2 kg of ice at $-{20}^0\mathrm{C}$ is mixed with 5 kg of water at ${20}^0\mathrm{C}$ in an insulating vessel having a negligilble heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg per ${ }^0\mathrm{C}$ and 0.5 kcal/kg/${ }^0\mathrm{C}$ while the latent heat of fusion of ice is 80 kcal/kg

Initially ice will absorb heat to raise its temperature to $0^0\mathrm{C}$ then its melting takes place If ${\mathrm{m}}_{\mathrm{i}}= \mathrm{Initial} \mathrm{mass} \mathrm{of} \mathrm{ice}, {{\mathrm{m}}_{\mathrm{i}}}^{\text{'}} = \mathrm{Massof} \mathrm{ice} \mathrm{that} \mathrm{melts}$ and ${\text{m}}_{\mathrm{w}} = \mathrm{initial} \mathrm{mass} \mathrm{of} \mathrm{water}$

By Law of mixture Heat gained by ice = Heat lost by water $\Rightarrow {\mathrm{m}}_{\mathrm{i}}\times \mathrm{c}\times \left(20\right)+{{\mathrm{m}}_{\mathrm{i}}}^{\text{'}}\times 80 = 5\times 1\times 20 \Rightarrow {{\mathrm{m}}_{\mathrm{i}}}^{\text{'}} = 1 \mathrm{kg}$

${{\text{⇒ 2×0.5(20) + m}}_1}^{\text{'}}\times 80 = 5 \times 1\times 20 \Rightarrow {m_1}^{\text{'}} = 1 kg$

So final mass of water = Initial mass of water + Mass of ice that melts = 5+1 = 6 kg.