GENERAL PHYSICS II Electromagnetism & Thermal Physics 5/6/2008 Chapter XVI The Second Law of Thermodynamics §1 Reversible Carnot cycles §2 The second law of thermodynamics §3 Entropy and quantitative formulation of the second law §4 Heat engines and refrigerators 5/6/2008 The first law of thermodynamics gives the quantiative relations between the internal energy of a system and the quantities of heat and work that the system exchange with surroundings It express the conservation of energy The first law of thermodynamics is true, but not enough ! Why ? Many thermodynamic processes which don’t violate the 1st law, but don’t happen in nature ! • Examples: In an isolating system hot heat flow cold a heat transfer from the cold part to the hot part doesn’t happen ! No one has succeded in building a machine that converts heat completely into mechanical energy It is necessary to have a supplemental law which says about the directions of thermodynamic processes → the 2nd law 5/6/2008 §1 Reversible Carnot cycle: We are interested in the directions of thermodynamic processes First we introduce the concept of reversible and irreversible processes 1.1 Revisible and irrevisible processes:  Definition: A reversible thermodynamic process is a process which can takes place in the direct as well as in the reverse way In the reverse process the system passes through all the immediate states that it undergoes in the direct process Otherwise the process is called irreversible  It is observed by experiment that in nature thermodynamic process are all irreversible process But for a theoretical consideration it is convenient to deal with reversible processes, as idealized models 5/6/2008 1.2 Reversible Carnot cycle: The reversible Carnot cycle is an idealized model for processe in heat engines  Consider a cyclic process which is represented by the closed path P → Q → P The mechanical work done in the cycle is In the P-V diagram, this work is pesented by the area within the closed path 5/6/2008  The Carnot cycle consists of steps: 1) A → B: isothermal expansion to volume VB at temperature T2 2) B → C: adiabatic expansion to volume VC , tempeature drops to T1 < T2 3) C → D: isothermal compression to volume VD at constant temperature T 4) D → A: adiabatic compression to the original state of the system  Exchanges of heat and work in each step: • During s-1: the system absorbs heat Q (Q2 > 0) and does work WAB (WAB > 0) An ideal gas is used for working substance 5/6/2008 Q2  AB W P-V diagram of a Carnot cycle VB  ln nRT VA • During s-3: the system rejects heat Q , Q1 < and the work is done on the system, WCD < 0: Q1  CD W VC  nRT1 ln  VD • For two adiabatic steps s-2 & s-4:  1 T2VB  1VC & T2VA   1VD  T  T 1   VB VC  1 1  VA VD  VB VC  VA VD 1  VC / VD ) Q1 T ln( T1      ln(V / V )  Q2 T T2 2  B A P-V diagram of a Carnot cycle | Q2 | | Q1 |  T2 T1 The ratio of the heat rejected and the heat absorbed is just equal to the ratio of the temperatures 5/6/2008 What is the results of a cycle ABCDA ? - The system absorbs heat Q from the surroundings at temperature T2 (hot reservoir) , deposits heat Q1 in the surroundings at temperature T1 (cold reservoir) - The internal energy of the system remains unchanged - The net mechanical work done by the system in one cycle is represented by the area enclosed and is positive From the 1st law: Wnet = |Q2| - |Q1| > (since T2 > T1 ) P-V diagram of a Carnot cycle By continously repeating the cycle it is possible to mechanical work by continously taking heat energy from a hot reservoir and depositing a smaller amount of heat energy in a cold reservoir Such a device is called a heat engine 5/6/2008 Consider the cycle in the reverse direction A→D→C→B→A: The result is as follows: Mechanical work is put into the system in order to take heat energy from the cold reservoir and deposit heat in a higher-temperature reservoir In this case the device is a refrigerator We see that heat energy can’t flow from the cold reservoir to the hot reservoir without action of an external mechanical work 1.3 Efficiency of a Carnot engine: For a heat engine the efficiency represents the fraction of heat that is converted to work Mechanical work done by engine Efficiency = 5/6/2008 Heat energy supplied from the hot reservoir From the general definition we can calculate the efficiency of a Carnot engine: eCarnot W | Q2 |  Q1 | | | Q1 | T1     Q2 | Q2 | | Q2 | T2 Remark: The efficiency of Carnot engine depends only on the temperatures of two heat reservoirs The efficiency is large when the temperature difference is large, and the efficiency is very small when the temperatures are nearly equal The efficiency can never be exactly unity unless T1 = This means heat can never be converted completely into mechanical work Although the Carnot engine is an idealized model, but the conclusions derived from it give an good orientation for investigation of other heat engines that we usually encounter 5/6/2008 10 3.2 Entropy changes in a reversible process:  First consider a Carnot cycle It is a idealized reversible process Recall for a reversible Carnot cycle we have the following equation: |Q | |Q1|  T2 T1 Since Q1  Q1  | Q1 |  Q1 Q   T1 T2 In two adibatic steps there is no heat exchange → the net change in entropy in one cycle is   S1  S2  S   There is no net entropy change in one reversible Carnot cycle ! 5/6/2008 17  Now consider an arbitrary reversible cyclic process for an ideal gas: It can beconsidered to be equivalent to a large number of “thin” Carnot cycles For any of “thin” Carnot cycle we have Q1 Q   T1 T2 The total entropy change is Qi T 0 i i The approximation becomes better as if the number of subcycles is larger, and every subcycle is thiner, so we obtain for an arbitrary reversible cyclic process: dQ  0 T 5/6/2008 18  This result leads also to the consequence: The entropy change of a system in a reversible process from A to B is independent of the path P B dQ  S  T doesn’t depend on the path ! A V 3.3 A quantitative statement of the second law:  practise, reversible processes are only idealized situations In  the real world, friction can never be completely eliminated, and in the In reverse direction of a process, it is difficult for the system to collect back completely the energy which was lost by friction in the direct process → real thermodynamic processes are irrevisible 5/6/2008 19 m m m m Consider an example of irreversible process: The system consists of two identical metal blocks The whole system is thermally isolated from the surroundings The initial state a): the temperatures of two blocks are different T1 & T2 The final state b): the two blocks in contact → T = (½)(T1 + T2 )   T  (T1  )      dQ dQ dT dT T T     S  mc  mc  ln  mc ln  2mc ln  mc        T T2 T T1 T T2 T T T  T1T2  1  2  T1     T 5/6/2008 T T T 20 Entropy increases Investigation of various processes in isolated systems show that For an isolated thermodynamic system Δ ≥0 , S and the equality sign refers to reversible processes This is a quantitative statement of the second law We have expressed the second law of thermodynamics by an equation for entropy 5/6/2008 21 3.4 Microscopic interpretation of entropy: In the system in picture: Remove partition → molecules spontanously distribute themselves throughout the container This is an irrevsersible process: molecules never can spontaneously gather themselves in one corner as at the beginning So Δ > , entropy increases S Molecules are more ordered We can conclude that entropy can use as a quantitative measure of the degree of disorder in a system Molecules are more In the example, entropy increases, at the same disordered time the degree of disorder of molecules increases also  Notice that the concepts of order/disorder have the meaning for systems which consist of a very large number of particles (molecules) For systems of few particles the concept of entropy plays no any role 5/6/2008 22 §4 Heat engines and refrigerators: 4.1 Heat engines:  The function: Turn heat into work  The standard heat engine works on a cyclic process: Hot reservoir at Th Qh Recycle over and over Wby Engine: Qc Cold reservoir at Tc 5/6/2008 extract heat from a hot reservoir, perform work, dump excess heat into a cold reservoir (often the environment) A “reservoir” is a large body whose temperature doesn’t change when it absorbs or gives up heat 23  Gasoline (internal-combution) engine : p Combustion b pb P-V diagram for a idealized model Is shown in the picture It is called Otto cycle c pa exhaust / intake a adiabats d V1 V2 V Calculate the theoretical efficiency of an Otto cycle 5/6/2008 24 p Combustion b pb Wby e  Qin c pa exhaust / intake a Qin Cv (Tb  a ) T adiabats d Wby  bc  Wd  a | W | V1 V2 V Qb c  (adiabatic)  Wb c  b  c  v (Tb  c ) U U C T similarly Wd a  d  a  v (Td  a ) U U C T Wby  v (Tb  c )  v (Ta  d ) C T C T Cv (Tb  c )  v (Ta  d ) T C T (Tc  d ) T e  Cv (Tb  a ) T (Tb  a ) T 5/6/2008 25 p Combustion b pb (Tc  d ) T e (Tb  a ) T pa a adiabats Tc  =TbV1  Tc =Tb (V1/V2 )1/ V   d 1/ Td V2 =Ta V1  Td =Ta (V1 /V2 ) V1 1/  (Tc  d ) (Tb  a )(V1 / V2 ) T T   (Tb  a ) T (Tb  a ) T 1  2  V e      1   r   V 1  exhaust / intake c 1/    1/   2  V V       V2    V  r ≡V2 / V1 = 10 V V2 1  2  V    V 1  (compression ratio) = 1.4 (diatomic gas) e = 60% (in reality about 30%, due to friction, turbulence, ) 5/6/2008 26  The diesel engine: Combustion p adiabat exhaust / intake adiabat V V2 V3 V1 Diesel engine is similar in operation to the gasoline enegine, the differences are: There is no fuel at the beginning of compressin stroke → the compression ratio can be much higher than in gasoline engines, and pre-ignition can’t occur This improves the efficiency 5/6/2008 27 4.2 Refrigerators: Kitchen,295K  Function: Take heat from a cool place ad give it off a warmer place, using a net input work Refrigerator and air conditioner operate on the same principle  From the 1st law: Hot reservoir at Th Qh Won Fridge: QC Food, 263K Cold reservoir at TC QH + QC – W = Heat Leak Since W < 0, QC > 0, QH < → |QH| = QC + |W| A frigerator is characterized by the coefficient of performance K : QC QC K  W QH QC 5/6/2008 28  For a conditioner the quantities of greatest practical importance are the rate of heat removal H and the power input P : QC H t H K   W P.t P The technical unit of H is Btu / h , (1W = 3.413 Btu / h) For a conditioner 10.000 Btu / h , the electric power input ~ 1.200 W Then K ~ (dimenisionless) In the technical units H / P are measured by (Btu / h) / W , and are called “energy efficiency rating” (EER) If K ~ → EER ~ 10 5/6/2008 29 Summary  Three equivalent form of the second law: “engine” statement; “refrigerator” statement; and Δ ≥0 S dQ T S    In any reversible cyclic process:   dQ  Entropy change in a reversible process   S  doesn’t depend T on the path W  Efficiency of a heat engine: e  Qhot For a heat egine operating on the Carnot cycle: 5/6/2008 eCarnot | Q1 | T1   | Q2 | T2 30 For a gasoline engine: 1  r : compression ratio 2  V 1  e     r  capacity ratio  heat  V1    Coefficient of performance for a refrigerator: Q QC K C  W QH QC and for a air conditioner: 5/6/2008 QC H t H K   W P.t P 31 .. .Chapter XVI The Second Law of Thermodynamics §1 Reversible Carnot cycles §2 The second law of thermodynamics §3 Entropy and quantitative formulation of the second law §4 Heat engines... thermal efficiency This impossibility is the basis of the following statement of The second law of thermodynamics 2.1 “Engine” statement of the second law: “It is impossible for any system to... refrigerators 5/6/2008 ? ?The first law of thermodynamics gives the quantiative relations between the internal energy of a system and the quantities of heat and work that the system exchange with