A refrigerator of coefficient of performance | vedclass

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(D) The coefficient of performance $(\beta)$ of a refrigerator is given by the formula: $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted fro

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$300$

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(D) The coefficient of performance $(\beta)$ of a refrigerator is given by the formula: $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.Given: $\beta = 5$ and $Q_2 = 250 \ J$.Substituting the values: $5 = \frac{250}{W} \implies W = \frac{250}{5} = 50 \ J$.The heat released to the room $(Q_1)$ is the sum of the heat extracted and the work done: $Q_1 = Q_2 + W$.$Q_1 = 250 \ J + 50 \ J = 300 \ J$.Therefore, the heat released per cycle to the room is $300 \ J$.

$A$ refrigerator of coefficient of performance $5$ that extracts heat from the cooling compartment at the rate of $250 \ J$ per cycle is placed in a room. The heat released per cycle to the room by the refrigerator is (in $J$)

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