Q 11-10-101JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
$M$ grams of steam at $100\ ^\circ\text{C}$ is mixed with $200$ g of ice at its melting point in a thermally insulated container. If it produces liquid water at $40\ ^\circ\text{C}$ [heat of vaporization of water is $540$ cal/g and heat of fusion of ice is $80$ cal/g], the value of $M$ is ______.
Numerical value type. Enter your answer.
Answer: 40
Heat given by steam (condensing, then cooling from $100$ to $40\ ^\circ$C):
$$M(540 + 60) = 600M\ \text{cal}$$
Heat taken by ice (melting, then warming from $0$ to $40\ ^\circ$C):
$$200(80 + 40) = 24000\ \text{cal}$$
$600M = 24000 \Rightarrow M = 40$ g.
Solution by Sreeraj P, M.Sc Physics