Answer
120 W.
Work Step by Step
We want the minimum power requirement, so assume that the freezer is ideal.
Heat enters the freezer due to conductivity. Calculate it using equation 14–5.
$$\frac{Q_L}{t}=kA\frac{T_H-T_L}{\ell}$$
This heat must be removed from the freezer to stabilize the temperature inside. Apply the equation for the COP, 15–6a. The work is provided by the cooling motor.
$$COP=\frac{Q_L}{W}=\frac{T_L}{T_H-T_L}$$
This motor is to remove the heat in $15\%$ of the time that it took for the heat to enter the freezer, because it is only allowed to fun for $15\%$ of the time.
$$COP=\frac{Q_L/t}{W/(0.15t)}=\frac{T_L}{T_H-T_L}$$
Solve for the cooling motor’s power.
$$\frac{W}{t}=\frac{Q_L/t}{0.15}\frac{T_H-T_L}{T_L}$$
$$=\frac{ kA\frac{T_H-T_L}{\ell}}{0.15}\frac{T_H-T_L}{T_L}$$
$$=\frac{ (0.050W/(m \cdot K))(8.0m^2)\frac{37K}{0.12m}}{0.15}\frac{37K}{258K}$$
$$=117.9W\approx 120W$$
