Arun
Last Activity: 5 Years 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