Question

If 1 gm of steam at 100⁰ C is sent into 1 gm of ice at 0⁰ C, final condition of the mixture is

• A. $\frac{4}{3}$ gm of Water and $\frac{1}{3}$ gm of Steam both at ${100}^{\circ }$ C
• B. $\frac{4}{3}$ gm of ice and $\frac{1}{3}$ gm of water both at $0°C$
• C. $\frac{4}{3}$ gm of water and $\frac{2}{3}$ gm of Steam both at ${100}^{\circ }$ C
• D. $2$ gm of ice at $0°C$

Answer 1

The correct option is C $\frac{4}{3}$ gm of water and $\frac{2}{3}$ gm of Steam both at ${100}^{\circ }$ C

Heat required to melt ice and to convert into water at ${100}^{\circ }C=1\times 80+1\times 1\times 100=180$ cal. This heat is provided by steam of mass $m$ by condensing then, 180 cal $=m\times 540$ cal $m=\frac{1}{3}gm$ i.e. $\frac{2}{3}gm$ of steam is uncondensed. So the final mixture is $\frac{4}{3}gm$ of water at ${100}^{\circ }C$.