What amount of heat was released during the condensation of 50 g of water vapor with a temperature of 100

What amount of heat was released during the condensation of 50 g of water vapor with a temperature of 100 degrees and cooling the resulting water to a temperature of 60 degrees?

Initial data: m (mass of water vapor) = 50 g = 0.05 kg; t0 (steam temperature) = 100 ºС; t (final temperature of the formed water) = 60 ºС.

Constants: L (specific heat of condensation of water vapor) = 2.3 * 10 ^ 6 J / kg; C (specific heat of water) = 4200 J / (kg * K).

Heat quantity: Q = Q1 + Q2 = L * m + C * m * (t0 – t) = m * (L + C * (t0 – t)).

Calculation: Q = 0.05 * (2.3 * 10 ^ 6 + 4200 * (100 – 60)) = 123,400 J = 123.4 kJ.

Answer: For the entire thermal process, 123.4 kJ of heat was released.