91Ó°ÊÓ

The heat gain by gain of ice and water is equals to the heat lost by steam.

${\mathrm{Q}}_{\mathrm{ice}}+{\mathrm{Q}}_{\mathrm{water}}=-{\mathrm{Q}}_{\mathrm{steam}}$

Where, ${\mathrm{Q}}_{\mathrm{ice}}$and ${\mathrm{Q}}_{\mathrm{water}}$are heat gain by ice and water and ${\mathrm{Q}}_{\mathrm{steam}}$is heat lost by steam.

The heat gain by ice is

$225.6×{10}^3\mathrm{J}/\mathrm{kg}$

${\mathrm{Q}}_{\mathrm{ice}}={\mathrm{m}}_1{\mathrm{L}}_{\mathrm{f}}+{\mathrm{m}}_1{\mathrm{c}}_1∆{\mathrm{T}}_1$

Where, ${\mathrm{L}}_{\mathrm{f}}$latent heat of fusion of ice and ${\mathrm{c}}_1$is specific heat capacity of water having values $33.4×{10}^3\mathrm{J}/\mathrm{kg}$and $4190\mathrm{J}/\mathrm{kg}.\mathrm{K}$respectively.

The heat gain by water is

${\mathrm{Q}}_{\mathrm{water}}={\mathrm{m}}_2{\mathrm{c}}_1∆{\mathrm{T}}_2$

Where, ${\mathrm{c}}_1$is specific heat capacity of water which is equals to $4190\mathrm{J}/\mathrm{kg}.\mathrm{K}$.

The heat lost by steam is

${\mathrm{Q}}_{\mathrm{steam}}={\mathrm{m}}_3{\mathrm{L}}_{\mathrm{v}}+{\mathrm{m}}_3{\mathrm{c}}_1∆{\mathrm{T}}_3$

Where, ${\mathrm{m}}_3$is the mass of steam and ${\mathrm{L}}_{\mathrm{v}}$is the latent heat of vaporization water which is equals to .

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{ice}}+{\mathrm{Q}}_{\mathrm{water}}=-{\mathrm{Q}}_{\mathrm{steam}} \\ {\mathrm{m}}_1{\mathrm{L}}_1+{\mathrm{m}}_1{\mathrm{c}}_1∆\mathrm{T}+{\mathrm{m}}_2{\mathrm{c}}_1∆{\mathrm{T}}_2=-\left({\mathrm{m}}_3{\mathrm{L}}_{\mathrm{v}}+{\mathrm{m}}_3{\mathrm{c}}_1∆{\mathrm{T}}_3\right) \\ {\mathrm{m}}_3=\frac{-\left({\mathrm{m}}_1{\mathrm{L}}_{\mathrm{f}}+{\mathrm{m}}_1{\mathrm{c}}_1∆{\mathrm{T}}_1+{\mathrm{m}}_2{\mathrm{c}}_1∆{\mathrm{T}}_2\right)}{{\mathrm{L}}_{\mathrm{v}}+{\mathrm{c}}_1∆{\mathrm{T}}_3} \end{gathered}$$

Now, putting the values of constants in above equation

$$\begin{gathered} {\mathrm{m}}_3=\frac{-\left(0.45×33.4×{10}^3+0.45×4190×\left(28°-0°\right)+2.4×4190×\left(28°-0°\right)\right)}{225.6×{10}^3+4190×\left(28°-100°\right)} \\ {\mathrm{m}}_3=0.190\mathrm{kg} \\ {\mathrm{m}}_3=190\mathrm{g} \end{gathered}$$

Thus, the mass of steam is 190g