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.

Options

A. 14 J
B. 12 J
C. 19 J
D. 20 J

Answer

C

Explanation

Efficiency of a perfect engine working between `-3^@C` and `27^@C` (i.e., `T_2`= 270 K and `T_1` = 300 K )
`eta_"engine"=1-T_2/T_1=1-"270 K"/"300 K"=0.1`
Since efficiency of 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