Question

$20gm$ ice at $-10^{\circ }C$ is mixed with $mgm$ steam at $100^{\circ }C$. The minimum value of $m$ so that finally all ice and steam converts into water is,
(Use, $s_{\text{ice }}=0.5$ cal $g{m}^{-1}{}^{\circ }C,s_{\text{water }}=1$ cal $g{m}^{-1}{}^{\circ }C,L$ $($ melting $)=80$ cal $g{m}^{-1}$ and $L$ (vaporization) $=540$ cal $g{m}^{-1}.)$

• A. $\frac{185}{27}gm$
• B. $\frac{135}{17}gm$
• C. $\frac{85}{32}gm$
• D. $\frac{113}{17}gm$

Answer 1

Answer: (C)

For minimum value of $m$ , the final temperature of the mixture must be $0^{\circ }C$ .
$\therefore 20\times \frac{1}{2}\times 10+20\times 80=m\times 540+m\times 1\times 100$
$\therefore m=\frac{1700}{640}=\frac{85}{32}gm$ .