Question

A refrigerator has, to transfer an average of 263 J of heat per second from temperature – 10°C to 25°C. Calculate the average power consumed assuming ideal reversible cycle and no other losses.
thankyou ~​

Answer 1

$\huge \mathfrak \red{Answer} \\ \\ \: \: \implies \: \sf 35 \: watt \\ \\ \\ \bf \: Given \: : \\ \\ \: \: \tt Q_{1} = 263j \\ \: \tt \: T_{1} = - 10 {}^{ \circ} C \: = - 10 + 273 = 263 \: k \\ \: \tt \: T_{2} = 25^{ \circ} C = 25 + 273 = 298 \:k \\ \\ \\ \bf \: To \: Find \: : \: \: P ?\: \\ \\ \: \sf \green{Solution :} \\ \\ \: \: \large \: \bf \: P = Q _{2} - Q_{1} \\ \\ \: \: \: \tt \: \rightarrow \frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} \\ \\ \:\rightarrow \tt \: \: \: Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1} \\ \\ \: \tt \: \: \rightarrow \: Q_{2}= \frac{298}{ \cancel 263} \times \cancel 263 \\ \\ \: \: \tt \:\rightarrow \: Q_{2} =298\: \: J / s\\ \\ \: \: \: \sf \: \bf \: \: P= Q_{2} - Q_{1} \\ \tt \: \: \: \: \: \: \: \: \: \:=298 - 263 \\ \\ \: \: \: \tt \: \: \: \: \: \: \rightarrow \: \fbox{ \blue{35 \: watt}}$