Determine the final result when 200g of water and 20 g of ice at `0^@C` are in a calorimeter having a water equivalent of 30g and 50 g of steam is passed into it at `100^@C`

Explanation

When steam is passed, the final temperature can be `0^@C`, between `0^@C` and `100^@C`, or `100^@C`.
We will consider all three pssibilities.
CASE I: Final temperature`=0^@C`
In this case, all the steam condenses and then cools down to `0^@C`. Heat given out by steam
`=50xx540+50xx1xx(100-0)=32000cal`
Mass of ice which will melt by this heat`=(32000)/(80)=400gm`
But there is only 20 g of ice in the calorimeter. Hence final temperature cannot be `0^@C`.
CASE II: Final temperature`=theta` and `0ltthetalt100`
Heat lost by steam`=` heat gained by (ice + water + calorimeter)
`implies50xx540+50xx1xx(100-theta)=20xx80+(20+200+30)xx1xx(theta-0)impliestheta=101.3^@C`
The assumption `(0 lt theta lt 100)` is proved to be wrong. Hence, the final temperature cannot be between `0^@C` and `100^@C`
`implies` the final temperature will be `200^@C`
CASE III: Let m=mass of steam condensed
Heat lost by steam=heat gained by ice to melt + heat gained by (water+water+calorimeter) to reach `100^@C`
`impliesm(540)=20xx80+(20+200+30)xx(100-0)`
`implies26600//540 ~~ 49g`
`implies`49g` of steam gets condensed and the final temperature is `100^@C`