A refrigerator works between 4^{circ}C and 30 | vedclass

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(B) Given: Cold reservoir temperature $T_2 = 4^{\circ}C = 277 \, K$,Hot reservoir temperature $T_1 = 30^{\circ}C = 303 \, K$.Heat removed per second $

Vedclass · Accepted Answer

$236.5$

Vedclass · Answer

(B) Given: Cold reservoir temperature $T_2 = 4^{\circ}C = 277 \, K$,Hot reservoir temperature $T_1 = 30^{\circ}C = 303 \, K$.Heat removed per second $Q_2 = 600 \, cal/s$.Coefficient of performance $\alpha = \frac{T_2}{T_1 - T_2}$.$\alpha = \frac{277}{303 - 277} = \frac{277}{26}$.We know that $\alpha = \frac{Q_2}{W}$,where $W$ is the work done per second (power).Therefore,$W = \frac{Q_2}{\alpha} = \frac{600 	imes 26}{277} \, cal/s$.Converting to Joules: $W = \frac{600 	imes 26}{277} 	imes 4.2 \, J/s$.$W \approx 236.5 \, W$.

$A$ refrigerator works between $4^{\circ}C$ and $30^{\circ}C$. It is required to remove $600 \, cal$ of heat every second in order to keep the temperature of the refrigerated space constant. The power required is ....... $W$ (Take $1 \, cal = 4.2 \, J$)

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