How much energy is required to change a 20 g ice cube from ice at -40°C to steam at 100°C?

How much energy is required to change a 20 g ice cube from ice at -40°C to steam at 100°C?

How much energy is required to change a 20 g ice cube from ice at -40°C to steam at 100°C?
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 -40C 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 100C


4. The latent heat of vaporization to change the water from liquid form to gas form (steam) at 100 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 * 40C = 3.34 kJ


2) 334 kJ / kg * 0.02 kg = 6.68 kJ


3) 4.18 kJ / kgC * 0.02kg * 100C = 8.36 kJ


4) 2260 kJ / kg * 0.02 kg = 45.2 kJ





Add 'em all up





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