In the previous question, if the specific latent heat of vaporization of water at `0^@C` is `eta` times the specific latent heat of freezing of water at `0^@C`, the fraction of water that will ultimately freeze is

Options

A. `(1)/(eta)`
B. `(eta)/(eta+1)`
C. `(eta-1)/(eta)`
D. `(eta-1)/(eta+1)`

Answer

B

Explanation

`DeltaQ_("top") = DeltaQ_("freezing")`
`m.(etaL) = M(L) implies M = etam`
L = latent heat of freezing
m = mass of vapour
`M` = mass of freezed
`therefore ` Fraction of water which freezed
= `(M)/(m+M) = (etaM)/(m + etam) = (n)/(n+1)`