First slide

Refrigerator

Question

In a cold storage, ice melts at the rate of 2 kg/h when the external temperature is $20^{0} C$ . Find the minimum power output of the motor used to drive the refrigerator which just prevents the ice from melting. Latent heat of fusion of ice = 80 cal/g.

Moderate

A

28.5 W
B

9.75 W
C

16.4 W
D

3.25 W

Solution

$ω = \frac{T_{2}}{T_{1} - T_{2}} = \frac{273}{293 - 273} = \frac{273}{20}$
Mass of ice melting per second = $\frac{2 \times 1000}{3600} = (\frac{5}{9}) g$
To prevent the melting of ice, heat to be driven out per second, i.e., $Q_{2} = (\frac{5}{9}) \times 80 = \frac{400}{9} cal = 44.44 J$
As, $ω = \frac{Q_{2}}{W}, W = \frac{Q_{2}}{ω} = \frac{44.44}{(\frac{273}{20})} = 3.25 W$
Power of the motor = 3.25 W

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