Question

A lake is covered with ice $2cm$ thick. The temperature of ambient air is $-{15}^{\circ }C$. Find the rate of thickening of ice. For ice $K=4\times {10}^{-4}k-cal{m}^{-1}{s}^{-1}{(}^{\circ }C{)}^{-1}$, density $=0.9\times {10}^{3}kg/m^3$ and latent heat of ice $\left(L\right)=80kcal./kg$.

Answer 1

Heat energy flowing per sec is given by
$H=\frac{dQ}{dt}=KA\frac{△\theta }{x}.....\left(i\right)$
If $dm$ is the mass of ice formed in time $dt$, then
$\frac{dm}{dt}=\frac{Adx\rho }{dt}=A.\rho .\frac{dX}{dt}$
Since, $H=\left(\frac{dm}{dt}\right)L$
$\therefore H=A\rho \frac{dx}{dt}L.....\left(ii\right)$
From eq. (i) and (ii)
$A\rho L\frac{dx}{dt}=-KA\frac{△\theta }{x}$
Rate of thickening of ice $=\frac{dx}{dt}$
$\therefore \frac{dx}{dt}=-\frac{KA}{\rho AL}\frac{△\theta }{x}=\frac{-K}{\rho L}\frac{△\theta }{x}=\frac{4\times {10}^{-4}}{0.9\times {10}^3\times 80}\times \left(\frac{0-(-15)}{2\times {10}^{-2}}\right)=4.166\times {10}^{-6}m/s$
$=1.45cm/hour$.