M grams of steam at $100°C$ is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at  $40°C$, the value of M is 
[ Latent heat of vaporisation of water is $540cal/g$ and Latent heat of fusion of ice is $80cal/g$] 

M grams of steam at$100°C$  is mixed with 200g of ice at $0°C$ and mixture obtained is at $40°C$ . So, heat lost by steam in condensation at  $100°C$+ Heat lost by water formed to reach at $40°C$ = Heat required by ice for melting +Heat gained by water formed at $0°C$ to reach up to a temperature of $40°C$.

$\Rightarrow {\mathrm{M}}_{\mathrm{steam}}\times {\mathrm{L}}_{\mathrm{V}}+{\mathrm{M}}_{\mathrm{steam}}\times {\mathrm{S}}_{\mathrm{W}}\times \mathrm{\Delta T}={\mathrm{M}}_{\mathrm{ice}}\times {\mathrm{L}}_{\mathrm{f}}+{\mathrm{M}}_{\mathrm{ice}}\times {\mathrm{S}}_{\mathrm{w}}\times \mathrm{\Delta T}$

$\Rightarrow \mathrm{M}\times 540+\mathrm{M}\times 1\times \left(100-40\right)=200\times 80+200\times 1\times \left(40-0\right)$

$\Rightarrow \mathrm{M}\times 600=24000\Rightarrow \mathrm{M}=40\mathrm{g}$