Question

5. The amount of heat required to convert
1 gm of ice at 0°C into steam at 100°C is
(Lice = 80 cal/gm, Sw = 1 cal/gm k, L steam
= 540 cal/gm)​

Answer 1

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Answer 2

Answer:

725cal

Explanation:

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 0to 100 C + Heat required to convert water at 100 C to vapor at 100 C

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 0to 100 C + Heat required to convert water at 100 C to vapor at 100 C=1×0.5[0−(−10)]+1×80+1×1×100+1×540

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 0to 100 C + Heat required to convert water at 100 C to vapor at 100 C=1×0.5[0−(−10)]+1×80+1×1×100+1×540=5+80+100+540

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 0to 100 C + Heat required to convert water at 100 C to vapor at 100 C=1×0.5[0−(−10)]+1×80+1×1×100+1×540=5+80+100+540=725cal