In a cold storage,ice melts at the rate of 2 | vedclass

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(B) The rate of heat entering the cold storage is $Q_2 = \frac{mL}{t} = \frac{2 	imes 10^3 \ g 	imes 80 \ cal/g}{3600 \ s} = \frac{160000}{3600} \ c

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$13.6$

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(B) The rate of heat entering the cold storage is $Q_2 = \frac{mL}{t} = \frac{2 	imes 10^3 \ g 	imes 80 \ cal/g}{3600 \ s} = \frac{160000}{3600} \ cal/s = \frac{400}{9} \ cal/s$.Converting to Joules per second (Watts): $Q_2 = \frac{400}{9} 	imes 4.2 \ J/s = 186.67 \ W$.The coefficient of performance $(COP)$ of a Carnot refrigerator is given by $	ext{COP} = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W}$.Here,$T_2 = 0^{\circ} C = 273 \ K$ and $T_1 = 20^{\circ} C = 293 \ K$.So,$\frac{273}{293 - 273} = \frac{186.67}{W}$.$\frac{273}{20} = \frac{186.67}{W}$.$W = \frac{186.67 	imes 20}{273} \approx 13.67 \ W$.Thus,the minimum power output is approximately $13.6 \ W$.

In a cold storage,ice melts at the rate of $2 \ kg$ per hour when the external temperature is $20^{\circ} C$. The minimum power output of the motor used to drive the refrigerator which just prevents the ice from melting is (latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $W$)

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