Q101 / 180 · NEET 2019 (Odisha) · Physics

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

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

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

Hence, the correct answer is option: 'C'.

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