Chapter 24 Entropy and the second law of thermodynamics
Chapter 24 Entropy and the second law of thermodynamics
24-1 Two kinds of processes and entropy 1. Irreversible processes(不可逆过程) (a)All naturally occuring processes proceed in one direction only. They never, of their own accord, proceed in the opposite direction See动画库\力学夹\5-08过程的方向性 Such spontaneous one-way processes are irreversible (b) Although the wrong-way events do not occur, none of them would violate the law of conservation of energy
24-1 Two kinds of processes and entropy 1. Irreversible processes（不可逆过程） (a) “All naturally occuring processes proceed in one direction only. They never, of their own accord, proceed in the opposite direction.” Such spontaneous one-way processes are “irreversible”. (b) Although the “wrong-way” events do not occur, none of them would violate the law of conservation of energy. See动画库\力学夹\5-08过程的方向性
2. Reversible process.(可逆过程) See动画库\力学夹\5-09可逆过程 In reversible process, we make a small change in a system and its environment by reversing that change, the system and environment will return to their original conditions In a truly reversible process, there would be no friction, turbulence(a#), or other dissipative effects which will cause non-compensatory losses of energy
2. Reversible process（可逆过程） In reversible process, we make a small change in a system and its environment; by reversing that change, the system and environment will return to their original conditions. In a truly reversible process, there would be no friction, turbulence(紊乱), or other dissipative effects, which will cause non-compensatory losses of energy. See动画库\力学夹\5-09可逆过程
3. Entropy(s Entropy is a physical quantity that controls the direction of irreversible processes It is a property of the state of a system; like t, P, v, eint Entropy principle: If an irreversible process occurs in a closed system the entropy of that system al ways increases; it never decreases
3. Entropy (S) Entropy is a physical quantity that controls the direction of irreversible processes. It is a property of the state of a system; like T, P, V, Eint. Entropy principle: “If an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases
24-2 Entropy change for reversible processes 1.The definition of entropy change for a reversible process △S f d@(reversible)(24-1) T Here dQ is the increment of heat energy that is transferred into (or out)of the closed system at temperature T. If the process is isothermal △S (24-2) e>0,AS>0,the entropy of that system increases. if Q<0, AS<0)
24-2 Entropy change for reversible processes 1.The definition of entropy change for a reversible process: (reversible) (24-1) = f i T dQ S Here dQ is the increment of heat energy that is transferred into (or out) of the closed system at temperature T. If the process is isothermal, (24-2) , , the entropy of that system increases.( if Q<0, ). T Q S = Q 0 S 0 S 0
2. Entropy as a state property What 's the entropy change for ideal gases? dQ+dw=dEnt from the first lay ∫aW=-Pl de =nc dt -/y Int nRT do= Pav+nc. aT dv+nc.dT Divided by t, we obtain nr-+nC dT Integrate between an arbitrary initial state i and an final state f △S nRin+nCy Only related to initial and final states
Divided by T, we obtain dQ + dW = dEint dW = −PdV dEint = nCv dT dV nC dT V nRT dQ = PdV + nCv dT = + v T dT nC V dV nR T dQ = + v 2. Entropy as a state property Integrate between an arbitrary initial state i and an final state f: i f V i f f i T T nC V V nR T dQ S = = ln + ln from the first law ※ What’s the entropy change for ideal gases? P=nRT/V Only related to initial and final states
Sample problem 24-1 A vessel containing 1. 8kg of water is placed on a hot plate. Both the water and hot plate being initially ato c The temperature of the hot plate is raised very slowy to at which point the water begins to bog. Find of the water during this process? Solution@= mcdT △S Meet dT : TmcIn =1820J/K
Sampleproblem 24-1 A vessel containing 1.8kg of water is placed on a hot plate. Both the water and hot plate being initially at . The temperature of the hot plate is raised very slowly to , at which point the water begins to boil. Find of the water during this process? Solution: c 20 c 100 S dQ = mcdT J/K T T mc T dT mc T dQ S i f T T T T f i f i Δ = = = ln =1820
24-3 Entropy change for irreversible processes Due to unavoidable non-compensatory losses of energy in an irreversible process, it is difficult to calculate AS directly for it How to obtain As for irreversible processes? Two steps 1. Design a reversible process that has the same initial and final states as those in the irreversible process 2. Use Eq(24-1) to calculate ASfor this reversible process The obtained As is actually the one for the original irreversible process since entropy change is as a state property
• Due to unavoidable non-compensatory losses of energy in an irreversible process, it is difficult to calculate directly for it. 24-3 Entropy change for irreversible processes S 1. Design a reversible process that has the same initial and final states as those in the irreversible process. 2. Use Eq(24-1) to calculate for this reversible process. The obtained is actually the one for the original irreversible process since entropy change is as a state property. S How to obtain for irreversible processes? S •Two steps: S
An example Irreversible process reversible process Insulation nsulation T T N waterN Nwater Water water Insulating slab Thermal reservoir (a)Initial state (b)Final state(a)Initial state (b)Final state Fi24-1 Fg24-2
h water m (a) Initial state (b) Final state Fig 24-1 Ti Tf Insulating slab An example: water h water Thermal reservoir m Ti Tf water (a) Initial state (b) Final state Fig 24-2 Irreversible process reversible process Insulation Insulation
Sample problem 24-2 AS Fig 24-1, m=1.5kg h=2.5m m=4, 5kgt= 300K la what is the temperature riser of the system water+ stone? b) find the entropy change As of this system (c What would be the as for the reverse process-that is, for the system to cool down, transferring its energy to the stone in kinetic form, allowing it to leap 2.5m into the air The specific heat of water is C=4190 g· kand that of the stone material is c=790 kg·k
Sample problem 24-2 As Fig 24-1, , , , (a) What is the temperature rise of the system water + stone? (b) Find the entropy change of this system. (c) What would be the for the reverse process—that is, for the system to cool down, transferring its energy to the stone in kinetic form, allowing it to leap 2.5m into the air. The specific heat of water is and that of the stone material is . ms = 1.5kg h = 2.5m mw = 4.5kg Ti = 300K S T S kg K C J w = 4190 kg K C J s = 790
