Explanation

(c) Given that, the temperature of freezer, $T_{2} = - 13℃$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{2} = - 13 + 273 = 260\ K$
Coefficient of performance, $\beta$=5
The coefficient of performance is defined as,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta = \frac{T_{2}}{T_{1} - T_{2}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 = \frac{260}{T_{1} - 260}$
$\therefore\ \ T_{1} - 260 = \frac{260}{5}$
$or\ \ T_{1} - 260 = 52$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{1} = (52 + 260)K = 312\ K$
${\ or\ \ \ \ \ \ \ \ \ \ \ \ \ T}_{1} = (312 - 273)℃$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ T_{1} = 39℃$