How much heat is needed to obtain 10 kg of 100-degree steam from water taken at 20 degrees?

How much heat is needed to obtain 10 kg of 100-degree steam from water taken at 20 degrees? The specific heat capacity of water is 4.2 J / kg * t, the specific heat of vaporization is 2300 kJ / kg.

Given: m (mass of water vapor) = 10 kg; t0 (initial temperature) = 20 ºС.

Constants: tpap (temperature of the beginning of vaporization) = 100 ºС; by condition C (specific heat capacity of water) = 4200 J / (kg * ºС); L (specific heat of vaporization) = 2300 kJ / kg = 2300 * 10 ^ 3 J / kg.

1) Water heating: Q1 = C * m * (tp – t0) = 4200 * 10 * (100 – 20) = 4200 * 10 * 80 = 3360000 J = 3.36 MJ.

2) The transition of water to steam: Q2 = L * m = 2300 * 10 ^ 3 * 10 = 2300 * 10 ^ 4 J = 23 MJ.

3) Heat consumption: Q = Q1 + Q2 = 3.36 + 23 = 26.36 MJ.