`1 kg` ice at `-20^(@)C` is mixed with `1 kg` steam at `200^(@)C`. The equilibrium temperature and mixture content is

Explanation

Let equilibrium temperature is `100^(@)C` heat required to convert `1kg` ice at `-20^(@)C` to `1kg` water at `100^(@)c` is equal to
`H_(1) = 1 xx (1)/(2) xx 20 +1 xx 80 +1 xx1xx 100 = 190 Kcal`
heat released by steam to convert `1kg` steam at `200^(@)C` to `1kg` water at `100^(@)C` is equal to
`H_(2) = 1 xx (1)/(2) xx 100 +1 xx 540 = 590 Kcal`
`1kg` ice at `-20^(@)C = H_(1) = 1kg` water at `100^(@)C ....(1)`
`1kg` steam at `200^(@)C = H_(@) +1kg` water at `100^(@)C ...(2)`
by adding equation (1) and (2)
`1kg` ice at `-20^(@)C+1kg` steam at `200^(@)C = H_(1) = H_(2) +2kg` water at `100^(@)C`.
Here heat required to ice is less than heat supplied by steam so mixture equilibrium temperature is `100^(@)C` then steam is not completely converted into water.
So mixture has water and steam which is possible only at `100^(@)C` mass of steam which converted into water is equal to
`m = (190-1xx(1)/(2)xx100)/(540) = (7)/(27) kg`
so mixture content
mass of steam `= 1-(7)/(27) = (20)/(27) kg`
mass of water `=1+(7)/(27) =(34)/(27) kg`