80 grams of water at 30° C are poured on a large block of ice at 0° C. The mass of ice melted is:(Latent Heat of fusion of ice =80 cal $g^{-1}$)

from principle of method of mixtures

$\begin{array}{rcl}\begin{array}{rcl}\mathrm{heat} \mathrm{lost} \mathrm{by} \mathrm{water} & =& \mathrm{heat} \mathrm{gained} \mathrm{by} \mathrm{ice}\end{array}\mathrm{ms} ∆\mathrm{T}& =& {\mathrm{m}}_{\mathrm{ice} \mathrm{melted}}\mathrm{L}---\left(1\right)\\ & & \\ \mathrm{here} \mathrm{s} \mathrm{is} \mathrm{specific} \mathrm{heat} \mathrm{of} \mathrm{water} & =& 1\\ \mathrm{m} \mathrm{is} \mathrm{mass} \mathrm{of} \mathrm{water}& =& 80\mathrm{g} \left(\mathrm{given}\right)\\ ∆\mathrm{T} \mathrm{is} \mathrm{change} \mathrm{in} \mathrm{temperature}& =& 30-0={30}^0\mathrm{C}\\ \mathrm{L} \mathrm{is} \mathrm{latent} \mathrm{heat} \mathrm{of} \mathrm{fusion} \mathrm{of} \mathrm{ice} & =& 80 \mathrm{calorie} {\mathrm{g}}^{-1}\\ \begin{array}{cl} & s\mathrm{ubstitute} \mathrm{above} \mathrm{values} \mathrm{in} \mathrm{equation} \left(1\right)\end{array}& & \\ & & \\ 80\left(1\right)\left(30\right)& =& m\left(80\right)\\ & & \begin{array}{rcl}\mathrm{mass} \mathrm{of} \mathrm{ice} \mathrm{melted} & =& 30 \mathrm{g}\end{array}\end{array}$