A mixture of 250 g of water and 200 g of ice at `0^@C` is kept in a calorimeter which has a water equivalent of 50 g. If 200 g of steam at `100^@C` is passed through this mixture, calculate the final temperature and the weight of the contents of the calorimeter.

Explanation

If 200 g of steam condensed, the heat released by it
`Q_1=mL_("steam")=200xx540=108000cal`
Heat required by ice and water to change into water at `100^@C`
`Q_2=200xx80+(200+250+50)xx1xx(100-0)`
`=16000+5000=66000cal`
`Q_2 lt Q_1`, whole of steam will not condense, hence equilibrium temperature is `100^@C`.
If `m_0` is the mass of steam condensed.
`m_0xx540=66000` i.e., `m_0=122.2g`
Contents of mixture:
Steam: `m-m_0=200-122.2=77.8g`
Water: `200+250+122.2=572.2g`