College Physics, Young,Geller CH16-CH20

College Physics, Young,Geller CH16-CH20 - THE SECOND LAW OF...

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Unformatted text preview: THE SECOND LAW OF THERMODYNAMICS 16 9. C 10. B, D 11. A, D 12. A, B 13. A, C Answers to Multiple-Choice Problems 1. B 2. B 3. B 14. A, D 15. B 4. A, B, C, D 5. A 6. A 7. B, D 8. B Solutions to Problems 16.1. Set Up: e 5 W / Q H and W 5 Q H 1 Q C . W and Q H are positive; Q C is negative. Q H 5 150 J. Solve: W 5 eQ H 5 1 0.33 2 1 150 J 2 5 50 J. Q C 5 W 2 Q H 5 50 J 2 150 J 5 2 100 J. 100 J of heat is wasted. 16.2. Set Up: W 5 Q H 1 Q C . W and Q H are positive; Q C is negative. Q H 5 1 325 J and Q C 5 2 250 J. e5 QC W 511 . QH QH 2 250 J Solve: e 5 1 1 5 0.231 5 23.1%. W 5 Q H 1 Q C 5 2 250 J 1 325 J 5 1 75 J 325 J W . Q H . 0, Q C , 0. QH Solve: (a) W 5 2200 J. 0 Q C 0 5 4300 J. Q H 5 W 1 0 Q C 0 5 6500 J. 2200 J 5 0.34 5 34%. (b) e 5 6500 J Reflect: Since the engine operates on a cycle, the net Q equal the net W. But to calculate the efficiency we use the heat energy input, Q H 16.3. Set Up: For a heat engine, W 5 0 Q H 0 2 0 Q C 0 . e 5 16.4. Set Up: For a heat engine, W 5 0 Q H 0 2 0 Q C 0 . e 5 Solve: (a) W 5 9000 J 2 6400 J 5 2600 J. 2600 J W 5 5 0.29 5 29%. (b) e 5 QH 9000 J W . Q . 0, Q C , 0. QH H 16.5. Set Up: For a heat engine, W 5 0 Q H 0 2 0 Q C 0 . e 5 Solve: (a) e 5 W . Q . 0, Q C , 0. 1 hp 5 746 W. QH H 3700 J W 5 5 0.230 5 23.0%. QH 1.61 3 104 J (b) 0 Q C 0 5 0 Q H 0 2 W 5 1.61 3 104 J 2 3700 J 5 1.24 3 104 J. 1.61 3 104 J (c) m 5 5 0.350 g. 4.60 3 104 J / g 1 60.0 2 1 3700 J 2 5 2.22 3 105 W 5 222 kW 5 298 hp. (d) P 5 1.00 s 16-1 ...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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