Question

Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ }C$ are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at ${100}^{\circ }C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

• A. 270 gm ice at $0^{\circ }C$
• B. 250 gm ice, 20 gm water at $0^{\circ }C$
• C. 270 gm water at ${75}^{\circ }C$
• D. 1 gm steam, 269 gm water at ${100}^{\circ }C$

Answer 1

The correct option is D

1 gm steam, 269 gm water at ${100}^{\circ }C$

When steam is passed, the final temperature can be $0^{\circ }C$ , between $0^{\circ }C$ and ${100}^{\circ }C$ , ${100}^{\circ }C$ . We will consider all three possibilities

Case I:

Final temperature = $0^{\circ }C$

In this case, all the steam condenses and then cools down to $0^{\circ }C$.

Heat given out by steam

= $50\times 540+50\times 1\times (100-0)$ = 32000 cal.

Mass of ice which will melt by this heat = $\frac{32000}{80}$ = 40 gm

But there are only 20 gm of ice in the calorimeter.

Hence final temperature cannot be $0^{\circ }C$.

Case II : Final temperature = $\theta$ and 0 < $\theta$ < 100

Heat lost by steam = heat gained by (ice + water + calorimeter)

$\Rightarrow$ $50\times 540+50\times 1\times (100-\theta )$ = $20\times 80+(20+200+30)\times 1\times (\theta -0)$

$\Rightarrow$ $\theta$ = ${101.3}^{\circ }C$

the assumption (0 <$\theta$< 100) is proved to be wrong. Hence final temperature cannot be between $0^{\circ }$C and ${100}^{\circ }$C

$\Rightarrow$ The final temperature will be ${100}^{\circ }C$

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 ${100}^{\circ }C$

$\Rightarrow$ $m$(540) = $20\times 80+(20+200+30)\times (100-0)$

$\Rightarrow$ $m$ = $\frac{26600}{540}$ ≈ 49 gm

$\Rightarrow$ 49 gm of steam condense and the final temperature is ${100}^{\circ }C$