A block of ice of mass 120 g at temperature 0°C is put in 300 g water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is_______. [Use specific heat capacity of water = 4200 J kg^-1K^-1, Latent heat of ice = 3.5 X 10⁵ J kg^-1].

$\begin{array}{l}{\mathrm{m}}_{\text{ice }}=120\mathrm{g}=0.12\mathrm{kg},{\mathrm{T}}_{\text{ice }}=0^o\mathrm{C}\\ {\mathrm{m}}_{\text{water }}=300\mathrm{g}=0.3\mathrm{kg},{\mathrm{T}}_{\mathrm{w}}={25}^{\circ }\mathrm{C}\\ {\mathrm{c}}_{\mathrm{w}}=4200\mathrm{J}/\mathrm{kg}/\mathrm{K},{\mathrm{L}}_{\text{ice }}=3.5\times {10}^5\mathrm{J}/\mathrm{kg}\end{array}$
Heat lost by water ${\mathrm{H}}_{\mathrm{l}}={\mathrm{m}}_{\text{water }}\times {\mathrm{c}}_{\mathrm{w}}\times \mathrm{\Delta T}=0.3\times 4200\times (25-0);{\mathrm{H}}_{\mathrm{I}}=31500\mathrm{J}$
Heat gained by xg ice, ${\mathrm{H}}_2=\frac{\mathrm{x}\times \mathrm{L}}{1000}=\frac{\mathrm{x}}{1000}\times 3.5\times {10}^5=350\mathrm{x}$

By calorimetry law total heat lost by the body is equal to Total heat gained by body,

 31500 = 350x 

 x = 90