60g of ice at $0^{0}C$ is mixed with 60g of steam at ${100}^{0}C$ At thermal equilibrium, the mixture contains

$m{L}_S=Q_1$

$60\times 540=32400\to steam$

$mLi+ms\Delta t=Q_2$

$60\times 80+60\times 1\times 100$

$=10800\to \mathrm{water}\left(\mathrm{ice} \mathrm{from} 0^{\mathrm{o} }\mathrm{C}\right)$

$Q_1=Q_2;32400x=10800$

$x=\frac{1}{3}$for 120gm $\frac{120}{3}=40gm$

40gm of steam and 80gm of water