Question

20 gm ice at $-{10}^{o}C$ is mixed with m gm steam at ${100}^{o}C$. Minimum value of m so that finally all ice and steam converts into water. (Use $S_{ice}=0.5$ $cal/g{m}^{o}C$, $S_{water}=1$ $cal/g{m}^{o}C$, L(melting)=80 cal/gm and L (vapourization)=540 cal/gm)

• A. $\frac{85}{32}gm$
• B. $\frac{85}{64}gm$
• C. $\frac{32}{85}gm$
• D. $\frac{64}{85}gm$

Answer 1

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

$Q_{absordedbyice}=Q_{releasedbysteam}$

$\Rightarrow$ heat absorbed by ice to raise the temp upto 0 degree+Heat absorbed for melting+heat absorbed to raise the temp. upto T degree=Heat released by steam to condensate+ Heat released to drop temp upto T degree

$\Rightarrow 1700+20T=640m-mT$

$\Rightarrow m=\frac{1700+20T}{640-T}$

$\Rightarrow m=-20+\frac{14500}{640-T}$

from the above expression we can say that- to minimize the mass 'm', 'T' should be the minimum i.e T=0

So ,

$\Rightarrow m=-20+\frac{14500}{640-0}$

$\Rightarrow m=\frac{85}{32}$g