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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 pwer, 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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