If the coefficient of performance of a refrig | vedclass

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(C) The coefficient of performance $(COP)$ is given by $\beta = 5$.The room temperature (sink temperature) is $T_1 = 27 + 273 = 300\,K$.Let the temper

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$-23\,^{\circ}C$

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(C) The coefficient of performance $(COP)$ is given by $\beta = 5$.The room temperature (sink temperature) is $T_1 = 27 + 273 = 300\,K$.Let the temperature inside the refrigerator be $T_2$.The formula for the coefficient of performance of a refrigerator is $\beta = \frac{T_2}{T_1 - T_2}$.Substituting the given values: $5 = \frac{T_2}{300 - T_2}$.Cross-multiplying gives: $5(300 - T_2) = T_2$.$1500 - 5T_2 = T_2$.$1500 = 6T_2$.$T_2 = \frac{1500}{6} = 250\,K$.Converting back to Celsius: $t_2 = 250 - 273 = -23\,^{\circ}C$.

If the coefficient of performance of a refrigerator is $5$ and it operates at a room temperature of $27\,^{\circ}C$,find the temperature inside the refrigerator.

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