Question

Suppose that heat leaks into your kitchen refrigerator at an average rate of 300 watts. Assuming ideal operation, how much power must it draw from the wall?

Short answer

The power drawn from wall is 57.69 W.

Step-by-step solution

Given information

Assume temperatures T1 = 250 K and T2 = 298 K
Power leak = 300 W

Explanation

COP can be found by using $\mathrm{COP}=\frac{T_1}{T_2-T_1}$

Substitute the values

$$\begin{gathered} \mathrm{COP}=\frac{250}{298-250} \\ COP=5.2 \end{gathered}$$

The amount of power drawn from the wall is:

$$\begin{gathered} P=\frac{P_{\text{avg}}}{COP} \\ P=\frac{300\mathrm{W}}{5.2} \\ P=57.69\mathrm{W} \end{gathered}$$