A mass m of steam at $100°\mathrm{C}$  is to be passed into a vessel containing 10g of ice and 100g of water at $0°\mathrm{C}$  so that all the ice is melted and the temperature is raised to $5°C$ . Neglecting heat absorbed by the vessel, we get,   $[L_f=80cal/g,L_V=540cal/g]$       

By laws of calorimetry , $\left(\text{Total heat lost by steam}\right)=\left(\text{Total heat gained by water + ice}\right)$ 
Let m gram of steam be required for the purpose. Then $540\mathrm{m}+\mathrm{m}\left(1\right)\left(100-5\right)$

$=\underset{\begin{array}{l}\text{heat required to raise }\\ \text{temoperature of 100 g }\\ \text{of water from 0}°\text{C to 5}°\text{C}\end{array}}{\underbrace{100\left(1\right)\left(5-0\right)}}+\underset{\begin{array}{l}\text{heat required to covert 10g}\\ \text{of ice to 10g of water at 0}°\text{C }\\ \text{and then raise temperature of }\\ \text{10g of water at 0}°\text{C to 5C}\end{array}}{\underbrace{10\left(80\right)+10\left(1\right)\left(5-0\right)}}$  

$540m+95m=500+850\Rightarrow m=2.1g$