Chapter 19 - PHYS 132, WEEK 11 November 29-December 3 19.3....

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PHYS 132, WEEK 11 November 29-December 3 19.3. Solve: (a) During each cycle, the heat transferred into the engine is Q H = 55 kJ , and the heat exhausted is Q C = 40 kJ . The thermal efficiency of the heat engine is h = 1 - Q C Q H = 1 - 40 kJ 55 kJ = 0.273 (b) The work done by the engine per cycle is W out = Q H - Q C = 55 kJ - 40 kJ = 15 kJ 19.7. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: K = Q C W in 4.0 = Q C 50 J Q C = 200 J (b) The heat exhausted to the hot reservoir is Q H = Q C + W in = 200 J + 50 J = 250 J 19.8. Model: Assume that the car engine follows a closed cycle. Solve: (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is W out = 500 kJ s 1 s 40 cycles = 12.5 kJ cycle (b) The heat input per cycle is calculated as follows: = W out Q H Q H = 12.5 kJ 0.20 = 62.5 kJ The heat exhausted per cycle is Q C = Q H - W in = 62.5 kJ - 12.5 kJ = 50 kJ 19.9. Solve: The amount of heat discharged per second is calculated as follows: = W out Q H = W out Q C + W out Q C = W out 1 - 1 ˆ ˜ = 900 MW ( ) 1 0.32 - 1 ˆ ˜ = 1.913 10 9 W That is, each second the electric power plant discharges 1.913 10 9 J of energy into the ocean. Since a typical American house needs 2.0 10 4 J
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Chapter 19 - PHYS 132, WEEK 11 November 29-December 3 19.3....

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