Question

A thermally insulted vessel contains 145 g of water at 0degree celcius. Then the air from the vessel is pumped out adiabatically . A fraction of water turns into ice and the rest evaporates at 0degree celcius itself. The mass of evaporated water will be closed to,

Answer 1

Answer:

A thermally insulated vessel contains 150g of water at 0∘C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0∘C itself. The mass of evaporated water will be closest to:

(Latent heat of vaporization of water =2.10×106J kg−1 and Latent heat of Fusion of water =3.36×105J kg−1)

A .

130 g

B .

35 g

C .

20 g

D .

150 g

ANSWER

Suppose ′m′ gram of water evaporates then, heat required

△Qreq=mLv

Mass that converts into ice =(150−m)

So, heat released in this process

△Qrel=(150−m)Lf

Now,

△Qrel=△Qreq

(150−m)Lf=mLV

m(Lf+Lv)=150Lf

m=Lf+Lv150L

Answer 2

Answer: guys, how do you insult a vessel?

Explanation: