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) 24gm

Show Answer

Answer:

Correct Answer: C

Solution:

Heat released by steam inconversion to water at $ 100{}^\circ C $ is $ Q _{1}=mL=4x540=2160cal.

$ Heat required to raise temperature of water from 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=22gm $



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