1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture is

Explanation

Heat available on steam (changes into steam to water)
`=mL=1xx540=540cal`
Heat gained by ice to change into water and then rise its temperature to `100^@C`
`=m_("ice")L+m_("wat")cDeltaT`
`=1xx80+1xx1xx(100-0)`
`=180cal`
The above calculation show that some part of steam will condense to change the ice into water at `100^@C`. Let m is the mass of steam condensed, then
`mxx540=180`
or `m=(80)/(540)=(1)/(3)g`
Final contents: ice`=0g`
water`=(1+1)/(3)=(4)/(3g)`
steam`=(1-1)/(3)=(2)/(3g)`