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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 Jkg1K1

Latent heat of fusion of ice=3.36×105 Jkg1

The specific heat capacity of water=4200 Jkg1K1

Latent heat of vaporization of water=2.25×106 Jkg1


A
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B
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C
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D
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Solution

The correct option is A

Heat required to take the ice from -10°C to 0°C
=(1 kg)(2100 Jkg1K1)(10K)=21000 J.
The heat required to melt the ice at 0°C to water
=(1 kg)(3.36×105 Jkg1)=336000 J.
Heat required to take 1 kg of water from 0°C to 100°C
=(1 kg)(4200 Jkg1K1)(100K)=420000 J.
Heat required to convert 1 kg of water of 100°C into steam
=(1 kg)(2.25×106 Jkg1)=2250000 J

So, total heat required =3.03×106J


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