In a refrigerator,one removes heat from a lower temperature and deposits it 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 $-3\,^{\circ}C$ to $27\,^{\circ}C$,find the heat taken out of the refrigerator per second,assuming its efficiency is $50\%$ of a perfect engine.

(N/A) The temperature of the source (surroundings) is $T_1 = 27^{\circ}C = 27 + 273 = 300\,K$.
The temperature of the sink (refrigerator interior) is $T_2 = -3^{\circ}C = -3 + 273 = 270\,K$.
The efficiency of a perfect Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
$\eta = 1 - \frac{270}{300} = 1 - 0.9 = 0.1$.
The actual coefficient of performance $(COP)$ is related to the efficiency of a perfect engine. For a refrigerator,the maximum $COP$ is $\beta_{max} = \frac{T_2}{T_1 - T_2} = \frac{270}{300 - 270} = \frac{270}{30} = 9$.
Given that the refrigerator's efficiency is $50\%$ of the perfect engine,the actual $COP$ is $\beta = 0.5 \times \beta_{max} = 0.5 \times 9 = 4.5$.
Since $\beta = \frac{Q_2}{W}$,where $Q_2$ is the heat removed and $W$ is the work done per second $(1\,kW)$,
$Q_2 = \beta \times W = 4.5 \times 1\,kW = 4.5\,kJ/s$.
Therefore,the heat taken out of the refrigerator per second is $4.5\,kJ$.