A deep rectangular pond of surface area A , containing water...

A deep rectangular pond of surface area

\mathrm{A}

, containing water (density

=\rho

, specific heat capacity

=s

), is located in a region where the outside air temperature is a steady value at the

-26^{\circ} \mathrm{C}

. The thickness of the frozen ice layer in this pond, at a certain instant is

\mathrm{x}

. Taking the thermal conductivity of ice as

\mathrm{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

Options

A $26 \mathrm{~K} / \rho r(\mathrm{~L}-4 \mathrm{~s})$
B $26 \mathrm{~K} /\left(\rho x^2-L\right)$
C $26 K /(\rho x L)$
D $26 K / \rho r(L+4 s)$

Correct answer

C. $26 K /(\rho x L)$

Step-by-step solution

Key Idea If area of cross-section of a surface is not uniform or if the steady state condition is not reached, the heat flow equation can be applied to a thin layer of material perpendicular to direction of heat flow. The rate of heat flow by conduction for growth of ice is given by,

\frac{d \theta}{d t}=\frac{\left.K A\left(\theta_0-\theta_0\right)\right]}{x}

where,

d \theta=\rho A d x L, \theta_0=0

and

\theta_1=-\theta

Given,

\theta_0=0^{{}^{\circ}} \mathrm{C}, \theta_1=-26^{{}^{\circ}} \mathrm{C}

The rate of increase of thickness can be calculated from Eq.

\begin{aligned}\frac{\mathrm{d} \theta}{\mathrm{dt}}& =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\mathrm{x}}\\ \Rightarrow \quad \frac{\rho \mathrm{AdxL}}{\mathrm{dt}}& =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\mathrm{x}}\\ \Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{dt}}& =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\rho \mathrm{AxL}}\\ & =\frac{\mathrm{K}[0-(-26)]}{\rho \times \mathrm{L}}=\frac{26 \mathrm{~K}}{\rho \times \mathrm{L}} \end{aligned}

Practice Thermal Properties of Matter on Quantrex Academy →

More from Thermal Properties of Matter