A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at `40 .^(@)C`. When m gram of ice at `-10^(@)C` is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be `20^(@)C`. It is known that specific heat capacity of the liquid changes with temperature as `S = (1+(theta)/(500)) cal g^(-1) .^(@)C^(-1)` where `theta` is temperature in `.^(@)C`. The specific heat capacity of ice, water and the calorimeter remains constant and values are `S_("ice") = 0.5 cal g^(-1) .^(@)C^(-1), S_("water") = 1.0 cal g^(-1) .^(@)C^(-1)` and latent heat of fusion of ice is `L_(f) = 80 cal g^(-1)`. Assume no heat loss to the surrounding and calculate the value of m.
