Question

A mixture of $250 \, g$ of water and $200 \, g$ of ice at $0^\circ C$ is kept in a calorimeter of water equivalent $50g$ . If $200g$ of steam at $100^\circ C$ is passed through the mixture then the final amount of water in the mixture will be (Latent Heat of ice = $80 \, $ $cal \, g^{- 1}$ , Latent Heat of vaporisation of water = $540 \, $ $cal \, g^{- 1}$ and specific heat of water = $1 \, cal \, g^{- 1}/^\circ C$ )

• A. $450 \, g$
• B. $622 \, g$
• C. $572 \, g$
• D. $650 \, g$

Answer 1

Answer: (C)

Heat required by ice to melt $=\left[200 \times 80\right]=16000\text{ cal}$
Heat required by $250 \, g$ water
(ice $+$ initial water $+$ calorimeter) to raise up to $100^\circ C$
$=\left[500 \times 1 \times 100\right]=50000\text{ cal}$
$\Rightarrow $ Total heat required $=16000+50000=66000\text{ cal}$
Let $m\text{ g}$ of steam condenses, then
$66000=m\times 540$
$\Rightarrow \, \, m\approx 122g$
So final amount of water $=450+122=572 \, g$