Heat of Vaporization-the
amount of heat required to convert unit mass of a liquid into
the vapor without a change in temperature.
For
water at its normal boiling point of 100 ºC, the heat of
vaporization is 2260 J g-1. This means that to
convert 1 g of water at 100 ºC to 1 g of steam at 100 ºC,
2260 J of heat must be absorbed by the water. Conversely, when 1
g of steam at 100 ºC condenses to give 1 g of water at 100 ºC,
2260 J of heat will be released to the surroundings.
Heat
of Vaporization of Water Hv = 2260 J /g
q=
m Hv
Sample
Questions
Answers
1. How much energy
is required to vaporize 10.g of water at its boiling
point?
q= m Hv
q =
10.g x 2260 J/g = 22600j
or 22.6kJ
2. How much energy is released when 20. g
of steam is condensed at 100oC?
q= m Hv
q =
20.g x 2260 J/g = 45200j
or 45.2kJ
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
Note- Heat of Vaporization
questions occur in section D only and there is
no
temperature change. It stays at 100
Celsius.
