2kg of ice at $-{20}^{\circ }\mathrm{C}$ is mixed with 5kg of water at 20^o C . The water content of final mixture is (Latent heat of ice = 80 kcal/kg, specific heat of water $=1\mathrm{k} \mathrm{cal} {{\mathrm{kg}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$specific heat of ice $=0.5\mathrm{k} \mathrm{cal} {{\mathrm{kg}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$)

 $$\begin{gathered} \mathrm{Heat} \mathrm{lost} \mathrm{by} \mathrm{water}=\mathrm{ms} \mathrm{\theta },\text{ heat gained by ice }=\mathrm{MS} \mathrm{\theta }+\mathrm{M}\text{'}\mathrm{L} \\ \mathrm{maximum} \mathrm{heat} \mathrm{loss} \mathrm{by} \mathrm{water}=5\times 20\times 1\mathrm{kcal}=100\mathrm{kcal} \\ \mathrm{heat} \mathrm{gained} \mathrm{by} \mathrm{ice} \mathrm{to} \mathrm{reach} 0^0\mathrm{C}=2\times 20\times 0.5=20 \mathrm{kcal} \\ \mathrm{remaining} \mathrm{heat} 100-20=80\mathrm{Kcal} \\ \mathrm{it} \mathrm{is} \mathrm{just} \mathrm{enough} \mathrm{to} \mathrm{melt} 1 \mathrm{kg} \mathrm{of} \mathrm{ice}. \mathrm{so} \mathrm{total} \mathrm{water} \mathrm{in} \mathrm{the} \mathrm{mixture}=5+1=6\mathrm{kg} \end{gathered}$$