Properties Of Solids And Liquids Question 120
Question: Steam is passed into 22 gm of water at 20°C. The mass of water that will be present when the water acquires a temperature of 90°C (Latent heat of steam is 540 cal/gm) is [SCRA 1994]
Options:
A) 24.8 gm
B) 24 gm
C) 36.6 gm
D) 30 gm
Show Answer
Answer:
Correct Answer: A
Solution:
Let m gm of steam get condensed into water (By heat loss).
This happens in following two steps.
Heat gained by water (20°C) to raise it’s temperature upto 90° $ =22\times 1\times (90-20) $
Hence, in equilibrium heat lost = Heat gain
therefore $ m\times 540+m\times 1\times (100-90)=22\times 1\times (90-20) $
therefore $ m=2.8 $ gm
The net mass of the water present in the mixture $ =22+2.8=24.8gm. $
BETA
