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The amount of heat required to convert 1 gm of ice at 0℃ into steam at 100℃ is

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10 months ago

```							m = 1 g = 0.001 kg  T1 = 0℃ and T2 = 100℃  T2 - T1 = 100 ℃  As The heat required to melt 1g of ice to water can be calculated as  q = Lm m  = 334000 (J/kg) 0.001 (kg)  = 334 (J)  = 334 (J)  The heat required to convert 1g of water at 0℃ to 1g of water at 100 ℃ can be calculated as  q = m c (T2 - T1)  = 0.001 kg * 4185.5 J/(kg⋅K) * 100 K  = 418.55 J  The heat required to convert 1g of water to steam can be calculated as  q = Lf m  = 2258000 (J/kg) 0.001 (kg)  =2258 (J)  = 2258 (J)  So, total amount of heat required  = 334 + 418.55 + 2258 J  = 3010.55 J  = 3.01055 kJ of heat
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10 months ago
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