ThermalreseelkIWhyWOombinedsyste(system and eyclic device)Considera systemconnected to a thermal energy reservoir at aconstanttemperatureofTthroughareversiblecyclicdeviceThecyclicdevicereceivesheatSQfromreservoir;suppliesheat to the system sg , and producing work swrey.The system produces work sw as a result of heat transfer,andtheboundarytemperatureisT(variable)Wehave a combined systemas the dashed linesyields:SORSQ8Q(SWrey+SWsys)=SWc=8QR-dEcSWC=TRdEcTTRTKelvinPlanck statement:nosystem canproduce a net amount ofwork while operatingina cycle and exchanging heat with a singleSOthermal energy reservoir.For a cycleWc=Wccan not be a work output, can not be a6positivequantity
• Why • • Consider a system connected to a thermal energy reservoir at a constant temperature of TR through a reversible cyclic device. • The cyclic device receives heat from reservoir; supplies heat to the system , and producing work • The system produces work as a result of heat transfer， and the boundary temperature is T (variable) • We have a combined system as the dashed lines yields: 6 For a cycle Kelvin Planck statement: no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir. WC can not be a work output, can not be a positive quantity
0: For reversible cycles: all quantities have thesame magnitude but the opposite sign, thus800三Tintrev The equality in the Clausius inequality holds forreversible cyclesThe inequality for the irreversible cycles
• For reversible cycles: all quantities have the same magnitude but the opposite sign, thus: • The equality in the Clausius inequality holds for reversible cycles • The inequality for the irreversible cycles
. Entropy'(%)?intrevCyclic integral is zero:- W, no;- Q, no;- net changein volume,yes.-? Depends on state only, not the process path.So,itisaproperty,statefunctionIn 1865,Clausius named this property as ENTROPY, S.theentropychangeisdefinedasAds(KJ/K)ntrev
• Entropy: • Cyclic integral is zero: – W, no; – Q, no; – net change in volume, yes. – ? • Depends on state only, not the process path. – So, it is a property, state function – In 1865, Clausius named this property as ENTROPY, S. – the entropy change is defined as ?
橘名称的由来克劳修斯（1865）创造entropy()的由来：因为它具有能量的含义，所以用字头en，又因它代表热功转换能力，则用希腊字转变的tropy作字段，所以构成新字en-tropy1923年普朗克到南京作报告《热力学第二定律及之观念》讲 entropy,物理学家胡刚复教授任翻泽，由于它概念太复杂，就想了个简单的办法，根据dS=dQ/T，认为S是热“滴”量与温度之商，而且与火有关，从而构成新字
熵名称的由来 • 克劳修斯(1865)创造 entropy(熵)的由来：因为它具有能 量的含义，所以用字头 en,又因它代表热功转換能力，则 用希腊字转变的 tropy 作字段, 所以构成新字 en-tropy • 1923年普朗克到南京作报告《热力学第二定律及熵之观念 》讲 entropy,物理学家胡剛复教授任翻泽，由于它概念太 复杂，就想了个简单的办法,根据 dS=dQ/T，认为S 是热 量与温度之商，而且与火有关，从而构成新字 “熵” 9
(kJ/K): Then, the entropy change of a system during aprocess can be determined by integrationbetween the initial and the final states:(kJ/K)AS=S2-STintrevWeconcernmoreonchangeofS,butnot S itself,sowehaveareferencestate(S=0),The relation of Q and T is often not available, so we get entropy mostly fromtables.Entropyisaproperty,isfixedatfixedstates.Theentropychange betweentwospecifiedstatesisthesamenomatterwhatprocesspathitisfollowed.Thecalculationisonly validforreversiblepath betweenthetwo states;and usedascomparedvalueforirreversibleprocess.10
• Then, the entropy change of a system during a process can be determined by integration between the initial and the final states: – We concern more on change of S, but not S itself, so we have a reference state (S=0). – The relation of Q and T is often not available, so we get entropy mostly from tables. – Entropy is a property, is fixed at fixed states. The entropy change between two specified states is the same no matter what process path it is followed. – The calculation is only valid for reversible path between the two states; and used as compared value for irreversible process. 10