During the condensation of steam taken at the condensation temperature and subsequent cooling

During the condensation of steam taken at the condensation temperature and subsequent cooling of the resulting water to 40 degrees, the amount of heat is released 5104 kJ. what is the mass of steam

Initial data: t (water temperature after cooling) = 40 ºС; Q (released amount of heat) = 5104 kJ = 5104 * 10 ^ 3 J.

Reference data: tк (condensation temperature) = 100 ºС; L (beats heat of vaporization) = 23 * 10 ^ 5 J / kg; C (specific heat capacity) = 4200 J / (kg * K).

We express the mass of water vapor from the formula: Q = L * m + C * m * (tc – t) = m * (L + C * (tc – t)), whence m = Q / (L + C * (tc – t)).

Calculation: m = Q / (L + C * (tk – t)) = 5104 * 10 ^ 3 / (23 * 10 ^ 5 + 4200 * (100 – 40)) = 2 kg.