The mass of steam at 100⁰C is to be passed into water of mass 100 g at 20⁰C to rise its temperature by 5⁰C is

$\begin{array}{l}{\mathrm{m}}_{\mathrm{s}}=?; {\mathrm{t}}_3={100}^{\circ }\mathrm{C}; {\mathrm{m}}_{\mathrm{w}}=100\mathrm{gm};\\ {\mathrm{t}}_1={20}^{\circ }\mathrm{C}; {\mathrm{t}}_2={25}^{\circ }\mathrm{C}; {\mathrm{L}}_{\mathrm{s}}=540\mathrm{cal}/\mathrm{gm}\\ {\mathrm{m}}_{\mathrm{s}}{\mathrm{L}}_{\mathrm{s}}+{\mathrm{m}}_{\mathrm{s}}{\mathrm{s}}_{\mathrm{w}}\left({\mathrm{t}}_3-{\mathrm{t}}_2\right)={\mathrm{m}}_{\mathrm{w}}{\mathrm{s}}_{\mathrm{w}}\left({\mathrm{t}}_2-{\mathrm{t}}_1\right)\\ {\mathrm{m}}_{\mathrm{s}}\left({\mathrm{L}}_{\mathrm{s}}+{\mathrm{S}}_{\mathrm{w}}\left({\mathrm{t}}_3-{\mathrm{t}}_2\right)\right)={\mathrm{m}}_{\mathrm{w}}{\mathrm{s}}_{\mathrm{w}}\left({\mathrm{t}}_2-{\mathrm{t}}_1\right)\\ {\mathrm{m}}_{\mathrm{s}}(540+1(100-25\left)\right)=100.1.(25-20)\\ {\mathrm{m}}_{\mathrm{s}}\cdot 615=500\Rightarrow {\mathrm{m}}_{\mathrm{s}}=\frac{500}{615}=0.8\mathrm{gm}\end{array}$