Question

$5gm$ of steam at ${100}^{o}C$ is passed into $6gm$ of ice at $0^{o}C$. If the latent heats of steam and ice in cal per gm are $540$ and $80$ respectively, 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 absorbed by ice to melt $=mL=6\times 80=480$ $cal$

Heat absorbed by water at $0^{\circ }$ to reach ${100}^{\circ }=mc△T=6\times 1\times ({100}^{\circ }-0^{\circ })=600$ $cal$

Total heat absorbed by ice to reach ${100}^{\circ }=480+600=1480$ $cal$

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

Since heat released by steam to condense is greater than the heat absorbed by ice to reach ${100}^{\circ }$, the final temperature of mixture will be ${100}^{\circ }$