Lecture 26: Thermodynamics II l Heat Engines l Refrigerators l Entropy l 2 nd Law of Thermodynamics l Carnot Engines

THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR system l system goes through a closed cycle   U system = 0 l therefore, net heat absorbed = work done Q H - Q C = W (engine) Q C - Q H = -W (refrigerator) energy into system = energy leaving system Engines and Refrigerators

THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1 st Law: Q H -Q C = W Efficiency e  W/Q H = (Q H -Q C )/Q H = 1-Q C /Q H Heat Engine: Efficiency

THTH TCTC QHQH QCQC W REFRIGERATOR The objective: remove heat from cold reservoir The cost: work 1st Law: Q H = W + Q C Coeff. of performance: K r  Q C /W = Q C /(Q H - Q C ) Refrigerator: Coefficient of Performance

Entropy (S) l A measure of “disorder.” l A property of a system (just like P, V, T, U): è related to number of number of different “states” of system l Examples of increasing entropy: è ice cube melts è gases expand into vacuum l Change in entropy: è  S = Q/T  S > 0 if heat flows into system (Q > 0)  S < 0 if heat flows out of system (Q < 0)

Second Law of Thermodynamics l Total entropy NEVER decreases: è  S  0 è order to disorder l Consequences: è A “disordered” state cannot spontaneously transform into an “ordered” state. è No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine.”

Carnot Cycle l Idealized Heat Engine è No Friction   S = Q/T = 0 è Reversible Process »Isothermal Expansion »Adiabatic Expansion »Isothermal Compression »Adiabatic Compression è Carnot Efficiency »e c = 1 - T C / T H »e = 1 is forbidden! »e largest if T C << T H

Summary l First Law of Thermodynamics: Energy Conservation  Q =  U + W l Heat Engines: è e = 1-Q C /Q H l Refrigerators: è K p = Q C /(Q H - Q C ) l Entropy:   S = Q/T l 2 nd Law: è Entropy always increases! l Carnot Cycle: Reversible, Maximum Efficiency è e c = 1 – T c /T h

Example l An engine operates between T h = 560 K and T c = 300 K. If the actual efficiency is half its ideal efficiency and the intake heat is Q h = 1285 J every second, at what rate does it do work? First, to find the ideal efficiency we will use: e c = 1 – T c /T h Second, to find the output work we will use: e = W/Q h

Example l An engine operates between T h = 560 K and T c = 300 K. If the actual efficiency is half its ideal efficiency and the intake heat is Q h = 1285 J every second, at what rate does it do work? Find the ideal efficiency: e c = 1 – T c /T h = 1 – (300 K)/(560 K) = 46.4%

Example l An engine operates between T h = 560 K and T c = 300 K. If the actual efficiency is half its ideal efficiency and the intake heat is Q h = 1285 J every second, at what rate does it do work? Find the work done: e = W/Q h 23.3% = W/(1285 J) W = 298 J every second