Question

A deep rectangular pond of surface area $A$, containing water $($ density $=\rho$, specific heat capacity $=s$ ), is located in a region where the outside air temperature is a steady at $-26^{\circ }C$. The thickness of the frozen ice layer in this pond, at a certain instant is $x$. Taking the thermal conductivity of ice as $K$ and its specific latent heat of fusion as $L$, the rate of increase of the thickness of ice layer, at this instant would be given by

• A. $26K/\rho r(L-4s)$
• B. $26K/\rho x^2-L$
• C. $26K/pxL$
• D. $26K/\rho r(L+4s)$

Answer 1

Answer: (C)

If area of cross-section of a surface is not uniform or if the steady state condition is not reached, then the heat flow equation can be applied to a thin layer of material perpendicular to the direction of heat flow.
In this case, the thickness of a frozen layer in the pond at a certain instant is $x$.
So, the rate of heat flow by conduction for growth of ice is given by
$\frac{dQ}{dt}=\frac{KA\left({\theta }_0-{\theta }_1\right)}{x}...(i)$
where, $dQ=\rho AdxL,{\theta }_0=0$ and ${\theta }_1=-\theta$
Given, ${\theta }_0=0^{\circ }C$ and ${\theta }_1=-26^{\circ }C$
The rate of increase of thickness can be calculated from Eq. (i), we get
$\frac{dQ}{dt}=\frac{KA\left({\theta }_0-{\theta }_1\right)}{x}$
$\Rightarrow \frac{\rho AdxL}{dt}=\frac{KA\left({\theta }_0-{\theta }_1\right)}{x}$
$\Rightarrow \frac{dx}{dt}=\frac{KA\left({\theta }_0-{\theta }_1\right)}{\rho AxL}$
$=\frac{K[0-(-26)]}{\rho xL}=\frac{26K}{\rho xL}$