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NCERT Exemplar · Q20

Q.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 1 kW power, and heat is transferred from −3°-3°C to 27°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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The ideal (Carnot) coefficient of performance between −3°-3°C and 27°27°C is 9; scaling by the given 50% efficiency gives an actual COP of 4.5, so a 1 kW motor removes heat from the refrigerator at a rate of 4.5 kJ/s, i.e. 4500 J every second.

Converting temperatures

TC=−3+273=270 K,TH=27+273=300 KT_C = -3+273 = 270\text{ K}, \qquad T_H = 27+273 = 300\text{ K}

Ideal (Carnot) coefficient of performance

For a perfect refrigerator operating between these two temperatures:

COPCarnot=TCTH−TC=270300−270=27030=9\text{COP}_{Carnot} = \frac{T_C}{T_H-T_C} = \frac{270}{300-270} = \frac{270}{30} = 9

This means an ideal refrigerator would remove 9 J of heat from the cold side for every 1 J of work put in.

Actual coefficient of performance

The real refrigerator's efficiency is stated as 50% of the ideal:

COPactual=0.5×9=4.5\text{COP}_{actual} = 0.5\times9 = 4.5

Heat removed per second …

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