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1st Law of Thermodynamics

A refrigerato...

Question

A refrigerator converts $100 g$ of water at $20^{\circ} C$ to ice at $- 10^{\circ} C$ in $73.5 m i n$ . Calculate the average rate of heat extraction in watt. The specific heat capacity of water is $44.2 J g^{- 1} K^{- 1}$ , specific latent heat of ice is $336 J g^{- 1}$ the specific heat capacity of ice is $2.1 J g^{- 1} K^{- 1}$ .

0.1 W

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100 W

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1 W

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10 W

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Solution

The correct option is B $10 W$
Total heat extracted $= Q = 100 \times 4.2 \times 20 + 100 \times 336 + 100 \times 2.1 \times 10 = 44100 J$
$t = 73.5 m i n = 4410 s e c$
$P = \frac{44100}{4410} = 10 W$

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