The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 7,77,000 J. Calculate the specific latent heat of ice.

Specific heat capacity of ice = 2100 J $k g^{- 1} K^{- 1}$ , specific heat capacity of water=4200 J k $g^{- 1} K^{- 1}$ .

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Solution

$H e a t g a i n e d = m S_{i} Δ T_{i} + m L + m S_{w} Δ T_{w}$

$\Rightarrow 777000 = 1 \times 2100 \times 10 + 1 \times L + 1 \times 4200 \times 100$

$\Rightarrow 777000 = 21000 + L + 420000$

$∴ L = 336000 J / k g$

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