`1kg` ice at `0^(@)C` is mixed with `1kg` of steam at `100^(@)C`. What will be the composition of the system when thermal equilibrium is reached ? Latent heat of fusion of ice `= 3.36xx 10^(6)J kg^(-1)` and latent heat of vaporization of water `= 2.26 xx 10^(6)J kg^(-1)`

Explanation

Heat released when 1 kg of steam is condensed.
`Q_1=mL_("steam")=1000xx540=540000cal`
Heat required to melt ice and increasing its temperature to `100^@C`
`Q_2=1000xx80+1000xx1xx(100-0)=180000`
At `Q_2 lt Q_1`, all of steam is not condensed, hence, equilibrium temperature is `100^@C` Let `m_0` is mass of steam condensed
`m_0xx540=Q_2=180000`
`m_0=333.3g`
Contents of mixture:
steam `1000-333.3=666.7g`
Water: `1000+333.3=1333.3g`