500 gm of water at 80°C is mixed with 100 gm steam at 120°C. Find out the final mixture.

Explanation

`120^@ C " Steam " to 100^@` Steam
Req. heat `=100xx(1)/(2)xx20=1` kcal
`80^@ C` water `to 100^@ C` water
Req. heat `=500xx1xx20=10` kcal
100 gm steam `to` 100 gm water at `100^@ C`
Req. heat `=100xx540-54` kcal
Total heat = 55 kcal.
Remaining heat = 55 - 10 = 45 kcal
Now we have 600 gm water at `100^@ C`
`rArr 4500 = m xx 540 rArr m=(250)/(3) gm`
So at last we have `(250)/(3)` gm steam and `(600-(250)/(3))` gm of water.