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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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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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