Water weighing 5 kg is poured into the apparatus for obtaining distilled water at a temperature of 20

Water weighing 5 kg is poured into the apparatus for obtaining distilled water at a temperature of 20 degrees and it is heated to a boil, then 2 kg is evaporated. What energy is consumed in this case?

Initial data: m1 (mass of water) = 5 kg; m2 (mass of evaporated water) = 2 kg; t0 (initial water temperature) = 20 ºС.

Constants: t (temperature at which water evaporation starts) = 100 ºС; C (specific heat) = 4200 J / (kg * K), L (specific heat of vaporization) = 2.3 * 10 ^ 6 J / kg.

Energy to be consumed: Q = Q1 + Q2 = C * m1 * (t – t0) + L * m2.

Let’s perform the calculation: Q = 4200 * 5 * (100 – 20) + 2.3 * 10 ^ 6 * 2 = 6280000 J = 6.28 MJ.

Answer: 6.28 MJ of energy is consumed.