20 g of ice at 0° C absorbs 10,920 J of heat energy to melt and change to water at 50° C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg^-1 K^-1.

Given,

Mass (m) = 20 g

Heat energy absorbed (Q) = 10,920 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 (L) = ?

(i) Heat energy required to melt the ice at 0° C to water at 0° C (Q₁) = m x L

Substituting the values in the formula we get,

$Q_1 = 20 \times L$

(ii) Heat energy required to raise temperature from 0° C to 50° C = m x c x rise in temperature

Substituting the values in the formula we get,

$Q_2 = 20 \times 4.2 \times 50 \\[0.5em] = 4200 \text{ J}$

From relation,

$Q = Q$1 + Q2 \\[0.5em] 10,920 = (20 × \text{ L}) + 4200 \\[0.5em] 10,920 - 4200 = 20 × L\\[0.5em] 6720 = 20 × L\\[0.5em] \Rightarrow L = \dfrac{6720}{20} \\[0.5em] \Rightarrow L = 336 \text{ J g}^{-1} \\[0.5em]

Hence, Specific latent heat of fusion of ice = 336 J g^-1