A Carnot refrigerator operates between reservoirs at $-20°C$and $50°C$in a $25°C$room. The refrigerator is a $40cm\times 40cm\times 40cm$box. Five of the walls are perfect insulators, but the sixth is a $1.0-cm-$thick piece of stainless steel. What electric power does the refrigerator require to maintain the inside temperature at $-20°C$?

The temperature of cold reservoir is $-20°C$, the temperature of hot reservoir is$50°C$, the thickness of the piece of stainless steel is$1cm$and the dimension of refrigerator is$\left(40cm\times 40cm\times 40cm\right)$.

The formula used to calculate thermal efficiency of the engine is, $e=1-\frac{T_L}{T_H}$

$.e$is the efficiency of the refrigerator.

$.T_L$is the temperature of cold reservoir.

$.T_H$is the temperature of hot reservoir.

The formula to calculate power is,

$$\begin{gathered} p=\frac{dQ}{dt} \\ =\frac{kA\left(T-T_L\right)}{et} \end{gathered}$$

$.A$is the area of refrigerator.

$.T$is the room temperature.

$.t$is the thickness of piece of steel.

$.k$is the thermal conductivity of steel.

Substitute $\left(1-\frac{T_L}{T_H}\right)$for $e$in above equation.

$p=\frac{kA\left(T-T_L\right)}{\left(1-\frac{T_L}{T_H}\right)t}$

Substitute $50.2$for $k$, $\left(40cm\times 40cm\right)$for $A,25°C$for $T,-20°C$for $T_{L,}50°C$for and for in above equation to find $P$.

$P=\frac{(50.2)(40cm\times 40cm)\left(\right(25°C+273)K-(-20°C+273\left)K\right)}{(1-\frac{(-20°C+273)K}{(50°C+273)K})}$

localid="1649586972505" role="math" $$\begin{gathered} =\frac{(50.2)(1600cm)\left(\frac{{10}^{-2}m}{1cm}\right)\left(\right(298)K-(253\left)K\right)}{(1-\frac{\left(253\right)K}{\left(323\right)K})} \\ =28kW \end{gathered}$$

Conclusion :

Therefore, the electric power required by the refrigerator to maintain the inside temperature is localid="1649586987870" $28kW$.