Question

$10\text{kg}$ of ice completely melts when water at $100{}^{\circ }\text{C}$ is poured into a hole drilled in the ice. If the specific heat capacity of water is $4200{\text{J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}$ and the specific latent heat of melting of ice is $336\times {10}^3{\text{J kg}}^{-1}$, the amount of the water needed is

• A. 10 litres
• B. 12 litres
• C. 8 litres
• D. 4 litres

Answer 1

The correct option is C 8 litres

Given:
Mass of the ice, ${\text{m}}_{\text{i}}=10\text{kg}$
Specific heat capacity of water, ${\text{c}}_{\text{w}}=4200{\text{J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}$
Specific latent heat of melting of ice, ${\text{L}}_{\text{i}}=336\times {10}^3{\text{J kg}}^{-1}$
Temperature of water, ${\text{t}}_{\text{w}}={100}^{\circ }\text{C}$
Amount of heat energy required to melt the ice, ${\text{Q}}_{\text{i}}={\text{m}}_{\text{i}}\times {\text{L}}_{\text{i}}=10\times 336\times {10}^3\text{J}$$=336\times {10}^4\text{J}$ ---- (1)
Since the ice completely melts into water, the final temperature would be $0^{\circ }\text{C}$.
Change in the temperature, $\Delta \text{t}=100-0={100}^{\circ }\text{C}$
Let the mass of the water required be ${\text{m}}_{\text{w}}$.
Heat energy supplied by the boiling water, ${\text{Q}}_{\text{w}}={\text{m}}_{\text{w}}\times {\text{c}}_{\text{w}}\times \Delta \text{t}$
${\text{Q}}_{\text{w}}={\text{m}}_{\text{w}}\times 4200\times 100={\text{m}}_{\text{w}}\times {10}^4\text{J}$ ---- (2)
Assuming there is no heat loss in any other form, according to the principle of method of mixtures, ${\text{Q}}_{\text{i}}={\text{Q}}_{\text{w}}$
$⟹336\times {10}^4=42\times {10}^4\times {\text{m}}_{\text{w}}$
$⟹336=42\times {\text{m}}_{\text{w}}$
$⟹{\text{m}}_{\text{w}}=\frac{336}{42}=8\text{kg}$
Since, for water, $1\text{kg}=1$ litre, the amount of water needed would be $8$ litres.