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The specific heat of water = 4200 J kg−1 K−1 and the latent heat of ice = 3.4 × 10⁵ J kg−1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams):

1. 61.7
2. 69.3
3. 64.6
4. 63.8

Correct answer: 61.7

Solution

Heat released by water cooling 25 to 0 C = 0.2*4200*25 = 21000 J. Ice melted = 21000/3.4e5 = 0.0618 kg ~ 61.7 g.

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