1. Tardigrade
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  4. 20 gm ice at -10° C is mixed with m gm steam at 100° C. The minimum value of m so that finally all ice and steam converts into water is, (Use, s text ice =0.5 cal gm -1 ° C , s text water =1 cal gm -1 ° C , L ( melting )=80 cal gm -1 and L (vaporization) =540 cal .gm -1 .)

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Q. $20 \,gm$ ice at $-10^{\circ} C$ is mixed with $m gm$ 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 $gm ^{-1}{ }^{\circ} C , s_{\text {water }}=1 $ cal $gm ^{-1}{ }^{\circ} C , L$ $($ melting $)=80$ cal $gm ^{-1}$ and $L$ (vaporization) $=540$ cal $\left.gm ^{-1} .\right)$

NTA AbhyasNTA Abhyas 2022

Solution:

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