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  4. 10 gm ice at 0° C is mixed with 2 gm steam at 100° C then final temperature of mixture will be:-

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Q. $10\, gm$ ice at $0^{\circ} C$ is mixed with $2\, gm$ steam at $100^{\circ} C$ then final temperature of mixture will be:-

Solution:

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$Q _{\text {given }}< Q _{\text {taken }} T <100^{\circ} C$
$Q _{\text {extra }}=1800-1080=720\, Cal .$
$720=12 \times 1 \times(100- T )$
$60=100- T$
$T =40^{\circ} C$