Let the mass of ice be $={}^{\prime }{\mathit{m}}^{\prime }\mathit{g}$ .
As ice will gain heat from calorimeter and water, it will undergo conversion as :


So, heat gained $=\mathit{m}\mathrm{L}+\mathit{m}\mathit{c}∆\mathrm{T}$
$=\mathit{m}\times 330+\mathit{m}\times 4.2\times \left(5-0\right)$
$=330\mathit{m}+21\mathit{m}$
$=\left(351\mathrm{m}\right)\mathrm{J}.$
$\because$ Heat is lost by calorimeter and water both,
$\therefore$ Heat lost by calorimeter
$={\mathit{m}}_1{\mathit{c}}_1∆{\mathrm{T}}_1$
$=50\times 0.4\times \left(32-5\right)$
$=540\mathrm{J}$
Heat lost by hot water
$={\mathit{m}}_2{\mathrm{C}}_{\mathit{W}}∆{\mathrm{T}}_2$
$=150\times 4.2\times \left(32-5\right)$
$=150\times 4.2\times 27$
$=17010\mathrm{J}$
$\therefore$ Total heat lost $=540+17010=17550\mathrm{J}$
By the principle of calorimetry

$\therefore 351\mathrm{m}=17550$