In a thermally isolated system 1 gm of ice at 0⁰C,1 gm of water at 80⁰C and 1 gm of steam at 100⁰C are mixed.The resultant temperature of the mixture and amount of water existing at that temperature are

1 gm of ice requires 180 cal to melt and to raise temperature up to ${100}^{0}C$.

1 gm of water requires 20 cal to raise temperature up to ${100}^{0}C$

 $200cal=m\times L_V=m\times 540$

  $\Rightarrow m=\frac{20}{54}=0.37gm$

At equilibrium, 2.37 gm of water and 0.63 gm of steam at ${100}^{0}C$