The motor in a refrigerator has a power of 200 W. If the freezing compartment is at 270 K and the outside air is at 300 K , and assuming the efficiency of a Carnot refrigerator, what is the maximum amount of energy that can be extracted as heat from the freezing compartment in 10.0 min?

Power of motor in a refrigerator, $\mathrm{P}=200\mathrm{W}$

The temperature of the freezing compartment,${\mathrm{T}}_{\mathrm{L}}=270\mathrm{K}$

The temperature of the outside air, ${\mathrm{T}}_{\mathrm{H}}=300\mathrm{K}$

Time of extraction of heat from the compartment,$\mathrm{t}=10.0\min \left(\frac{60\mathrm{s}}{1\min }\right)=600\mathrm{s}$

The power of the motor and the time for which it is running is given. So, we can find the work done by the motor. Also, enough data to calculate the coefficient of performance of the refrigerator is given. Using the relation between coefficient of performance, heat energy extracted, and work done, we can find the required answer.

Formulae:

The work done by the Carnot refrigerator,

$\mathrm{W}=\frac{{\mathrm{Q}}_{\mathrm{L}}}{\mathrm{K}}$ (1)

The work done per cycle by the Carnot refrigerator,

$\mathrm{W}=\mathrm{Pt}$ (2)

The coefficient of performance of the Carnot refrigerator,

$\mathrm{K}=\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}$ (3)

The coefficient of performance using the given values in equation (3) is calculated as:

$\begin{array}{l}\mathrm{K}=\frac{270\mathrm{K}}{300\mathrm{K}-270\mathrm{K}}\\ =9\end{array}$

Substituting the given values and equation (2) in equation (1), we can get the maximum amount of energy as given:

$$\begin{gathered} \mathrm{Pt}=\frac{{\mathrm{Q}}_{\mathrm{L}}}{\mathrm{K}} \\ {\mathrm{Q}}_{\mathrm{L}}=\mathrm{PtK} \\ {\mathrm{Q}}_{\mathrm{L}}=200\times 600\times 9\mathrm{J} \\ {\mathrm{Q}}_{\mathrm{L}}=1.08\times {10}^6\mathrm{J} \end{gathered}$$

Hence, the value of the maximum energy is$1.08\times {10}^6\mathrm{J}$