Find the amount of heat that will release 14 kg of centigrade water vapor during

Find the amount of heat that will release 14 kg of centigrade water vapor during condensation and cooling to 30 degrees.

Data: m (mass of centigrade water vapor) = 14 kg; t1 (initial temperature of water vapor) = 100 ºС; t2 (temperature to which the steam will be cooled) = 30 ºС.

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

The amount of heat that water vapor will release: Q = Q1 + Q2 = L * m + C * m * (t1 – t2) = 2.3 * 10 ^ 6 * 14 + 4200 * 14 * (100 – 30) = 36316000 J ≈ 36.3 MJ.

Answer: During condensation and cooling of water vapor, 36.3 MJ will be released.