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^@ C to 27^@ C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.

In a refrigerator one removes heat from a lower temperature and deposi

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 1 kW 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 :Temperature of source is `27^@C` `therefore T_1=27+273`=300 K Temperature of sink `T_2=-3+273`=270 K The EFFICIENCY of a perfect ENGINE , `eta=1-T_2/T_1` `therefore eta=1-270/300=30/300=0.1` The efficiency of a REFRIGERATOR , `eta.`=50% efficiency of a perfect engine, `=0.1xx50/100` =0.05 Coefficient of performance of a refrigerator , `beta=Q_2/W=(1-eta.)/(eta.)` `therefore beta=(1-0.05)/0.05 =0.95/0.05`=19 `therefore Q_2=betaW` =19 x 1 kW =19 kW `=19 "KJ"/s` Therefore , heat is taken out of the refrigerator at a rate of 19 kJ per second.