Steam at$1\text{00°C}$  is passed into 22 g of water at$2\text{0°C}$ . When resultant temperature is$\text{90°C}$ , then weight of the water present is

let the mass of steam condensed in to water = x

${\text{t}}_1=100°\text{C}$

${\text{m}}_{\text{w}}=\text{22 g}$ of water

${\text{t}}_2=\text{ 2}0°\text{C}$

${\text{t}}_{\text{R}}=90°C$

We know heat lost by steam = heat gained by water

${\text{m}}_{\text{s}}{\text{L}}_{\text{v}}+{\text{m}}_{\text{s}}\times {\text{s}}_{\text{w}}\left({\text{t}}_{\text{1}}-{\text{t}}_{\text{R}}\right)={\text{m}}_{\text{w}}\times {\text{s}}_{\text{w}}\left({\text{t}}_{\text{R}}-{\text{t}}_{\text{2}}\right)$

$\Rightarrow x\times 540+x\left(100-90\right)=22\times 1\times 70$

$\Rightarrow 550x=22\times 70$

$\Rightarrow x=\frac{22\times 7}{55}2.8\text{g}$

$\therefore$ water in beaker$=\text{22}+\text{2}\text{.8}=\text{24}\text{.8g}$