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Second Law of ThermodynamicsEngines and Refrigerators

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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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how it works1. Working substance

absorbs heat from the hot reservoir

2. Performs some mechanical work

3. Discards the remaining energy in the form of heat into the cold reservoir

*cyclic processYoung

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1st law applied to WSQH>0: WS absorbs heat from HRQC<0: WS discards heat into CRW>0: WS performs mechanical workU=0 after one cycle

1st law: Q = U + W

Qnet = WQH + QC = W|QH| - |QC| = |W|

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thermal efficiency

e = W/QH = (|QH|-|QC|)/|QH|= 1 –(|QC|/QH|)

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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 refrigeratorhot reservoir – outside of refrigerator

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how it works1. fluid absorbs heat

from the cold reservoir

2. work is done on the fluid

3. energy from heat transfer and work done is discarded into the hot reservoir

*cyclic processYoung

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1. the fluid in the evaporator coil is colder than the inside of the ref, so it absorbs heat

2. the compressor takes in fluid and compresses it adiabatically (work is done on the fluid)

3. fluid is delivered to condenser at high pressure, fluid temperature is higher than that of surrounding air, fluid gives of heat and condenses

4. fluid expands adiabatically (expansion valve) into the evaporator and cools considerably

Young

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1st law applied to fluidQC>0: fluid absorbs heat from CRQH<0: fluid discards heat into HRW<0: work is done on fluidU=0 after one cycle

1st law: Q = U + W

Qnet = WQH + QC = W-|QH| + |QC| = -|W| |QC| + |W| = |QH|

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coefficient of performance

K = |QC|/|W| = |QC|/(|QH|-|QC|)

• 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?

A. 400 JB. 800 JC. 1200 JD. 1600 J

• 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?

A. ¼B. 1/3C. 3D. 4

• 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?

A. 300 JB. 600 JC. 900 JD. 1500 J

• 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?

A. 3/9B. 6/9C. 3/15D. 6/15

• 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?

A. 100 JB. 400 JC. 500 JD. 900 J

• 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?

A. 4/9B. 4/5C. 5/9D. 1

• 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?

A. ¼B. 1/3C. 2/5D. 3/5

• 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?

A. 150 JB. 300 JC. 450 JD. 600 J

• 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)?

A. 1/3B. 1/2C. 2D. 3

• 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?

A. 12 JB. 18 JC. 30 JD. 48 J

• 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?

A. 18/12B. 30/12C. 12/30D. 18/30