Question

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

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

Answer 1

The correct option is C $\frac{85}{32}gm$

Formula used: $Q=mL,Q=ms\Delta \theta$
GIven: $s_{ice}=0.5cal/g^{\circ }C,s_{water}=1cal/g^{\circ }C,L\left(\text{melting}\right)=80cal/g\text{and}{L}_v=540cal/g$

For minimum value of $m$, the final temperature of the mixture must be $0^{\circ }C$ as converting ice into water will be sufficient.
Heat gained by ice is equal to heat released by steam for reaching at $0^{\circ }C$ water,
$\therefore 20\times \frac{1}{2}\times 10+20\times 80=m\times 540+m\times 1\times 100$
$\therefore m=\frac{1700}{640}=\frac{85}{32}g$

FINAL ANSWER: (c).