What amount of heat will be released during the condensation of 200 g of water vapor

What amount of heat will be released during the condensation of 200 g of water vapor having a temperature of 100 ° C before water crystallization at 0 ° C?

Initial data: m (mass of water vapor (water)) = 200 g (0.2 kg); t (temperature that water vapor had) = 100 ºС.

Reference values: L (specific heat of condensation of water vapor) = 23 * 10 ^ 5 J / kg; C (specific heat of water) = 4200 J / (kg * K); λ (specific heat of crystallization of water) = 34 * 10 ^ 4 J / kg.

The amount of heat: Q = L * m + C * m * (t – 0) + λ * m = m * (L + C * (t – 0) + λ).

Calculation: Q = 0.2 * (23 * 10 ^ 5 + 4200 * (100 – 0) + 34 * 10 ^ 4) = 612,000 J (612 kJ).