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

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
$\frac{d Q}{d t}=\frac{K A\left(\theta_{0}-\theta_{1}\right)}{x}\,\,\,...(i)$
where, $d Q=\rho A d x L, \theta_{0}=0$ 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
$\frac{d Q}{d t} =\frac{K A\left(\theta_{0}-\theta_{1}\right)}{x}$
$\Rightarrow \frac{\rho A d x L}{d t} =\frac{K A\left(\theta_{0}-\theta_{1}\right)}{x} $
$\Rightarrow \frac{d x}{d t} =\frac{K A\left(\theta_{0}-\theta_{1}\right)}{\rho A x L}$
$=\frac{K[0-(-26)]}{\rho x L}=\frac{26 K}{\rho x L}$