Question

The heat of fusion for ice at 0°C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, and the heat of vaporization of water at 100°C is 2,256 kJ/kg. What is theenergy needed to vaporize 100 grams of ice at starting at 0°C?

Answer 1

Mass of ice/water = 100 g = 0.100 kg

Energy required to melt ice to water at 0^oC

= mass x heat of fusion

= 0.100 x 333.7 = 33.37 kJ

Energy required to heat water from 0 to 100^oC

= mass x specific heat of water x temperature change

= 0.100 x 4.186 x 100 = 41.86 kJ

Energy required to vaporize water to steam at 100^oC

= mass x heat of vaporization

= 0.100 x 2256 = 225.6 kJ

Total energy required = 33.37 + 41.86 + 225.6

= 300.83 kJ