Physics
The amount of heat energy required to convert 1 kg of ice at -10° C completely to water at 100° C is 7,77,000 J. Calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg-1 K-1, Specific heat capacity of water is 4200 J kg-1 K-1.
Calorimetry
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Answer
Given,
m = 1 kg
heat energy required = 7,77,000 J
Specific heat capacity of ice = 2100 J kg-1 K-1
Specific heat capacity of water = 4200 J kg-1 K-1
specific latent heat of ice (L) = ?
Heat energy taken by ice to raise temperature from – 10° C to to 0° C
= m x c x change in temperature
= 1 × 2100 × [0 - (-10)]
= 21000 J
Heat energy gained by ice at 0° C to convert into water at 0° C
= m x L
= 1 x L
= L
Heat energy taken by water to raise it's temperature from 0° C to 100° C
= m x c x change in temperature
= 1 x 4200 x (100 - 0)
= 1 × 4200 × 100
= 4,20,000 J
Total heat energy gained
= 21,000 + L + 4,20,000 = 4,41,000 + L
As,
4,41,000 + L = 7,77,000
L = 7,77,000 – 4,41,000
L = 3,36,000 J kg-1
Hence, specific latent heat of ice = 3,36,000 J kg-1
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