The specific heat of water $=4200Jk{g}^{-1}{K}^{-1}$  and the latent heat of ice  $=3.4\times {10}^{5}Jk{g}^{-1}$. 100 grams of ice at  $0°C$ is placed in 200 g of water at  $25°C$. The amount of ice that will melt as the temperature of water reaches  $0°C$ is close to (in grams) :

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40^oC. When m gram of ice at -10^oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^oC. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta }}{500}\right)\text{ cal }{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$where $\mathrm{\theta }$ is temperature in ^oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice }}=0.5\text{ cal }{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water }}=1.0\text{ cal }{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$ and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.