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Second Law of Thermodynamics Engines and Refrigerators

Second Law of Thermodynamics Engines and Refrigerators. Heat Engine. Any device that transforms heat partly into work or mechanical energy * working substance – matter inside the engine which undergoes inflow and outflow of heat, expansion and compression, and sometimes phase change.

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Second Law of Thermodynamics Engines and Refrigerators

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  1. Second Law of ThermodynamicsEngines and Refrigerators

  2. Heat Engine • Any device that transforms heat partly into work or mechanical energy *working substance – matter inside the engine which undergoes inflow and outflow of heat, expansion and compression, and sometimes phase change

  3. how it works • Working substance absorbs heat from the hot reservoir • Performs some mechanical work • Discards the remaining energy in the form of heat into the cold reservoir *cyclic process Young

  4. 1st law applied to WS QH>0: WS absorbs heat from HR QC<0: WS discards heat into CR W>0: WS performs mechanical work U=0 after one cycle 1st law: Q = U + W Qnet = W QH + QC = W |QH| - |QC| = |W|

  5. thermal efficiency e = W/QH = (|QH|-|QC|)/|QH| = 1 –(|QC|/QH|)

  6. Refrigerator • takes heat from a cold place and gives of heat to a warmer place • requires a net input of mechanical work working substance – refrigerant fluid cold reservoir – inside of refrigerator hot reservoir – outside of refrigerator

  7. how it works • fluid absorbs heat from the cold reservoir • work is done on the fluid • energy from heat transfer and work done is discarded into the hot reservoir • *cyclic process Young

  8. the fluid in the evaporator coil is colder than the inside of the ref, so it absorbs heat • the compressor takes in fluid and compresses it adiabatically (work is done on the fluid) • fluid is delivered to condenser at high pressure, fluid temperature is higher than that of surrounding air, fluid gives of heat and condenses • fluid expands adiabatically (expansion valve) into the evaporator and cools considerably Young

  9. 1st law applied to fluid QC>0: fluid absorbs heat from CR QH<0: fluid discards heat into HR W<0: work is done on fluid U=0 after one cycle 1st law: Q = U + W Qnet = W QH + QC = W -|QH| + |QC| = -|W| |QC| + |W| = |QH|

  10. coefficient of performance K = |QC|/|W| = |QC|/(|QH|-|QC|)

  11. A heat engine takes in 1200 J of heat from the high-temperature heat source in each cycle and does 400 J of work in each cycle. How much heat is released into the environment in each cycle? 400 J 800 J 1200 J 1600 J

  12. A heat engine takes in 1200 J of heat from the high-temperature heat source in each cycle and does 400 J of work in each cycle. What is the efficiency of this engine? ¼ 1/3 3 4

  13. In one cycle, a heat engine takes in 900 J of heat from a high-temperature reservoir and releases 600 J of heat to a lower-temperature reservoir. How much work is done by the engine in each cycle? 300 J 600 J 900 J 1500 J

  14. In one cycle, a heat engine takes in 900 J of heat from a high-temperature reservoir and releases 600 J of heat to a lower-temperature reservoir. What is its efficiency? 3/9 6/9 3/15 6/15

  15. In one cycle a heat engine does 400 J of work and releases 500 J of heat to a lower-temperature reservoir. How much heat does it take in from the higher-temperature reservoir? 100 J 400 J 500 J 900 J

  16. In one cycle a heat engine does 400 J of work and releases 500 J of heat to a lower-temperature reservoir. What is the efficiency of the engine? 4/9 4/5 5/9 1

  17. In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-temperature reservoir, and does 400 J of work. What is its efficiency? ¼ 1/3 2/5 3/5

  18. A heat pump takes in 300 J of heat from a low-temperature reservoir in each cycle and uses 150 J of work per cycle to move the heat to a higher-temperature reservoir. How much heat is released to the higher-temperature reservoir in each cycle? 150 J 300 J 450 J 600 J

  19. A heat pump takes in 300 J of heat from a low-temperature reservoir in each cycle and uses 150 J of work per cycle to move the heat to a higher-temperature reservoir. What is the coefficient of performance (COP)? 1/3 1/2 2 3

  20. In each cycle of its operation, a refrigerator removes 18 J of heat from the inside of the refrigerator and releases 30 J of heat into the room. How much work per cycle is required to operate this refrigerator? 12 J 18 J 30 J 48 J

  21. In each cycle of its operation, a refrigerator removes 18 J of heat from the inside of the refrigerator and releases 30 J of heat into the room. What is the COP of this refrigerator? 18/12 30/12 12/30 18/30

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