How much heat is needed to turn 5 kg of water into steam at a temperature of 0 degrees.

Given:
m = 5 kg is the mass of water;
t1 = 0 ° (Celsius scale) – initial water temperature;
t2 = 100 ° (Celsius scale) – boiling point of water;
c = 4200 J / (kg * C) – specific heat of water;
q = 2300000 J / kg is the specific heat of vaporization of water.
It is required to determine Q (Joule) – the amount of heat required to heat and turn into steam water of mass m.
To heat water to its melting point, energy is required equal to:
Q1 = c * m * (t2 – t1) = 4200 * 5 * (100 – 0) = 4200 * 5 * 100 = 4200 * 500 = 2100000 Joules.
To evaporate heated water requires energy:
Q2 = q * m = 2,300,000 * 5 = 11,500,000 Joules.
In total, energy is required, equal to:
Q = Q1 + Q2 = 2100000 + 11500000 = 13600000 Joules = 13.6 MJ.
Answer: you need a heat equal to 13.6 MJ.