Question

A refrigerator removes heat from a refrigerated space at 0°C at a rate of 2.2 kJ/s and rejects it to an environment at 20°C. what is the minimum required power input?

Answer 1

Answer:

The minimum power required by the refrigerator is 161 W to remove heat from a refrigerated space.

Explanation:

To calculate the power required to refrigerate we know that COP of refrigerator is represented as $C O P=\frac{Q_{\text {coid}}}{W}C O P=\frac{T_{\text {cold }}}{T_{\text {hot }}-T_{\text {cold}}}$. Given that, $T_{\text {cold}}=0^{\circ} \mathrm{C}, T_{\text {hot}}=22^{\circ} \mathrm{C} \text { and } Q_{\text {cold}}=2.2 \mathrm{KJ} / \mathrm{s}$

$O P=\frac{T_{\text {cold }}}{T_{\text {hot }}-T_{\text {cold }}}=\frac{0+273}{(22+273)-(0+273)}=\frac{273}{293-273}=\frac{273}{20}=13.65$

$C O P=\frac{Q_{\text {cold}}}{w}=>13.65=\frac{2.2}{w} W=\frac{2.2}{13.65}=0.161 \mathrm{KJ} / \mathrm{s}=161 \mathrm{W}(0.1 \mathrm{KJ} / \mathrm{s}=100 \mathrm{W})$

Thus we can say that minimum power required by refrigerator if a refrigerator removes heat to an environment at 20°C from a refrigerated space at 0°C at 2.2 kJ/s is 161W.