Question

In a container of negligible mass, $20$ grams of steam at ${100}^{\circ }C$ is added to$100$ gm of water that has temperature ${20}^{\circ }C$. If no heat is lost to the surroundings at equilibrium, match the following.

Column -I	Column-II
(a)	mass of steam in the mixture (in gm)	(p)	114.8
(b)	mass of water in the mixture (in gm)	(q)	76.4
(c)	final temperature of the mixture (in ${}^{\circ }C$)	(r)	5.2
		(s)	100

• A. a-r, b-p, c-q
• B. a-r, b-p, c-s
• C. a-r, b-p, c-p
• D. a-r, b-q, c-p

Answer 1

The correct option is B a-r, b-p, c-s

$\left(a\right)\to \left(r\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(s\right)$
$Q_1=ms\Delta T=100\times 1\times 80=8000cal$
$Q_2=m{L}_v=m\times 540cal$
If m = 20 gm and $Q_2$ = 10800 cal
$\therefore$ mass of steam in the mixture = $\frac{2800}{540}$
= 5.2 g
$\therefore$ mass of water = 120 - 5.2 = 114.8 g
Since steam and water coexist, final temperature must be ${100}^{\circ }C$