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(A) The coefficient of performance $(COP)$ of a refrigerator is given by the ratio of heat extracted $(Q_L)$ to the work done $(W)$: $COP = \frac{T_L}

Vedclass · Accepted Answer

$263$

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(A) The coefficient of performance $(COP)$ of a refrigerator is given by the ratio of heat extracted $(Q_L)$ to the work done $(W)$: $COP = \frac{T_L}{T_H - T_L} = \frac{dQ_L/dt}{dW/dt}$.Here,$T_L = -10^{\circ}C = 263 \ K$ and $T_H = 25^{\circ}C = 298 \ K$.The power consumed is $dW/dt = 35 \ W$.Substituting the values: $COP = \frac{263}{298 - 263} = \frac{263}{35}$.Now,$\frac{dQ_L}{dt} = COP 	imes \frac{dW}{dt} = \frac{263}{35} 	imes 35 = 263 \ J/s$.

$A$ refrigerator consumes an average $35 \ W$ power to operate between temperatures $-10^{\circ}C$ and $25^{\circ}C$. If there is no loss of energy,how much average heat per second does it transfer? (in $J/s$)

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