An electrical refrigerator with $\beta=5$ extracts $5000 \mathrm{~J}$ from the contents of the refrigerator.…
An electrical refrigerator with $\beta=5$ extracts $5000 \mathrm{~J}$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.
$1 \mathrm{~kJ}$
$0.5 \mathrm{~kJ}$
$0.8 \mathrm{~kJ}$
$1.2 \mathrm{~kJ}$
Solution
Coefficient of performance of refrigerator,
$
\beta=5
$
Amount of heat removed,
$
Q=5000 \mathrm{~J}
$
Electrical energy utilised by the motor
$
\begin{aligned}
& & =\text { Work done by the motor }(W) \\
\therefore & & \beta=\frac{Q}{W} \\
\Rightarrow & W & =\frac{Q}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}=1 \mathrm{~kJ}
\end{aligned}
$