How much heat will be released during the condensation of 2 kg of water vapor at a temperature

How much heat will be released during the condensation of 2 kg of water vapor at a temperature of 100 degrees and cooling the resulting water to 20 degrees?

The released heat can be determined by the formula: Q = Q1 (water vapor condensation) + Q2 (water cooling) = L * m + C * m * (t1 – t2) = (L + C * (t1 – t2)) * m, where L (specific heat of condensation of water vapor) = 2.3 * 10 ^ 6 J / kg; C (specific heat of water) = 4200 J / (kg * K); t1 (initial temperature of water vapor) = 100 ºС; t2 (required water temperature) = 20 ºС; m (mass of water vapor) = 2 kg.

Let’s calculate: Q = (2.3 * 10 ^ 6 + 4200 * (100 – 20)) * 2 = 5,272,000 J (5.272 MJ).