An ideal refrigerator has a freezer at a temperature of $ - 13^\circ 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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(c) Coefficient of performance
$K = \frac{{{T_2}}}{{{T_1} - {T_2}}}$==>$5 = \frac{{(273 - 13)}}{{{T_1} - (273 - 13)}} = \frac{{260}}{{{T_1} - 260}}$
==> $5{T_1} - 1300 = 260$==> $5{T_1} = 1560$
==> ${T_1} = 312K \to 39^\circ C$
$K = \frac{{{T_2}}}{{{T_1} - {T_2}}}$==>$5 = \frac{{(273 - 13)}}{{{T_1} - (273 - 13)}} = \frac{{260}}{{{T_1} - 260}}$
==> $5{T_1} - 1300 = 260$==> $5{T_1} = 1560$
==> ${T_1} = 312K \to 39^\circ C$
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