Question

Freezing compartment of a refrigerator is at $0^{\circ} C$ and room temperature is $27.3^{\circ} C$. Work done by the refrigerator to freeze $1\, g$ of water at $0^{\circ} C$ is $\left(L_{ ice }=80\, cal\, g ^{-1}\right)$

• A. 336 J
• B. 33.6 J
• C. 3.36 J
• D. 40 J

Answer 1

The Correct Option is B

Coefficient of performance of refrigerator is
$\beta=\frac{T_{2}}{T_{1}-T_{2}}$
$T_{2}=0^{\circ} C =0+273=273\, K $
$T_{1}=27.3^{\circ} C \approx 300\, K$
$\therefore \beta=\frac{273}{300-273} \approx 10$
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 make it ice
$=m L_{1}=80\, cal =80 \times 4.2\, J =336\, J$
So, $W=336 / 10=336\, J$