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

Options

A. `80^(@)C`, and mixture content `2 kg`
B. `110^(@)C`, and mixture content `2 kg` steam
C. `100^(@)C`, and mixture content `(10)/(17) kg` steam and `(24)/(17) kg` water
D. `100^(@)C`, and mixture content `(20)/(27) kg` steam and `(34)/(27) kg`

Answer

D

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)=1xx1/2xx20+1xx80+1xx1xx100=190Kcal`
heat release by to convert `1kg` steam at `200^(@)C` to `1kg` water at `100^(@)C` is equal to
`H_(2)=1xx1/2xx100+1xx540=590Kcal`
`1 kg` ice at `-20^(@)C`
`=H_(1)+1kg` water at `100^(@)C` ...(1)
`1kg` steam at `200^(@)C`
`H_(2)+1kg` water at `100^(@)C` ...(2)
by adding equaltion (1) and (2)
`1 kg` ice at `-20^(@)C + 1 kg` 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 compeletely 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=(190-1xx1/2xx100)/(540) = 7/27 kg`
In mixture content
mass of steam = `1-7/27=20/27 kg`
mass of water `=1+7/27=34/27 kg`.