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 (chages 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 of `100^@C`. Let m is the mass of steam condensed, then
`mxx540=180`
or `m=(80)/(540)=(1)/(3)g`
Final constent, `ice=0g`
water `=1+(1)/(3)=(4)/(3)g`
steam`=1-(1)/(3)=(2)/(3)g`