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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.
A: Rise in
temperature as ice absorbs heat.
B:
Absorption of heat of fusion.
C:
Rise in temperature as liquid water absorbs heat.
D:
Water boils and absorbs heat of vaporization.
E:
Steam absorbs heat and thus increases its temperature.
from-http://www.physchem.co.za/Heat/Latent.htm
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***UNITS NOTE
J
g-1 ºC-1 = J / g ºC; any
exponent to the -1 is placed in the denominator. The specific heat capacity of water is 4.18 J g-1 ºC-1.
The specific heat capacity for Ice and Steam are 2.09 J g-1 ºC-1.
The heat of fusion of ice is 334 J g-1,
and the heat of vaporization of water is 2260 J g-1.
EXAMPLE
Calculate the amount of heat required
to completely convert 50 g of ice at -10 ºC to steam at
120 ºC.
Heat is taken
up in five stages:
1. The heating of the
ice
2. The melting of the ice,
3. The heating of
the water,
4. The vaporization of the
water and
5. The heating of the
steam.
The heat taken
up in the complete process is the sum of the heat taken up in
each stage.
1.Heat taken up
heating the ice
from -10 ºC to the melting point, 0 ºC. |
mass of water x specific heat x temperature change
= 50g x 2.09 (J g-1 ºC-1)x
10 ºC
= 1,045 J |
2. Heat taken up
for converting ice
at ºC to water at ºC. |
mass of water x latent heat of
fusion
= 50g x 334 (J g-1)
= 16,700J |
3. Heat taken up
heating the water
from 0 ºC to the boiling point, 100 ºC. |
mass of water x specific heat x temperature change
= 50g x 4.18 (J g-1 ºC-1)x 100
ºC
= 20,900 J |
4.
Heat taken up
vapourizing the
water. |
mass of water x heat of
vaporization
50g x 2260 J g-1
= 113,000 J |
5.Heat taken up
heating the steam
from 100 ºC to 120 ºC. |
mass of water x specific heat x temperature change
= 50g x 2.09 (J g-1 ºC-1)x
20 ºC
= 2,090 J |
| The sum of these is: |
1,045 J +16,700J
+ 20,900 J+ 113,000 J +2,090 J
= 153,735 roughly 154 kJ |
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