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Chapter 18 The Second Law of Thermodynamics

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Irreversible Processes Irreversible Processes: always found to proceed in one direction Examples: free expansion of ideal gas heat flow (for finite T): from hot to cold friction: converts mechanical energy to heat

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Reversible Processes Reverisble processes: just an idealization, but we can come close put system in equilibrium (with itself and its surroundings) make slow, infinitesimal changes in order to reverse a process

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Heat Engine Heat engine: transforms heat into mechanical energy, W Reservoirs: hot reservoir at T H cold reservoir at T C

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Heat Engine engine absorbs heat Q H from hot reservoir engine discards heat Q C to cold reservoir engine does work W

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Heat Engine Working substance: material in engine undergoing the heat transfer Cyclic process: reuse same substance, return it to initial state at end of each cycle

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Q = net heat input For each cycle: heat Q H enters engine heat Q C leaves engine net heat in per cycle: Q = Q H + Q C = Q H – |Q C |

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Heat Engine cyclic process: U = 0 for engine during one cycle 1st Law: Q = W + U Thus for each cycle: Q = W

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Heat Engine For each cycle: Q = Q H – |Q C | So by 1st Law: W = Q = Q H – |Q C |

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Engine Efficiency W= Q H – |Q C | Perfect engine: |Q C | = 0 All experiments: |Q C | > 0 Efficiency: e measures how close an engine gets to |Q C | = 0

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Engine Efficiency

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Ideal Engine = limiting behavior of real engines engine: converts heat to mechanical energy recall: converting mechanical energy to heat is ‘irreversible’

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Ideal Engine avoid irreversibility carry out engine cycle slowly and with reversible heat flow the reversibility of heat flow motivates the allowed processes

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Ideal Engine For reversible heat flow, there can be no finite T between engine and reservoir so during heat transfer we need an isothermal process

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Ideal Engine If there is a finite T between engine and reservoir, any heat flow would be irreversible So when there is a finite T we need an adiabatic process

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Ideal Engine so we are motivated to consider a cycle with only two processes: isothermal adiabatic called a ‘Carnot cycle’

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Ideal Engine efficiency: e = ? working substance = ? We’ll calculate e for substance = ideal gas Actually, e is independent of the working substance

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Carnot Cycle

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Example engine: internal combustion engine

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T H is provided by combustion of air-fuel mixture T C is provided by exhaust gases venting to outside air

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fuel = gasoline working substance = mixture of air and burned fuel this engine doesn’t actually recycle the same mixture in each cycle, but we can idealize with simple models

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Problem 18-44

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Refrigerator Refrigerator: like a heat engine, but operating in reverse Reservoirs: hot reservoir at T H cold reservoir at T C

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Refrigerator refrigerator absorbs heat Q C refrigerator discards heat Q H work W done on refrigerator

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Refrigerator cyclic process: U = 0 during one cycle 1st Law: Q = W + U Thus for each cycle: Q = W

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Refrigerator So by 1st Law: W = Q = Q H + Q C – |W| = – |Q H | + Q C |Q H | = Q C + |W|

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Announcements Today: finish Chapter 18 Wednesday: review Thursday: no class (office hours 12:30-2:30) HW 6 (Ch. 15):returned at front Midterms: returned at front (midterm scores: classweb, solutions: E-Res)

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The Second Law of Thermodynamics

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2nd Law of Thermodynamics Four equivalent approaches, in terms of: Engines Refrigerators Carnot Cycle Entropy

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Heat Engine engine absorbs heat Q H engine discards heat Q C engine does work W engine efficiency :

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2nd Law: Engines There are no perfect engines (real engines have e < 1) It is impossible for a system to change heat completely into work, with no other change to the system taking place

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2nd Law: Engine Version If 2nd Law weren’t true: ships could move by cooling the ocean cars could move by cooling surrounding air

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Refrigerator refrigerator absorbs heat Q C refrigerator discards heat Q H work W done on refrigerator

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2nd Law: Refrigerator Version There are no perfect refrigerators (for real refrigerators |W|> 0). It is impossible for heat to flow from hot to cold with no other change to the system taking place

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‘Engine’ and ‘Refrigerator’ versions of 2nd Law are equivalent

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(a)If you can build a perfect refrigerator, then you can build a perfect engine (perfect refrigerator) + (real engine) = perfect engine See notes

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(b)If you can build a perfect engine, then you can build a perfect refrigerator (perfect engine) + (real refrigerator) = perfect refrigerator See notes

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Engine Efficiency Revisited 2nd Law: Real engines have efficiency e < 1. How close can we get to e = 1? Is there a maximum possible efficiency? Yes!

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Carnot Cycle

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2nd Law: Carnot Cycle Version e < e Carnot The efficiency e of a real engine operating between temperatures T H and T C can never exceed that of a Carnot engine operating between the same T H and T C : Proof

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Problem 18-44, Revisited

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Defining Entropy For Carnot cycle: See notes

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Any reversible cycle = sum of many Carnot cycles For cycle, in limit of infinitesimally close isotherms:

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Entropy (S) Result for a reversible cycle: This defines a new state variable, entropy (S):

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Entropy measures the disorder of a system The more heat dQ you add, the higher the entropy increase dS that results The smaller the temperature T, the larger the entropy increase dS that results

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Entropy (S) For any reversible path from state 1 to 2: This must also equal S for an irreversible path from 1 to 2, since S is a state variable Exercises 18-20, 18-26 Example 18-8, Problem 18-50

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Example 18-8: Free Expansion of Ideal Gas

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Entropy (S) For any reversible path from state 1 to 2: This must also equal S for an irreversible path from 1 to 2, since S is a state variable Exercises 18-20, 18-26 Example 18-8, Problem 18-50

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2nd Law: Entropy Version If we consider all systems taking part in a process, For a reversible process: S total = 0 For an irreversible process: S total > 0

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Laws of Thermodynamics Law state variable Zeroth Law temperature: T 1st Law U = Q – W internal energy: U 2nd Law entropy: S

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Announcements Today: finish Chapter 18 Wednesday: review Thursday: no class (office hours 12:30-2:30) HW 6 (Ch. 15):returned at front Midterms: returned at front (midterm scores: classweb, solutions: E-Res)

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