Let the mass of ice be $= {}' m' g$ .As ice will gain heat from calorimeter and water, it will undergo conversion as : $\;\text{Ice}\;\rightarrow\;\text{Water}\;\rightarrow\;\text{Water}\;$ $\;\text{(at }0^{\circ}\text{C}\text{) (at }0^{\circ}\text{C}\text{) (at }5^{\circ}\text{C}\text{)}\;$So, heat gained $=m L + m c \Delta T$ $\;= m \times 330 + m \times 4.2 \times (5-0)$ $\;=330 m + 21 m$ $\;= (351 m) \,\text{J}.$ $\because$ Heat is lost by calorimeter and water both, $\therefore$ Heat lost by calorimeter $\;= m_1 c_1 \Delta T_1$ $\;=50 \times 0.4 \times (32-5)$ $\;=540 \,\text{J}$Heat lost by hot water $\;= m_2 C_W \Delta T_2$ $\;=150 \times 4.2 \times (32-5)$ $\;=150 \times 4.2 \times 27$ $\;=17010 \,\text{J}$$\therefore$ Total heat lost $=540+17010=17550 \,\text{J}$By the principle of calorimetry $\;\;\text{Heat gained = Heat lost}\;$$\;\therefore\;\; 351 m = 17550$$\;\;\text{or}\;\; m=\frac{17550}{351}=50\,\text{g}$