Question

How much energy is required to change a 20-g ice cube from ice at

−15°C

to steam at 135°C?
MJ

Answer 1

There are 4 separate energies that must be added together to get the total energy.
1. The heat necessary to get the water (ice) from -15C to 0C
2. The latent heat of fusion to change the water from a solid to a liquid at 0C (no temperature change just a phase change only)
3. The heat necessary to get the water from 0 C to 135C
4. The latent heat of vaporization to change the water from liquid form to gas form (steam) at 135 C (no temperature change, phase change only)

All these 4 numbers should be calculated based on a mass of 20 grams.

For water the latent heat of fusion is: 334 kJ/kg
and the latent heat of evaporation is: 2260 kJ/kg
Simply to heat the water in its solid or liquid form takes 4.18 kJ / kg C

The mass is 0.02 kg

1) 4.18 kJ / kgC * 0.02kg * 15C = 1.25 kJ
2) 334 kJ / kg * 0.02 kg = 6.68 kJ
3) 4.18 kJ / kgC * 0.02kg * 135C = 10.86 kJ
4) 2260 kJ / kg * 0.02 kg = 45.2 kJ