A calorimeter of water equivalent 20 g contains 180 g of water at 25⁰C, ‘m’ kilograms of steam at 100^oC is mixed in it till the temperature of the mixture is 31^oC. The value of ‘m’ is close to (Latent heat of water = 540 cal g^-1, specific heat
of water = 1 cal g^-1oC^-1)

Water equivalent, w = 20g
Amount of water = 180g at 25^oC, m gram steam at 100^oC
Final temperature, T = 31^oC
L_w = 540 cal/g, s_w = 1cal/g^oC
Heat lost by steam = Heat gained by (water+calorimeter)
L_w = 540 cal/g
$\begin{array}{l}\mathrm{m}\times {\mathrm{L}}_{\mathrm{v}}+{\mathrm{ms}}_{\mathrm{w}}\times {\mathrm{\Delta T}}_1=\left({\mathrm{m}}_{\mathrm{w}}{\mathrm{s}}_{\mathrm{w}}+{\mathrm{m}}_{\mathrm{c}}{\mathrm{s}}_{\mathrm{c}}\right){\mathrm{\Delta T}}_2\\ \mathrm{m}\times 540+\mathrm{m}\times 1(100-31)=(180\times 1+20\times 1)(31-25)540\mathrm{m}+69\mathrm{m}=200\times 6\\ \Rightarrow 609\mathrm{m}=1200\Rightarrow \mathrm{m}=\frac{1200}{609}=1.97\simeq 2\mathrm{g}\end{array}$