How much energy needs to be spent to make water weighing 5 kg. Bring it to a boil at a temperature of 0 degrees and evaporate it?

Quantity of heat:
Q = Q1 + Q2.
Q = C1 * m * (tc – tn), where C1 is the specific heat capacity of water (C1 = 4200 J / (kg * K)), m is the mass of water (m = 5 kg), tc is the final temperature (tc = 100 ºС), tн – initial temperature (tн = 0 ºС).
Q2 = L * m, where L is the specific heat of vaporization of water (L = 2.3 * 10 ^ 6 J / kg).
Q = C1 * m * (tc – tn) + L * m = 4200 * 5 * (100 – 0) + (2.3 * 10 ^ 6) * 5 = 2100000 + 13000000 = 13600000 J = 13.6 MJ.
Answer: It is necessary to spend 13.6 MJ of thermal energy.