Question

$2\text{g}$ of steam at $100{}^{\circ }\text{C}$ is passed into $6\text{g}$ of ice at $0{}^{\circ }\text{C}$. If the latent heat of steam and ice are $540\text{cal/g}$ and $80\text{cal/g}$ respectively and specific heat capacity of water is $1\text{cal}/\text{g}{}^{\circ }\text{C}$, then the final temperature of the mixture is

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

Answer 1

The correct option is B ${100}^{\circ }C$

Let us assume here that resulting temperature
$={100}^{\circ }C$
$\therefore Q_1$ = heat supplied by $2\text{gm}$ of steam if it was to condense totally into water at ${100}^{\circ }C$
$=2\times 540=1080\text{cal}$
and $Q_2$ = Heat required to melt $6\text{gm}$ of ice at $0^{\circ }C$ into water at $0^{\circ }C$ + heat required to raise the temperature of $6\text{gm}$ of water from $0^{\circ }C$ to ${100}^{\circ }C$
$=6\times 80+6\times 1\times 100$
$=480+600=1080\text{cal}$
As $Q_1=Q_2$, hence whole of steam will condense.
Thus, the temperature of the mixture is ${100}^{\circ }C$
Hence, the correct answer is option (b).