Heat of Fusion of Water

Heat of Fusion-the amount of heat required to convert unit mass of a solid into the liquid without a change in temperature.

For water at its normal boiling point of 0 ºC, the specific heat of vaporization is 334 J g-1. This means that to convert 1 g of ice at 0 ºC to 1 g of  water at 0 ºC, 334 J of heat must be absorbed by the water. Conversely, when 1 g of water at 0 ºC freezes to give 1 g of ice  at 0 ºC, 334 J of heat will be released to the surroundings.

Heat of Fusion of Water Hf = 334 J /g

q= m Hf

The diagram on the left shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC to steam at temperaturesabove 100 ºC, affects the temperature of the sample.

A: Rise in temperature as ice absorbs heat.
B: Absorption of latent heat of fusion.
C: Rise in temperature as liquid water absorbs heat.
D: Water boils and absorbs latent heat of vaporization.
E: Steam absorbs heat and thus increases its temperature.

from-http://www.physchem.co.za/Heat/Latent.htm

Note- Heat of Fusion questions occur in section B only and there is no temperature change. It stays at 0 Celsius.

 

Sample Questions Answers
1. How much energy is required to melt 10.g of ice at its melting point? q= m Hf

q = 10.g x 334 J/g = 3340J or 3.34kJ

2. How much energy is released when 20. g of water is frozen at 0oC?  q= m Hf

q = 20.g x 334 J/g = 6680j or 6.68kJ