An ideal refrigerator has a freezer at a temp | vedclass

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(C) The coefficient of performance $(K)$ of an ideal refrigerator is given by the formula: $K = \frac{T_2}{T_1 - T_2}$,where $T_2$ is the temperature

Vedclass · Accepted Answer

$39^{\circ}C$

Vedclass · Answer

(C) The coefficient of performance $(K)$ of an ideal refrigerator is given by the formula: $K = \frac{T_2}{T_1 - T_2}$,where $T_2$ is the temperature of the freezer and $T_1$ is the temperature of the surroundings.Given: $T_2 = -13^{\circ}C = 273 - 13 = 260K$ and $K = 5$.Substituting the values into the formula: $5 = \frac{260}{T_1 - 260}$.Multiplying both sides by $(T_1 - 260)$: $5(T_1 - 260) = 260$.$5T_1 - 1300 = 260$.$5T_1 = 1560$.$T_1 = \frac{1560}{5} = 312K$.Converting back to Celsius: $T_1 = 312 - 273 = 39^{\circ}C$.

An ideal refrigerator has a freezer at a temperature of $-13^{\circ}C$. The coefficient of performance of the refrigerator is $5$. The temperature of the air (to which heat is rejected) will be:

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