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

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

Step-by-step 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}</strong>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 }}}=(0+273)/((22+273)-(0+273))=(273)/(293-273)=(273)/(20)=13.65`

`C O P=\frac{Q_{\text {cold}}}{w}=>13.65=(2.2)/(w) W=(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.