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 capacity 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.