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Question
100 g of ice at -10°C is heated. It is converted into steam. Calculate the quantity of heat which it has consumed. (Sp. heat of ice = 2100J/kg°C, sp. heat of water= 4200 J/kgK, sp. heat of water = 42000 J/kgK, sp. latent heat of ice = 2260000 J/kg).
Solution
Amount of heat required to oonvert ice into steam is as given below:
(ice from -10°C to 0°C) = 0.1 x 2100 x 10 = 2100 J
(ice at 0°C to water at 0°C) = 0.1 x 336000 = 33600 J
(water from 0°C to 100°C) = 0.1 x 4200 x 100 = 42000 J
(water at 100°C to steam at 100°C) = 0.1 x 2260000 = 226000 J
Total amount of heat required = 2100 + 33600 + 42000 + 226000 = 30370 J
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