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 (H_{f} = 334
J /g)

q=
m H_{f}

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 H_{f}

_{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 0^{o}C?

q= m H_{f}

_{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.