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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.
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Verified Answer
The correct answer is:
$1 \mathrm{~kJ}$
Coefficient of performance of refrigerator,
$$
\beta=5
$$
Amount of heat removed,
$$
Q=5000 \mathrm{~J}
$$
Electrical energy utilised by the motor
$$
\begin{aligned}
& & =\text { Work done by the motor }(W) \\
\therefore & & \beta=\frac{Q}{W} \\
\Rightarrow & W & =\frac{Q}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}=1 \mathrm{~kJ}
\end{aligned}
$$
$$
\beta=5
$$
Amount of heat removed,
$$
Q=5000 \mathrm{~J}
$$
Electrical energy utilised by the motor
$$
\begin{aligned}
& & =\text { Work done by the motor }(W) \\
\therefore & & \beta=\frac{Q}{W} \\
\Rightarrow & W & =\frac{Q}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}=1 \mathrm{~kJ}
\end{aligned}
$$
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