$10\mathrm{gm}$ of ice in a closed vessel are melted by sending in steam slowly at ${100}^{\circ }\mathrm{C}$ .The quantity of water present when the ice has just melted is      
(Latent heat of water $=3.36\times {10}^5{\mathrm{Jkg}}^{-1}$
Latent heat of steam $=2.25\times {10}^6{\mathrm{Jkg}}^{-1}$
Specific heat of water $\left.=4.2\times {10}^3{\mathrm{Jkg}}^{-1}{ }^{\mathrm{o}}{\mathrm{C}}^{-1}.\right)$ 
 

Heat gained by 10g  or 0.01kg  of ice to be completely melted 
$=0.01\times 3.36\times {10}^5$
=3360 J
Heat lost by m kg  of steam $=2.25\times {10}^6\mathrm{mJ}$
Heat lost by m kg  of water in cooling from 100°C to 0°C 
$\begin{array}{c}=4.2\times {10}^3\times 100\mathrm{m}\\ =4.2\times {10}^5\mathrm{mJ}\end{array}$
Heat lost = Heat gained
$=2.25\times {10}^6\times \mathrm{m}+4.2\times {10}^5\mathrm{m}=3,360$
Whence $\mathrm{m}=1.26\times {10}^{-3}\mathrm{kg}$
$\mathrm{Or} \mathrm{m}=1.26 \mathrm{g}$
Quantity of water present when the ice has just melted
$=10+1.26=11.26 \mathrm{g}$