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In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from –3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

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n = 1 - 270/300 = 1/10

Efficiency of refrigerator = 0.5n = 1/20

If Q is the heat/s transferred at higher temperture then W/Q = 1/20

or Q = 20W = 20kJ,

and heat removed from lower temperture = 19 kJ.

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