60 g of ice at 0^oC is mixed with 60 g of steam at 100^oC. At thermal equilibrium, the mixture contains

if we take resultant temperature as ${100}^{0}C$ Then steam can liberate the heat of
60 x 540  = 32400 =$Q_1$

this steam will melt the ice at $0^{0}C$ and will be heating melted wataer to ${100}^{0}C$

mLi + ms$∆$t = Q₂ 
60 x 80 + 60 x 1 x 100 = 10800 $\to$ water (ice) 
hence steam is dominating, only part of stem is melted .Let x gm of steam is melted

if 10,800 calories heat  is given by steam then mass of the stem condensed will be x

540x  = 10800
x = 20 gm
$\therefore$ 40 gm of steam is remained and 20 gm is condensed and total 80gm of water is finally present