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In 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. |
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Answer» 14 J `eta_"engine"=1-T_2/T_1=1-"270 K"/"300 K"=0.1` Since efficiencyof the refrigerator `(eta_"REF")` is 50% of `eta_"engine"` `therefore eta_"ref."-0.5 eta_"engine"=0.05` If `Q_1` is the heat transferred per SECOND at higher temperature by doing work W, then `eta_"ref."=W/Q_1` or `Q_1=W/eta_"ref."="1 kJ"/"0.05"`= 20 kJ (as W=1 kW x 1s = 1 kJ) Since `eta_"ref."`is 0.05 , heat REMOVED from the refrigerator per second, i.e., `Q_2=Q_1-eta_"ref." Q_1=Q_1(1-eta_"ref.")`=20 kJ (1-0.05) =19 kJ |
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