1.

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.

Answer»

Solution :`eta=1-(270)/(300)=(1)/(10)`
Efficiency of refrigerator`=0.5eta=(1)/(20)`
If Q is the HEAT/s transferred at HIGHER temperature then `(W)/(Q)=(1)/(20) or Q=20W=20kJ`.
and heat REMOVED from lower temperature `=(Q-W)`
`=(20k-1k)=19kj`.


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