The temperatures inside and outside a refrige | vedclass

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(A) Given: Temperature inside the refrigerator $T_2 = 273 \, K$,Temperature outside (surrounding) $T_1 = 300 \, K$.For a reversible refrigerator,the c

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$11$

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(A) Given: Temperature inside the refrigerator $T_2 = 273 \, K$,Temperature outside (surrounding) $T_1 = 300 \, K$.For a reversible refrigerator,the coefficient of performance $\beta$ is given by $\beta = \frac{T_2}{T_1 - T_2}$.Substituting the values: $\beta = \frac{273}{300 - 273} = \frac{273}{27} \approx 10.11$.The coefficient of performance is also defined as $\beta = \frac{Q_2}{W}$,where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done.Given $W = 1 \, J$,we have $Q_2 = \beta 	imes W = 10.11 	imes 1 = 10.11 \, J$.The heat delivered to the surrounding $Q_1$ is given by $Q_1 = Q_2 + W$.$Q_1 = 10.11 \, J + 1 \, J = 11.11 \, J$.Rounding to the nearest integer,the heat delivered is approximately $11 \, J$.

The temperatures inside and outside a refrigerator are $273 \, K$ and $300 \, K$ respectively. Assuming that the refrigerator cycle is reversible,for every joule of work done,the heat delivered to the surrounding will be nearly ........ $J$.

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