`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 is 2 kg
B. `110^@ C` and mixture content is 2 kg steam
C. `100^@ C` and mixture content is `10/17 kg`steam and `24/17 kg` water
D. `100^@ C` and mixture content is `20/27 kg`steam and `34/27 kg` water

Answer

D

Explanation

Let equilibrium temperature be `100^@`C, heat required to canvert 1 kg ice at `-20^@`C to 1 kg water at `100^@`C is equal to
`H_1=1xx(1)/(2)xx20+1xx80+1xx1xx100=190kcal`
Heat released by steam to convert 1 kg steam at `200^@` C to 1 kg water at `100^@` C is equal to
`H_2=1xx(1)/(2)xx100+1xx540=590kcal`
1 kg ice at `-20^@C=H_1+1kg` water at `100^@`C ..(i)
1 kg steam at `200^@`C`=H_2+1kg` water at `100^@`C ..(ii)
By adding Eqs. (i) and (ii)
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`