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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. The specific heat capacity of ice=2100 Jkg−1K−1 Latent heat of fusion of ice=3.36×105 Jkg−1 The specific heat capacity of water=4200 Jkg−1K−1 Latent heat of vaporization of water=2.25×106 Jkg−1

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Solution

## The correct option is B Heat required to take the ice from -10°C to 0°C =(1 kg)(2100 Jkg−1K−1)(10K)=21000 J. The heat required to melt the ice at 0°C to water =(1 kg)(3.36×105 Jkg−1)=336000 J. Heat required to take 1 kg of water from 0°C to 100°C =(1 kg)(4200 Jkg−1K−1)(100K)=420000 J. Heat required to convert 1 kg of water of 100°C into steam =(1 kg)(2.25×106 Jkg−1)=2250000 J So, total heat required =3.03×106J

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