How many kcal of heat are required to vaporize 25.0kg of water at 100°C?

Answer 1

Q=λ•m
1 J =0.2388 Cal
Heat of vaporization for water
λ = 2257•10^3•0.2388 =5.39•10^5 Cal/kg,
Q=λ•m =5.39•10^5•25 =1.35•10^7 Cal =
= 1.35•10^4 kCal