Given : mass of the water $=170\mathrm{g}=\frac{170}{1000}\mathrm{kg}$ ,
Initial temperature $={50}^{\circ }\mathrm{C}$ .
Let the mass of ice added be $=$ ' $\mathit{x}$ ' $\mathrm{kg}$
Given that the final temperature of mixture $=5^{\circ }\mathrm{C}$
Heat lost by water $=\mathit{m}\times \mathit{c}\times ∆\mathrm{T}$
$=\frac{170}{1000}\times 4200\times \left(50-5\right)$
$=32130\mathrm{J}$
Now the change in ice will be as
Ice at $0^{\circ }\mathrm{C}\to$ Water at $0^{\circ }\mathrm{C}\to$ Water at $5^{\circ }\mathrm{C}$
$\therefore$ Heat gained by ice
$={\mathit{m}}^{\prime }\mathit{L}+{\mathit{m}}^{\prime }\mathit{c}∆{\mathrm{T}}^{\prime }$
$=\mathit{x}\times 336000+\mathit{x}\times 4200\times \left(5-0\right)$
$=336000\mathit{x}+21000\mathit{x}$
$=357000\mathit{x}\mathrm{J}$
By the principle of calorimetry,
Heat lost by water $=$ Heat gained by ice
$\therefore 32130=357000\mathit{x}$
or $\mathit{x}=\frac{32130}{357000}$