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An ideal refrigerator has a freezer at a temperature of $-13^{\circ} \mathrm{C}$. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be
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$39^{\circ} \mathrm{C}$
Coefficient of performance
$\begin{gathered}K=\frac{T_2}{T_1-T_2} \Rightarrow 5=\frac{(273-13)}{T_1-(273-13)}=\frac{260}{T_1-260} \\ \Rightarrow 5 T_1-1300=260 \Rightarrow 5 T_1=1560 \\ \Rightarrow T_1=312 \mathrm{~K} \rightarrow 39^{\circ} \mathrm{C}\end{gathered}$
$\begin{gathered}K=\frac{T_2}{T_1-T_2} \Rightarrow 5=\frac{(273-13)}{T_1-(273-13)}=\frac{260}{T_1-260} \\ \Rightarrow 5 T_1-1300=260 \Rightarrow 5 T_1=1560 \\ \Rightarrow T_1=312 \mathrm{~K} \rightarrow 39^{\circ} \mathrm{C}\end{gathered}$
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