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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 isA. `(1)/(eta)`B. `(eta)/(eta+1)`C. `(eta-1)/(eta)`D. `(eta-1)/(eta+1)` |
Answer» Correct Answer - B `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)` |
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