Q. Steam is passed into $22\ g$ of water at $20℃.$ The mass of water that will be present when the water acquires a temperature of $90℃$ (Latent heat of steam is $540\ cal/g$) is
A) $24.8\ g$
B) $24\ g$
C) $36.6\ g$
D) $30\ g$
Explanation
(a) Let $m\ gm$ of steam get condensed into water (By heat loss). This happens in following two steps.
Heat gained by water $(20℃)$ to raise it’s temperature upto $90^{o} = 22 \times 1 \times (90 - 20)$
Hence, in equilibrium heat lost = Heat gain
$\Rightarrow m \times 540 + m \times 1 \times (100 - 90) = 22 \times 1 \times (90 - 20)$
$\Rightarrow m = 2.8\ g$
The net mass of the water present in the mixture
$= 22 + 2.8 = 24.8\ g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.