$10 g$ of Ice at - $20^{0} C$ is added to $10 g$ of water at $50^{0} C$ . The amount of ice and water, present at equilibrium, are respectively

0, 20 g

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5 g, 15 g

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5 g, 10 g

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10 g, 10 g

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Solution

The correct option is B $5 g, 15 g$
Specific heat of ice is $2.1 J / g^{o} C$
Total heat released by water is $10 \times 4.2 \times 50 = 2100 J$
Total heat absorbed by ice from $- 20^{o} C t o 0^{o} C = 10 \times 2.1 \times 20 = 420 J$
$Δ Q = 2100 - 420 = m L$
Melted ice, $m = (2100 - 420) / 336 \approx 5 g m$
Hence, at equilibrium, total water is $15 g$ , total ice is $5 g$ .

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