Freezing compartment of a refrigerator is at $0^{\circ} \mathrm{C}$ and room temperature is $27.3^{\circ}…
Freezing compartment of a refrigerator is at $0^{\circ} \mathrm{C}$ and room temperature is $27.3^{\circ} \mathrm{C}$. Work done by the refrigerator to freeze $1 \mathrm{~g}$ of water at $0^{\circ} \mathrm{C}$ is $\left(L_{\text {ice }}=80 \mathrm{cal} \mathrm{g}^{-1}\right)$
336 J
33.6 J
3.36 J
40 J
Solution
Coefficient of performance of refrigerator is
$
\begin{aligned}
\beta & =\frac{T_2}{T_1-T_2} \\
T_2 & =0^{\circ} \mathrm{C}=0+273=273 \mathrm{~K} \\
T_1 & =27.3^{\circ} \mathrm{C} \approx 300 \mathrm{~K} \\
\therefore \quad \beta & =\frac{273}{300-273} \approx 10
\end{aligned}
$
Now, if $Q_2$ is heat extracted and $W$ is work performed them,
$
\beta=\frac{Q_2}{W} \text { or } W=\frac{Q_2}{\beta}
$
where, $Q_2=$ heat extracted from $1 g$ of water to make it ice
$
=m L_1=80 \mathrm{cal}=80 \times 4.2 \mathrm{~J}=336 \mathrm{~J}
$
So, $W=336 / 10=33.6 \mathrm{~J}$