Given : mass of the water $=170\,\text{g}=\;\frac{170}{1000}\,\text{kg}$ ,Initial temperature $=50^{\circ}\,\text{C}$ .Let the mass of ice added be $=$ ' $x$ ' $\,\text{kg}$Given that the final temperature of mixture $=5^{\circ}\,\text{C}$Heat lost by water $=m \times c \times \Delta T$ $\;=\;\frac{170}{1000} \times 4200 \times (50-5)$ $\;=32130\,\text{J}$ Now the change in ice will be asIce at $0^{\circ}\,\text{C} \rightarrow$ Water at $0^{\circ}\,\text{C} \rightarrow$ Water at $5^{\circ}\,\text{C}$ $\therefore$ Heat gained by ice $\;=m' L+m' c \Delta T'$ $\;=x \times 336000+x \times 4200 \times (5-0)$ $\;=336000 x+21000 x$ $\;=357000 x \,\text{J}$By the principle of calorimetry,Heat lost by water $=$ Heat gained by ice $\therefore \;\; 32130=357000 x$or $x\;=\;\frac{32130}{357000}$ $=0.09\,\text{kg}\;\text{or}\;90\,\text{g}$