Question

The amount of heat required to convert 10gm
of ice at -5°C to water at 75°C is ſlatent heat
of fusion of ice = 80 cal/gm latent heat of
vapourisation of water = 540 cal/gm, specific
heat of water = lcal/gm/°C and specific heat
of ice = 0.5 cal/gm/°C]
1. 1575 cal
2. 1557 cal
3. 1755 cal
4. 5175 cal​

Answer 1

Answer:

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Explanation:

Heat required to convert 100 g ice at 0°C to water at 0°C = Mass x Specific latent heat of fusion of ice 100 x 333 = 33300 J Heat required to raise the temperature of 100 g water from 0°C to 100°C Mass x Specific heat capacity of water x Rise in temperature 100 x 4.2 x (100 - 0) = 42000 J Therefore, total heat required 33300 + 42000 = 75300 J %3D %3D %3D

So option D is correct.

Thanks