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(B) The temperatures are $T_L = -3^{\circ} C = 270 \ K$ and $T_H = 27^{\circ} C = 300 \ K$.The coefficient of performance $(COP)$ for an ideal Carnot

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$5$

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(B) The temperatures are $T_L = -3^{\circ} C = 270 \ K$ and $T_H = 27^{\circ} C = 300 \ K$.The coefficient of performance $(COP)$ for an ideal Carnot refrigerator is given by $COP_{ideal} = \frac{T_L}{T_H - T_L}$.$COP_{ideal} = \frac{270}{300 - 270} = \frac{270}{30} = 9$.The actual $COP$ is $50\%$ of the ideal $COP$: $COP_{actual} = 0.5 	imes 9 = 4.5$.We know that $COP = \frac{Q_L}{W}$,where $Q_L$ is the heat removed per second and $W$ is the work done per second (power).Given $W = 1 \ kW = 1 \ kJ/s$.Therefore,$Q_L = COP_{actual} 	imes W = 4.5 	imes 1 \ kJ/s = 4.5 \ kJ/s$.

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$T_{1} = 27^{\circ}C$ [outside fridge]
$T_{2} = -23^{\circ}C$ [inside fridge]

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