Question

28. The amount of heat required to convert
10gm of ice at -5°C to water at 75C
is [Latent heat of fusion of ice = 80 cal/gm
latent heat of vapourisation of water = 540
cal/gm, specific heat of water = 1 cal/gm/
°C and specific heat of ice = 0.5 cal/gm/
°C 1​

Answer 1

Answer:

Explanation:

Correct option is

Amount of heat required = Change the temp Of Ice from -10 C to 0 C + Heat required to melt the Ice + Heat required to increase the temperature of water from 0 to 100 C + Heat required to convert water at 100 C to vapor at 100 C = 1 × 0.5[0- (-10)] + 1 × 80 + 1x1x

100 + 1 x 540

= 5 + 80 + 100 + 540

= 725cal

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A 725 cal