Question
Considering a household refrigerator as a reversible engine operating between the melting point of ice and room temperature $27^{\circ} C$, estimate the energy provided joules ( $J$ ) to freeze 1 kg of water, the temperature of water is given in $0^{\circ} C , L = 8 0 Cal g ^{-1}$.

Answer

Given that :
$\begin{aligned}T_1 & =27+273=300 K \\T_2 & =0+273=273 K \\m & =1 kg=1000 g, L=80 calg^{-1} \\\text { Melting energy } \quad Q_2 & =m \times L \\& =1000 \times 80 cal=8 \times 10^4 cal\end{aligned}$
Using the relation
$\begin{aligned}\frac{Q_1}{Q_2} & =\frac{T_1}{T_2} \\Q_1 & =\left(\frac{T_1}{T_2}\right) \times Q_2=\frac{300}{273} \times 8 \times 10^4 \\& =8.79 \times 10^4\end{aligned}$
Desired deliverable energy
$\begin{aligned}W & =Q_1-Q_2 \\W & =8.79 \times 10^4-8 \times 10^4 \\& =0.79 \times 10^4 cal \\& =7.9 Kcal .\end{aligned}$

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