Properties Of Solids And Liquids Question 548
Question: 4 gms of steam at $ 100{}^\circ C $ is added to 20 gms of water at $ 46{}^\circ C $ in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation $ =540cal/gm $ . Specific heat of water $ =1cal/gm-{}^\circ C $ .
Options:
A) 18gm
B) 20gm
C) 22gm
D) 24 g
Show Answer
Answer:
Correct Answer: C
Solution:
Heat released by steam in conversion to water at $ 100{}^\circ C $ is $ Q _{1}=mL=4x540=2160\text{ cal}.
$ Heat required to raise temperature of water from 0°C is
$ Q _{2}=\text{mS}\Delta \theta\text{ =20}\times \text{1}\times \text{54=1080J} $
$ Q _{1}>Q _{2}\text{and}\frac{Q _{1}}{Q _{2}}=2 $
Hence all steam is not converted to water; only half steam shall be converted to water$ \therefore $
Final mass of water $ =20+2=22\text{g}$
BETA
