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प्रश्न
10g of ice at 0℃ absorbs 5460 J of heat energy to melt and change to water at 50℃. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1.
उत्तर
Given, Mass of ice = 10g
Heat energy absorbed (Q) = 5460 J
Specific heat capacity of water = 4200 J kg-1 K-1 = 4.2 J g-1 K-1
Specific latent heat of fusion of ice = ?
(i) Heat energy required to melt the ice at 0° C to water at 0° C (Q1) = m × L
Substituting the values in the formula we get,
Q1 = 10 × L
(ii) Heat energy required to raise temperature from 0° C to 50° C = m × c × rise in temperature
Substituting the values in the formula we get,
Q2 = 10 × 4.2 × 50
Q2 = 2100 J
Then,
Q = Q1 + Q2
∴ 5460 J = 10 × L + 2100 J
∴ 5460 J - 2100 J = 10 × L
∴ 3360 = 10 × L
∴ `3360/10` = L
∴ L = 336 Jg-1
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