQ equalsequals zero, and the first law indicates thatzero, and the first law indicates that ∆∆U = Q - W = -WU = Q - W = -W..  Thus, when work is done by a systemThus, when work is done by a system adiabatically, the internal energy of theadiabatically, the internal energy of the system decreases by exactly the amount ofsystem decreases by exactly the amount of the work done.the work done.
25. 25.  Thus, when work is done by a systemThus, when work is done by a system adiabatically, the internal energy of theadiabatically, the internal energy of the system decreases by exactly the amount ofsystem decreases by exactly the amount of the work done. When work is done on athe work done. When work is done on a system adiabatically, the internal energysystem adiabatically, the internal energy increases correspondingly.increases correspondingly. The Adiabatic ProcessThe Adiabatic Process
26. 26. Thermal Processes ThatThermal Processes That Utilize an Ideal GasUtilize an Ideal Gas
27. 27. Isothermal Expansion orIsothermal Expansion or CompressionCompression (a)(a) The ideal gas in the cylinder is expandingThe ideal gas in the cylinder is expanding isothermally at temperatureisothermally at temperature TT. The force holding. The force holding the piston in place is reduced slowly, so thethe piston in place is reduced slowly, so the expansion occurs quasi-statically.expansion occurs quasi-statically. (b)(b) The workThe work done by the gas is given by the colored area.done by the gas is given by the colored area.
28. 28.  When a system performs work isothermally,When a system performs work isothermally, the temperature remains constant.the temperature remains constant.  In FigureIn Figure aa, for instance, a metal cylinder, for instance, a metal cylinder containscontains nn moles of an ideal gas, and themoles of an ideal gas, and the large mass of hot water maintains thelarge mass of hot water maintains the cylinder and gas at a constant Kelvincylinder and gas at a constant Kelvin temperaturetemperature TT. The piston is held in place. The piston is held in place initially so the volume of the gas isinitially so the volume of the gas is VVii.. Isothermal Expansion orIsothermal Expansion or CompressionCompression
29. 29.  As the external force applied to the piston is reducedAs the external force applied to the piston is reduced quasi-statically, the gas expands to the final volumequasi-statically, the gas expands to the final volume VVff ..  Figure b gives a plot of pressure (Figure b gives a plot of pressure (P = nRTP = nRT // VV) versus) versus volume for the process.volume for the process.  The solid red line in the graph is called an isothermThe solid red line in the graph is called an isotherm (meaning “constant temperature”) because it(meaning “constant temperature”) because it represents the relation between pressure and volumerepresents the relation between pressure and volume when the temperature is held constant.when the temperature is held constant. Isothermal Expansion orIsothermal Expansion or CompressionCompression
30. 30.  The workThe work WW done by the gas isdone by the gas is notnot given bygiven by W = PW = P∆∆V = PV = P((VVff -- VVii) because the pressure) because the pressure is not constant.is not constant.  Nevertheless, the work is equal to the areaNevertheless, the work is equal to the area under the graph. The techniques of integralunder the graph. The techniques of integral calculus lead to the following resultcalculus lead to the following result forfor WW:: Isothermal Expansion orIsothermal Expansion or CompressionCompression       = i f V V nRTW ln
31. 31.  Where does the energy for this workWhere does the energy for this work originate?originate?  Since the internal energy of any ideal gas isSince the internal energy of any ideal gas is proportional to the Kelvin temperature (proportional to the Kelvin temperature (UU== ((33 //22)) nRTnRT for a monatomic ideal gas, forfor a monatomic ideal gas, for example), the internal energy remainsexample), the internal energy remains constant throughout an isothermal process,constant throughout an isothermal process, and the change in internal energy is zero.and the change in internal energy is zero. Isothermal Expansion orIsothermal Expansion or CompressionCompression
32. 32.  The first law of thermodynamics becomesThe first law of thermodynamics becomes ∆∆UU = 0 == 0 = Q - WQ - W. In other words,. In other words, Q = WQ = W,, and the energy for the work originates inand the energy for the work originates in the hot water.the hot water. Isothermal Expansion orIsothermal Expansion or CompressionCompression
33. 33. Adiabatic Expansion orAdiabatic Expansion or CompressionCompression a)a) The ideal gas in the cylinder is expanding adiabatically.The ideal gas in the cylinder is expanding adiabatically. The force holding the piston in place is reduced slowly, soThe force holding the piston in place is reduced slowly, so the expansion occurs quasi-statically.the expansion occurs quasi-statically. (b)(b) A plot ofA plot of pressure versus volume yields the adiabatic curve shown inpressure versus volume yields the adiabatic curve shown in red, which intersects the isotherms (blue) at the initialred, which intersects the isotherms (blue) at the initial temperaturetemperature TTii and the final temperatureand the final temperature TTff . The work done. The work done by the gas is given by the colored area.by the gas is given by the colored area.
34. 34.  When a system performs work adiabatically,When a system performs work adiabatically, no heat flows into or out of the system.no heat flows into or out of the system.  FigureFigure aa shows an arrangement in whichshows an arrangement in which nn moles of an ideal gas do work undermoles of an ideal gas do work under adiabatic conditions, expanding quasi-adiabatic conditions, expanding quasi- statically from an initial volumestatically from an initial volume VVii to a finalto a final volumevolume VVff..  The arrangement is similar to that in FigureThe arrangement is similar to that in Figure for isothermal expansion.for isothermal expansion. Adiabatic Expansion orAdiabatic Expansion or CompressionCompression
35. 35.  However, a different amount of work is done here,However, a different amount of work is done here, because the cylinder is now surrounded by insulatingbecause the cylinder is now surrounded by insulating material that prevents the flow of heat, somaterial that prevents the flow of heat, so QQ = 0.= 0. According to the first law of thermodynamics, theAccording to the first law of thermodynamics, the change in internal energy ischange in internal energy is ∆∆U = Q - W = -WU = Q - W = -W..  Since the internal energy of an ideal monatomic gas isSince the internal energy of an ideal monatomic gas is UU= (= (33 //22))nRTnRT , it follows that, it follows that ∆∆U =U = ((33 //22))nR(TnR(Tff – T– Tii)) ,, wherewhere TTii andand TTff are the initial and final Kelvinare the initial and final Kelvin temperatures. With this substitution, the relationtemperatures. With this substitution, the relation ∆∆UU = -= -WW becomesbecomes Adiabatic Expansion orAdiabatic Expansion or CompressionCompression
36. 36.  When an ideal gas expands adiabatically, it doesWhen an ideal gas expands adiabatically, it does positive work, sopositive work, so WW is positive.is positive.  Therefore, the termTherefore, the term TTii -- TTff is also positive, and theis also positive, and the final temperature of the gas must be less than thefinal temperature of the gas must be less than the initial temperature.initial temperature.  The internal energy of the gas is reduced toThe internal energy of the gas is reduced to provide the necessary energy to do the work, andprovide the necessary energy to do the work, and because the internal energy is proportional to thebecause the internal energy is proportional to the Kelvin temperature, the temperature decreases.Kelvin temperature, the temperature decreases. Adiabatic Expansion orAdiabatic Expansion or CompressionCompression
37. 37.  FigureFigure bb shows a plot of pressure versusshows a plot of pressure versus volume for the adiabatic process. Thevolume for the adiabatic process. The adiabatic curve (red) intersects theadiabatic curve (red) intersects the isotherms (blue) at the higher initialisotherms (blue) at the higher initial temperature [temperature [TTii == PPiiVVii/(/(nRnR)] and the lower)] and the lower final temperature [final temperature [TTff == PPffVVff/(/(nRnR)]. The)]. The colored area under the adiabatic curvecolored area under the adiabatic curve represents the work done.represents the work done. Adiabatic Expansion orAdiabatic Expansion or CompressionCompression
38. 38.  The equation that gives the adiabatic curveThe equation that gives the adiabatic curve (red) between the initial pressure and(red) between the initial pressure and volume (volume (PPii,, VVii) and the final pressure and) and the final pressure and volume (volume (PPff,, VVff) in Figure) in Figure bb can be derivedcan be derived using integral calculus. The result isusing integral calculus. The result is  where the exponentwhere the exponent γγ is the ratio of theis the ratio of the specific heat capacities at constant pressurespecific heat capacities at constant pressure and constant volume,and constant volume, γγ == ccPP//ccVV.. Adiabatic Expansion orAdiabatic Expansion or CompressionCompression
39. 39. THE SECOND LAW OFTHE SECOND LAW OF THERMODYNAMICSTHERMODYNAMICS
40. 40. THE SECOND LAW OFTHE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOWTHERMODYNAMICS: THE HEAT FLOW STATEMENTSTATEMENT Heat flows spontaneously from a substance at aHeat flows spontaneously from a substance at a higher temperature to a substance at a lowerhigher temperature to a substance at a lower temperature and does not flow spontaneously intemperature and does not flow spontaneously in the reverse directionthe reverse direction
41. 41. It is important to realize that the second lawIt is important to realize that the second law of thermodynamics deals with a differentof thermodynamics deals with a different aspect of nature than does the first law ofaspect of nature than does the first law of thermodynamics.thermodynamics. The second law is a statement about theThe second law is a statement about the natural tendency of heat to flow from hot tonatural tendency of heat to flow from hot to cold, whereas the first law deals with energycold, whereas the first law deals with energy conservation and focuses on both heat andconservation and focuses on both heat and work.work.
42. 42. AA heat engineheat engine is any device that uses heat to performis any device that uses heat to perform work. It has three essential features:work. It has three essential features: 1 . Heat is supplied to the engine at a relatively highHeat is supplied to the engine at a relatively high temperature from a place called thetemperature from a place called the hot reservoir.hot reservoir. 2 . Part of the input heat is used to perform work by thePart of the input heat is used to perform work by the working substanceworking substance of the engine, which is the materialof the engine, which is the material within the engine that actually does the work (e.g., thewithin the engine that actually does the work (e.g., the gasoline–air mixture in an automobile engine).gasoline–air mixture in an automobile engine). 3 . The remainder of the input heat is rejected at aThe remainder of the input heat is rejected at a temperature lower than the input temperature to a placetemperature lower than the input temperature to a place called thecalled the cold reservoircold reservoir
43. 43. This schematic representationThis schematic representation of a heat engine shows the inputof a heat engine shows the input heat (magnitude =heat (magnitude = QQHH) that) that originates from the hotoriginates from the hot reservoir, the work (magnitudereservoir, the work (magnitude == WW) that the engine does, and) that the engine does, and the heat (magnitude =the heat (magnitude = QQCC ) that) that the engine rejects to the coldthe engine rejects to the cold reservoir.reservoir.
44. 44. To be highly efficient, a heat engine must produceTo be highly efficient, a heat engine must produce a relatively large amount of work from as littlea relatively large amount of work from as little input heat as possible. Thus, theinput heat as possible. Thus, the efficiency eefficiency e of aof a heat engine is defined as the ratio of the workheat engine is defined as the ratio of the work WW done by the engine to the input heatdone by the engine to the input heat QQHH:: If the input heat were converted entirely intoIf the input heat were converted entirely into work, the engine would have an efficiency ofwork, the engine would have an efficiency of 1.00, since1.00, since W = QW = QHH; such an engine would be; such an engine would be 100% efficient.100% efficient.
45. 45. What is it that allows a heat engine toWhat is it that allows a heat engine to operate with maximum efficiency?operate with maximum efficiency? The French engineer Sadi Carnot (1796–1832)The French engineer Sadi Carnot (1796–1832) proposed that a heat engine has maximumproposed that a heat engine has maximum efficiency when the processes within the engineefficiency when the processes within the engine are reversible.are reversible. A reversible process is one inA reversible process is one in which both the system and its environmentwhich both the system and its environment can be returned to exactly the states they werecan be returned to exactly the states they were in before the process occurred.in before the process occurred.
46. 46. In a reversible process,In a reversible process, bothboth the system andthe system and its environment can be returned to their initialits environment can be returned to their initial states.states. A process that involves an energy-dissipatingA process that involves an energy-dissipating mechanism, such as friction, cannot bemechanism, such as friction, cannot be reversible because the energy wasted due toreversible because the energy wasted due to friction would alter the system or thefriction would alter the system or the environment or both.environment or both.
47. 47. CARNOT'S PRINCIPLE: ANCARNOT'S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THEALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICSSECOND LAW OF THERMODYNAMICS No irreversible engine operating between twoNo irreversible engine operating between two reservoirs at constant temperatures can have a greaterreservoirs at constant temperatures can have a greater efficiency than a reversible engine operating betweenefficiency than a reversible engine operating between the same temperatures. Furthermore, all reversiblethe same temperatures. Furthermore, all reversible engines operating between the same temperaturesengines operating between the same temperatures have the same efficiency.have the same efficiency.
48. 48. Natural Limits on the Efficiency of aNatural Limits on the Efficiency of a Heat EngineHeat Engine Consider a hypothetical engine that receives 1000 JConsider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and deliversof heat as input from a hot reservoir and delivers 1000 J of work, rejecting no heat to a cold reservoir1000 J of work, rejecting no heat to a cold reservoir whose temperature is above 0 K. Decide whetherwhose temperature is above 0 K. Decide whether this engine violates the first or the second law ofthis engine violates the first or the second law of thermodynamics, or both.thermodynamics, or both.
49. 49. The first law of thermodynamics is an expressionThe first law of thermodynamics is an expression of energy conservation. From the point of view ofof energy conservation. From the point of view of energy conservation, nothing is wrong with anenergy conservation, nothing is wrong with an engine that converts 1000 J of heat into 1000 J ofengine that converts 1000 J of heat into 1000 J of work.work. Energy has been neither created nor destroyed; itEnergy has been neither created nor destroyed; it has only been transformed from one form (heat) tohas only been transformed from one form (heat) to another (work).another (work).
50. 50. This engine does, however, violate the secondThis engine does, however, violate the second law of thermodynamics. Since all of the inputlaw of thermodynamics. Since all of the input heat is converted into work, the efficiency of theheat is converted into work, the efficiency of the engine is 1, or 100%.engine is 1, or 100%. Since we know thatSince we know that TTCC is above 0 K, it is clearis above 0 K, it is clear that the ratiothat the ratio TTCC//TTHH is greater than zero, so theis greater than zero, so the maximum possible efficiency is less than 1, ormaximum possible efficiency is less than 1, or less than 100%.less than 100%.
51. 51. A Heat PumpA Heat Pump An ideal or Carnot heat pump is used toAn ideal or Carnot heat pump is used to heat a house to a temperature ofheat a house to a temperature of TTHH = 294= 294 K (21 °C). How much work must be doneK (21 °C). How much work must be done by the pump to deliverby the pump to deliver QQHH = 3350 J of heat= 3350 J of heat into the house when the outdoorinto the house when the outdoor temperaturetemperature TTCC is (a) 273 K (0 °C) and (b)is (a) 273 K (0 °C) and (b) 252 K (-21 °C)?252 K (-21 °C)?
52. 52. The conservation of energy (The conservation of energy (QQHH == WW ++ QQCC)) applies to the heat pump. Thus, the work can beapplies to the heat pump. Thus, the work can be determined fromdetermined from W = QW = QHH -- QQCC, provided we can, provided we can obtain a value forobtain a value for QQCC, the heat taken by the pump, the heat taken by the pump from the outside. To determinefrom the outside. To determine QQCC, we use the fact, we use the fact that the pump is a Carnot heat pump and operatesthat the pump is a Carnot heat pump and operates reversibly. Therefore, the relationreversibly. Therefore, the relation QQCC//QQHH == TTCC//TTHH applies. Solving it forapplies. Solving it for QQCC, we obtain, we obtain QQCC == QQHH((TTCC//TTHH).). Using this result, we find thatUsing this result, we find that ReasoningReasoning
53. 53. SolutionSolution (a)(a) At an indoor temperature ofAt an indoor temperature of TTHH = 294 K and an= 294 K and an outdoor temperature ofoutdoor temperature of TTCC = 273 K, the work needed= 273 K, the work needed isis (b)(b) This solution is identical to that in part (a),This solution is identical to that in part (a), except that it is now cooler outside, soexcept that it is now cooler outside, so TTCC = 252 K.= 252 K. The necessary work is , which is more thanThe necessary work is , which is more than in part (a).in part (a).
54. 54. To introduce the idea of entropy we recall theTo introduce the idea of entropy we recall the relationrelation QQCC//QQHH == TTCC//TTHH that applies to a Carnotthat applies to a Carnot engine. This equation can be rearranged asengine. This equation can be rearranged as QQCC//TTCC == QQHH//TTHH, which focuses attention on the heat, which focuses attention on the heat QQ divideddivided by the Kelvin temperatureby the Kelvin temperature TT. The quantity. The quantity QQ//TT isis called the change in the entropycalled the change in the entropy ∆∆SS:: ENTROPYENTROPY the SI unit for entropy is a joule per kelvin (J/K).
55. 55. THE SECOND LAW OFTHE SECOND LAW OF THERMODYNAMICS STATED INTHERMODYNAMICS STATED IN TERMS OF ENTROPYTERMS OF ENTROPY The total entropy of the universe does notThe total entropy of the universe does not change when a reversible process occurschange when a reversible process occurs ((∆∆SSuniverseuniverse = 0 ) and increases when an= 0 ) and increases when an irreversible process occurs (irreversible process occurs (∆∆SSuniverseuniverse 0 ).0 ).
56. 56. The Entropy of the Universe IncreasesThe Entropy of the Universe Increases This Figure shows 1200 J ofThis Figure shows 1200 J of heat flowing spontaneouslyheat flowing spontaneously through a copper rod from a hotthrough a copper rod from a hot reservoir at 650 K to a coldreservoir at 650 K to a cold reservoir at 350 K. Determinereservoir at 350 K. Determine the amount by which thisthe amount by which this irreversible process changes theirreversible process changes the entropy of the universe,entropy of the universe, assuming that no other changesassuming that no other changes occur.occur.
57. 57. ReasoningReasoning The hot-to-cold heat flow is irreversible, soThe hot-to-cold heat flow is irreversible, so the relationthe relation ∆∆SS = (= (QQ//TT))RR is applied to ais applied to a hypothetical process whereby the 1200 J ofhypothetical process whereby the 1200 J of heat is taken reversibly from the hotheat is taken reversibly from the hot reservoir and added reversibly to the coldreservoir and added reversibly to the cold reservoir.reservoir.
58. 58. SolutionSolution The total entropy change of the universe is theThe total entropy change of the universe is the algebraic sum of the entropy changes for eachalgebraic sum of the entropy changes for each reservoir:reservoir:
59. 59. Order to DisorderOrder to Disorder Find the change in entropy that results when aFind the change in entropy that results when a 2.3-kg block of ice melts slowly (reversibly)2.3-kg block of ice melts slowly (reversibly) at 273 K (0 °C).at 273 K (0 °C).
60. 60. ReasoningReasoning Since the phase change occurs reversibly at a constantSince the phase change occurs reversibly at a constant temperature, the change in entropy can be found bytemperature, the change in entropy can be found by using,using, ∆∆SS = (= (QQ//TT))RR, where, where QQ is the heat absorbed byis the heat absorbed by the melting ice. This heat can be determined by usingthe melting ice. This heat can be determined by using the relationthe relation Q = mLQ = mLff, where, where mm is the mass andis the mass and LLff == 3.35 × 103.35 × 1055 J/kg is the latent heat of fusion of water.J/kg is the latent heat of fusion of water. & Solution& Solution
61. 61. THE THIRD LAW OF THERMODYNAMICSTHE THIRD LAW OF THERMODYNAMICS It is not possible to lower the temperature of anyIt is not possible to lower the temperature of any system to absolute zero in a finite number ofsystem to absolute zero in a finite number of steps.steps.