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Question

# Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg−1 K−1 , latent heat of fusion of ice = 3.36 × 105 J kg−1 , specific heat capacity of water = 4200 J kg −1 K−1 and latent heat of vaporization of water = 2.25 × 106 J kg−1

A
3.03×106J
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B
3.03×1012J
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C
3.303×106J
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D
3.303×105J
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Solution

## The correct option is A 3.03×106JHeat required to take the ice from -10°C to 0°C = (1 kg) (2100 J kg−1 K−1) (10K) = 21000 J. Heat required to melt the ice at 0°C to water = (1 kg) (3.36 × 105 J kg−1) = 336000 J. Heat required to take 1 kg of water from 0°C to 100°C = (1 kg) (4200 J kg−1 K−1) (100 K) = 420000 J. Heat required to convert 1 kg of water of 100°C into steam = (1 kg) (2.25 × 106 J kg−1) = 2.25 × 106J Total heat required = 3.03 × 106 J

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