Lecture27

Lecture27 - Lecture 27 Purdue University Physics 220 1...

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Unformatted text preview: Lecture 27 Purdue University, Physics 220 1 Lecture 27 Thermodynamics II Textbook Sections: 16.5-16.8 Physics 220 Lecture 26 Purdue University, Physics 220 2 T H T C Q H Q C W HEAT ENGINE T H T C Q H Q C W REFRIGERATOR System System taken in closed cycle ⇒ Δ U system = 0 Therefore, net heat absorbed = work done Q H- Q C = W (engine) Q C- Q H = -W (refrigerator) energy into blob = energy leaving blob Engines and Refrigerators Lecture 26 Purdue University, Physics 220 3 T H T C Q H Q C W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W Efficiency e ˹ W/Q H =W/Q H = (Q H-Q C )/Q H = 1-Q C /Q H Heat Engine: Efficiency Lecture 26 Purdue University, Physics 220 4 T H T C Q H Q C W REFRIGERATOR The objective: remove heat from cold reservoir The cost: work 1st Law: Q H = W + Q C Coefficient of performance K r ˹ Q C /W = Q C /W = Q C /(Q H - Q C ) Refrigerator: Coefficient of Performance Lecture 26 Purdue University, Physics 220 5 iClicker • An ideal heat engine absorbs 36 kJ of heat and exhausts 18 kJ of heat every cycle. What is the efficiency of the engine? Efficiency e ˹ W/Q H A. 1 B. 0.5 C. 2 D. 0.25 Lecture 27 Purdue University, Physics 220 6 Carnot Cycle • Idealized Heat Engine – No Friction – Reversible Process • Isothermal Expansion • Adiabatic Expansion...
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This note was uploaded on 04/23/2011 for the course PHYS 220 taught by Professor Chang during the Spring '09 term at Purdue.

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Lecture27 - Lecture 27 Purdue University Physics 220 1...

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