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 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. (or released for freezing)

For water at its normal freezing point of 0 ºC, the specific heat of Fusion 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

Note- The Heat of Fusion equation is used only at the melting/freezing transition, where the temperature remains the same only and that is why there is no temperature change (DT) in this formula. It stays at 0 Celsius for water.

 Sample Questions Highlight to reveal 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

Note #2-Energy is required to melt and released when it freezes 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 temperatures above 100 ºC, affects the temperature of the sample. E: Steam absorbs heat and thus increases its temperature.D: Water boils and absorbs latent heat of vaporization.C: Rise in temperature as liquid water absorbs heat.B: Absorption of latent heat of fusion.A: Rise in temperature as ice absorbs heat.from-http://www.physchem.co.za/Heat/Latent.htm