A calorimeter of water equivalent 20 g contains 180 g of water at  ${25}^{0}C$. ‘m’ grams of steam at ${100}^{0}C$  is mixed in it till the temperature of the mixture is  ${31}^{0}C$. The value of m is close to (gm)  (Take, latent heat of steam  $=540cal{g}^{-1},$ specific heat of water  $=1cal{g}^{-1}{}^0{C}^{-1}$)

Heat lost by m grams of steam is gained by calorimeter and water in it. (Steam at ${100}^{0}C$ ) $\stackrel{Condensation}{\to }$ ((Water at  ${100}^{0}C \stackrel{Cooling}{\to }$  (Water at ${31}^{0}C$ ) 
Heat lost by steam ( Colorimeter and water at  ${25}^{0}C$) $\stackrel{Heating}{\to }$  calorimeter and water at  ${31}^{0}C$
 $$\begin{gathered} =m{L}_1+m\times S_w\times (100-31) \\ =m[L_1+S_w\times (100-31)] \\ =m[540+1(100-31)]=m\times 609 \end{gathered}$$
Heat gained by calorimeter and water in calorimeter
$$\begin{gathered} =(M+W)\times S\times (31-25) \\ =(180+20)\times 1\times 6=200\times 6 \end{gathered}$$
As, heat lost = heat gained 
 $$\begin{gathered} \Rightarrow m\times 609=200\times 6 \\ \Rightarrow m=\frac{1200}{609}=1.97=2g \end{gathered}$$