Question

5 Gm of steam at ${100}^{o}C$ is passed into 6 gm of ice at $0^{o}C$. If the latent heats of steam and ice n cal per gm are 540 and 80 respectively, them the final temperature is:-

• A. $0^{o}C$
• B. ${100}^{o}C$
• C. ${50}^{o}C$
• D. ${30}^{o}C$

Answer 1

The correct option is B ${100}^{o}C$

Heat taken by ice to melt $=6\times 80=480cal$

Heat taken to raise the temperature of water by ${100}^{0}C=6\times 1\times 100=600cal$

Therefore, the total heat required by ice to raise up to ${100}^{0}C=600+480=1080cal$

Heat supplied by steam to condense $=mL=5\times 540=2700cal$

It means all the steam has not condensed

Let mass of the steam condensed $=m$

Then,

$m\times 540=1080$

$m=\frac{1080}{540}$

$m=2g$

Now, the temperature of the mixture $={100}^{0}C$

Hence, this is the required solution