1 Chapter 15 The second law of thermodynamics The zeroth law of thermodynamics: The expression of the fundamental experiment fact. The first law of thermodynamics: The a statement of conservation of energy for pure thermodynamic system. The second law of thermodynamics: Why we need the second law of thermodynamics? 2 §15.1 why do some things happen, while others do not? One of the sacred truths of physics is the principle of energy conservation. 1 A rock does not jump spontaneously up to the top of the cliff;----violate the CWE theorem 2 Pizza does not warm itself; ----violate the first law of thermodynamics 3 Water cannot become into ice automatically; 4A drop of ink spread throughout water never regroup into a drop-shaped clump. For instance: 1. Some examples that can not happen It is not the energy of the system that controls the direction of irreversible processes; it is another property that we introduce in this chapter. §15.1 why do some things happen, while others do not? 2. What is the irreversible processes The isothermal expansion of an ideal gas: 1frictionless, quasi-static 21 VV → 0ln 1 2 11 >== V V nRTQW 12 VV → 0ln 2 1 22 <== V V nRTQW result ： 0 21 =+WW 0 21 =+QQ Reversible! 3 §15.1 why do some things happen, while others do not? 2friction, quasi-static 21 VV → f WWQ += 11 f W V V nRT += 1 2 ln (>0) (>0) 12 VV → f WWQ + ′ = 12 f W V V nRT += 2 1 ln (< 0) (>0) 02 21 >=+ f WQQ Irreversible!result ： §15.1 why do some things happen, while others do not? 4 3frictionless, not quasi-static 1 2 11 lnd 2 1 V V RTnVPWQ V V ∫ =<= 1 2 22 lnd 1 2 V V nRTVPWQ V V =>= ∫ result ： 0 2121 <+=+ WWQQ The wok done on the system： 12 WW ? The heat transfer to the environment： 12 QQ ? Irreversible! §15.1 why do some things happen, while others do not? §15.2 heat engines and the second law of thermodynamics 1. Heat engines and refrigerator engines We use the word engine to mean a system such as a gas that performs a closed cycle on P-V diagram. If the cycle is performed clockwise, we called it a heat engine; If counterclockwise, we called it a refrigerator engine. The function of a heat engine is to transform disordered internal energy into macroscopic work using the heat transfer between reservoirs at different temperatures. 5 2. The efficiency of the heat engines C Q C T W For a cycle: CH QQWQ U ?== = total 0? V P 0>W §15.2 heat engines and the second law of thermodynamics Q H Q C W 1 W 2 H C H CH H Q Q Q QQ Q W ?= ? == 1 net ε §15.2 heat engines and the second law of thermodynamics 6 Define the efficiency of the heat engine: H Q W =ε C Q The first law of thermodynamics does not forbid ε =1 Perfect engine ε =1 H C H CH Q Q Q QQ ?= ? = 1ε §15.2 heat engines and the second law of thermodynamics 3. The second law of thermodynamics A perfect heat engines (ε =1) do not exist. 4. The examples for efficiency of heat engines (a) Clausius-Rankine cycle 0d21 =→ p )( 12 TTnCQ pH ?= 0d32 =→ Q 1 13 1 22 ???? = γγγγ pTpT §15.2 heat engines and the second law of thermodynamics 7 0d43 =→ p )( 34 TTnCQ pC ?= 0d14 =→ Q 1 21 1 14 ???? = γγγγ pTpT H C Q Q ?=1ε 12 43 1 TT TT ? ? ?= γ γ? ?= 1 1 2 )(1 p p 10 2 <∴≠ εPQ §15.2 heat engines and the second law of thermodynamics (b) Otto cycle H Q C Q W e b c d V i V f V P a §15.2 heat engines and the second law of thermodynamics 8 1 1 1 1)(1 11 ? ? ?=?= ?= ? ? ?= r r i f c b cd be V V T T TT TT α ε f i V V =α Ratio of compression H Q C Q W a b c d V i V f V P e §15.2 heat engines and the second law of thermodynamics )( )( beVC cdVH TTncQ TTncQ ?= ?= c →d e →b 11 11 ?? ?? = = γγ γγ f f VTVT VTVT cib ied d →e b →c (c) Diesel cycle a d c b V o p 1 V 3 V 2 V H Q C Q )1()( 1)( 1 2 3 1 2 1 2 3 ? ? ?= ? V V V V V V γ γ γ ε §15.2 heat engines and the second law of thermodynamics b→c 3 2 V V T T c b = 11 2 1 3 1 1 1 ?? ?? = = γγ γγ VTVT VTVT ab cd c→d a→b )( )( adVC bcPH TTncQ TTncQ ?= ?=b →c d →a 9 §15.3 The Carrnot heat engine and its efficiency 1. Carrnot cycle of ideal gases a b HH V V nRTQba ln=→ d c CC V V nRTQdc ln|| =→ 11 ?? =→ r cC r bH VTVTcb 11 ?? =→ r dC r aH VTVTad d c a b V V V V = H C H CH a b H d c C a b H H CH T T T TT V V nRT V V nRT V V nRT Q QQ ?= ? = ? = ? = 1 ln lnln || ε H C T T ?=1ε --the efficiency of Carrnot cycle for ideal gases 2. The efficiency of Carrnot cycle §15.3 The Carrnot heat engine and its efficiency 10 §15.4 refrigerator engines and the second law of thermodynamics 1. The refrigerator engines C Q C T V P 0<W WQQ QQWQ CH CH += ?== total Define the coefficient of performance: 1 1 ? = ? == C HCH CC Q QQQ Q W Q K A perfect refrigerator engine is one that has an infinite coefficient of performance. It does not violate the first law of thermodynamics 2. The second law of thermodynamics A perfect refrigerator engines (K=∞) do not exist. §15.4 refrigerator engines and the second law of thermodynamics 11 3. The Carnot refrigerator engine Isothermal processes: 1 2 ln12 V V nRTQ HH =→ 4 3 ln34 V V nRTQ CC =→ 1 3 1 2 23 ?? =→ γγ VTVT CH 1 4 1 1 41 ?? =→ γγ VTVT CH 4 3 1 2 V V V V = p V o CH TT > C T Adiabatic processes: §15.4 refrigerator engines and the second law of thermodynamics CH C TT T K ? = Carnot 4 3 1 2 4 3 lnln ln V V nRT V V nRT V V nRT QQ Q W Q K CH C CH CC ? = ? == §15.4 refrigerator engines and the second law of thermodynamics 12 §15.5 The efficiency of real heat engines and refrigerator engines 1. Heat engines The efficiency of a Carnot heat engine is the maximum efficiency of a heat engine operating between the two given reservoirs. What is the most efficient heat engine or refrigerator that can operate between two given reservoirs? Prove this statement by contradiction! C T H Q C Q HC Q CC Q Proto- type engine Assume: HCH Q W Q W > Then: HHC QQ > §15.5 The efficiency of real heat engines and refrigerator engines 13 Prototype heat engine: CH QQW ?= Carnot refrigerator engine: CCHC QQW ?= HHC QQ ? CCC QQ ? CCHCCH QQQQ ?=? CCCHHC QQQQ ?=? Then we have Violate the second law of thermodynamics! Perfect refrigerator! §15.5 The efficiency of real heat engines and refrigerator engines 2. Refrigerator engines A Carnot refrigerator engine has the maximum coefficient of performance of any refrigerator engine operating between the same two reservoirs. §15.5 The efficiency of real heat engines and refrigerator engines Similar arguments can be used to prove above statements. 14 同学们好！ §15.5 The efficiency of real heat engines and refrigerator engines 3. Absolute zero and the third law of thermodynamics The absolute zero of temperature is an unataintable temperature; reservirs with T C =0 Cannot exist. 15 §15.6 A new concept: entropy 1. Entropy is a state variable For a Carnot cycle: C C H H C H C H T Q T Q T T Q Q == or 0=+ C C H H T Q T Q For a arbitrary quasi-static cycle: 0 d or0 cycle reversible == ∫ ∑ T Q T Q i i Initial state final state Path 1 Path 2 V P 0 dd =+ ∫∫ i f f i T Q T Q Path 1 Path 2 0 dd =? ∫∫ f i f i T Q T Q Path 1 Path 2 ∫∫ = f i f i T Q T Q dd Path 1 Path 2 T Q S d d = Define: the differential change in the entropy §15.6 A new concept: entropy 16 For a reversible processes: ∫∫ == revrev 0 d d T Q S ∫∫ == f i f i T Q SS d d? ? ? ? ? ? ? ? = == << >> = cycle0 0d0 0d0 0d0 Q Q Q S? Isentropic process 2. The calculation of the entropy §15.6 A new concept: entropy For irreversible processes: Since the entropy is a state variable, the entropy change between two equilibrium state is the same regardless of how the change is executed. 3. Entropy changes in various thermodynamic processes (a) Ideal gas in a quasi-static reversible process ∫∫ ∫∫ += + == f i f i V V T T V f i f i V V nR T T nc T VPU T Q S dd ddd ? §15.6 A new concept: entropy 17 i f i f V V V nR T T ncS lnln +=? (b) Melting or boiling a mass m: T mL T Q S ==? T mL T Q S ?==? (c) Freezing or condensation of a mass m: §15.6 A new concept: entropy (d) Warming or cooling solid or liquid of mass m i f T T f i T T mc T Tmc T Q S f i ln dd === ∫∫ ? (e) Heat transfer to a reservoir at temperature T C C T Q S =? (f) Heat transfer from a reservoir at temperature T H H T Q S ?=? §15.6 A new concept: entropy 18 Example 1: the change of entropy in the process of adiabatic expansion of ideal gas. Solution: 000 ===+= UWUWQ ?? §15.6 A new concept: entropy This process is not a reversible process, how to calculate the entropy? Because the entropy is a state variable, we can use the reversible isothermal process to calculate the entropy. 2lnln lnln nR V V nR V V nR T T ncS i f i f i f V == +=? 0>S? §15.6 A new concept: entropy 19 Example 2: Find the entropy change of a system of two reservoirs when heat transfer |Q| occurs from the warmer reservoir at temperature T H to the cooler reservoir at temperature T C . Solution: C f i H f i T Q T Q S T Q T Q S == ?== ∫ ∫ d d ? ? Warmer reservoir: Cooler reservoir: Total entropy: 0 total >?= HC T Q T Q S? §15.6 A new concept: entropy Example 3: A mass m 1 at temperature T H is placed in thermal contact with another mass m 2 at temperature T C . The system is isolated, so the heat transfer occurs only between the two masses. The final equilibrium temperature is T, where T C < T < T H . Find the total entropy change associated with the heat transfer between the hot and cold masses.the specific heats of the masses are c 1 and c 2 respectively. Solution: H T T f i T T cm T T cm T Q S H ln dd 11111 === ∫∫ ? §15.6 A new concept: entropy 20 C T T f i T T cm T T cm T Q S C ln dd 22222 === ∫∫ ? CH T T cm T T cmSSS lnln 221121total +=+= ??? In every case one will find that the total entropy change is positive. Example 4: A 3.0 kg sample of water is warmed from 0oC to 100oC. what is the entropy change of the water? §15.6 A new concept: entropy Solution: i f T T f i T T mc T T mc T Q S f i waterwater dd === ∫∫ ? )J/K(1092.3) 273 373 (ln418600.3 3 ×=××=S? §15.6 A new concept: entropy 0>S? 21 §15.7 entropy and the second law of thermodynamics 1.General statement of second law of thermodynamics For an irreversible process: H C H C T T Q Q ?<?= 11ε 0<+ C C H H T Q T Q For a arbitrary cycle: 0 d or0 cycle leirreversib << ∫ ∑ T Q T Q i i Initial state final state Path 1(irrev) Path 2 V P 0 dd <+ ∫∫ i f f i T Q T Q Path 1 Path 2 0 dd <? ∫∫ f i f i T Q T Q Path 1 Path 2 ∫∫ > f i f i T Q T Q irrevrev dd Path 1Path 2 ∫ >? f i if T Q SS irrev d T Q S d d ≥ §15.7 entropy and the second law of thermodynamics 22 The total entropy change of an isolated system is always greater than or equal to zero 0 isolatedtotal ≥S? The total entropy change is equal to zero only for reversible processes. Notice: 1isolated system, the entropy change can be negative, positive, and zero for individual components of the system, the total entropy is always positive. The second law of thermodynamics: §15.7 entropy and the second law of thermodynamics 2the total entropy change of an isolated system is equal to zero only when the system undergoes a quasi-static reversible process. 3 the total entropy change of an isolated system in an irreversible process is always greater than zero. 2. Perfect heat engine and perfect refrigerator engine do not exist System: heat engine, high temperature reservoir, low temperature reservoir. For one cycle: (a) Perfect heat engine §15.7 entropy and the second law of thermodynamics 23 1 The entropy change of Heat engine: 0 1 =S? 2 The entropy change of high temperature reservoir: H H T Q S ?= 2 ? 3 The entropy change of low temperature reservoir: 0 3 == C C T Q S? 000 321 <+?=++= H H T Q SSSS ???? Violate the second law of thermodynamics! §15.7 entropy and the second law of thermodynamics 0= C Q (b) perfect refrigerator engine System: refrigerator engine, high temperature reservoir, low temperature reservoir. For one cycle: The entropy change of low temperature reservoir: C T Q S ?= 2 ? 3The entropy change of high temperature reservoir: H T Q S = 3 ? The entropy change of refrigerator engine: 0 1 =S? §15.7 entropy and the second law of thermodynamics C T 24 00 321 <+?=++= HC T Q T Q SSSS ???? Violate the second law of thermodynamics! Example : A Carnot engine, in executing one cycle, uses 2000J of heat transfer from a reservoir at 500K, performs work, and effects heat transfer to a low-temperature reservoir at 300K. (a)Find the efficiency of the Carnot heat engine. (b)determine the work done by the heat engine in one cycle. §15.7 entropy and the second law of thermodynamics (c)Find the total entropy change of the isolates system of the heat engine and two reservoirs. Solution: (a) 4.0 500 300 11 Carnot =?=?= H C T T ε (b) J80020004.0 =×==∴ = H H QW Q W ε εQ §15.7 entropy and the second law of thermodynamics 25 (c)heat engine 0 1 =S? High temperature reservoir H H T Q S ?= 2 ? J/K0 300 8002000 500 2000 0 321 = ? ?= ?+=++= C C H H T Q T Q SSSS ???? C C T Q S = 3 ?low temperature reservoir §15.7 entropy and the second law of thermodynamics §15.8 The direction of heat transfer and the second law of thermodynamics 1.Why heat transfer is always from hotter objects to cooler objects and not the other way? Q H T C T Q H T C T 0<?= += CH CH T Q T Q SSS ??? 0>+?= += CH CH T Q T Q SSS ??? 26 Example : verify that it is impossible that water of 2.00kg at temperature 25.0oC freeze to ice at temperature -20.0oC automatically. Solution: 2freezing the water at 0.0oCto ice at 0.0oC 1cooling the water from 25.0oC to 0.0oC J/K734 298 273 ln dd water 273 298 water 1 ?== == ∫∫ mc T Tmc T Q S f i T T ? §15.8 The direction of heat transfer and the second law of thermodynamics 3freezing the ice at 0.0oC to the ice at -20.0oC J/K1044.227310335.300.2 d 35 water 2 ×?=÷××?= ?=?== ∫ T mL T Q T Q S f i T T ? J/K312 273 253 ln dd ice 253 273 ice 3 ?== == ∫∫ mc T Tmc T Q S f i T T ? J/K3486)3122440734( system ?=++?=S? §15.8 The direction of heat transfer and the second law of thermodynamics 27 J1058.9 )]20(205010335.3)25(4186[0.2 )( 0)( 5 5 21 21 ×= ?×+×??××= +??= =+?+ TmcmLTmcQ TmcmLTmcQ icewaterwater icewaterwater ?? ?? J/K1021.3 298 J1058.9 3 5 room room ×= × == T Q S? 4the heat transfer to the room and the entropy change of the room 0J/K27632103486 total <?=+?=S? §15.8 The direction of heat transfer and the second law of thermodynamics H H C H H C Q T T QW T T )1( 1 ?== ?= ε ε T H T C Q C Q H W 2.entropy and the energy crisis §15.8 The direction of heat transfer and the second law of thermodynamics 28 WW Q T T QW H H C H < ′ ′ ?= ′ = ′ )1(ε T H T H ′ T C Q C Q H Q H W ′ H C T T ′ ?= ′ 1ε §15.8 The direction of heat transfer and the second law of thermodynamics