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

`1 kg` ice at `-20^(@)C` is mixed with `1 kg` steam at `200^(@)C`. The

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

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

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