b$)$ Naphthalene ball $3cm.$ in diameter is exposed to on atmosphere of air at $35C.$ Calculate using Chilton $-$ Colburn analogy, the amount of naphthatene evaporated per unit time.

Data:

${}^n-670\frac{N}{m^2},1h-140\frac{m}{m^2}k$

$CP=1k\frac{J}{k}\frac{g}{C},=1\mathrm{.}14k\frac{g}{m^3}$

$D_{AB}=9\mathrm{.}5{10}^{-6}\frac{m^2}{s},k=0\mathrm{.}027\frac{w}{m}*k$

mol. wt of naphthalene $=128.$