Steam at 100°C  is passed into 28 g of water at 20°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal $g^{-1o}C$ and latent heat of steam = 540 cal $g^{-1}$ ]

Here, $\text{ Specific heat of water, }{s}_w=1\text{ cal }{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$

$\text{ Latent heat of steam, }{L}_s=540\text{ cal }{\mathrm{g}}^{-1}$

$\begin{array}{r}\text{ Heat lost by }m \mathrm{gram}\text{ of steam at }{100}^{\circ }\mathrm{C}\text{ to change into water at }{80}^{\circ }\mathrm{C} \mathrm{is}\\ \end{array}$

$\begin{array}{r}{Q}_1=m{L}_s+m{s}_w\mathrm{\Delta }{T}_w\\ =m\times 540+m\times 1\times (100-80)=540m+20m=560m\end{array}$

$\text{ Heat gained by }20\mathrm{g}\text{ of water to change its temperature }$$\text{ from }{20}^{\circ }\mathrm{C}\text{ to }{80}^{\circ }\mathrm{C}\text{ is }$

$Q_2=m_w{s}_w\mathrm{\Delta }{T}_w=28\times 1\times (80-20)=28\times 60$

$\text{ According to principle of calorimetry }$

$\begin{array}{l}{Q}_1=Q_2\\ \therefore 560m=28\times 60\text{ or }m=3 \mathrm{gm}\end{array}$

Total mass of water present 

$=(28+m)g=(28+3)g=31 \mathrm{g}$