1g of ice at $0^{0}C$ is mixed 1g of steam at ${100}^{0}C$. The mass of water formed is

Heat required to covert 1 g of ice to 1 g of water is 80 Cal

Heat required to raise the temperature of 1g of water from 0⁰C to 100⁰C is 

$\mathrm{Q}=\mathrm{ms}∆\mathrm{T}=1\times 1\times 100=100\mathrm{Cal}$

Assuming x grams of steam is converted into water, 

$80+100=\mathrm{x}\times 540$

$\Rightarrow \mathrm{x}=\frac{540}{180}=0.33\mathrm{gms}$

Total mass of water at equilibrium is 

${\mathrm{m}}_{\mathrm{water}}=1+\mathrm{x}=1.33\mathrm{g}$