Question

On a cold winter day, the atmospheric temperature is $-\theta$ (on Celsius scale) which is below $0^{\circ }\text{C}$. A cylindrical drum of height $h$ made of a bad conductor is completely filled with water at $0^{\circ }\text{C}$ and is kept outside without any lid. Calculate the time taken for the whole mass of water to freeze. Thermal conductivity of ice is $K$ and its latent heat of fusion is $L$. Neglect expansion of water on freezing.

• A. $\frac{\rho L{h}^2}{2K\theta }$
• B. $\frac{3\rho L{h}^2}{2K\theta }$
• C. $\frac{\rho L{h}^2}{4K\theta }$
• D. $\frac{5\rho L{h}^2}{2K\theta }$

Answer 1

The correct option is A $\frac{\rho L{h}^2}{2K\theta }$

Let the outside temperature be $-{\theta }^{o}C$.

Since it's given that the walls of the container are poorly conducting, the heat lost by the water is due to radiation.

Since the water is at $0^{o}C$, it freezes when it releases heat equal to latent heat of fusion.

At any time, let $x$ be the level of water below the suface of the container.


$\frac{dQ}{dt}=\frac{dm}{dt}.L_f$
where $\frac{dm}{dt}$ is the rate at which ice freezes.

Thermal current from the water to the surroundings flows through the frozen ice.

$i_{th}=\frac{-kA(-\theta -0)}{x}$ where $x$ is the thickness of the ice layer above water level.

$i_{th}=\frac{kA\theta }{x}=\frac{dQ}{dt}=\frac{dm}{dt}.L_f$

$\frac{kA\theta }{x}=\rho A\frac{dx}{dt}.L_f$

$\Rightarrow {\int }_0^{h}xdx=\frac{kA\theta }{\rho Al}{\int }_0^{t}dt$

$\Rightarrow \frac{x^2}{2}=\frac{k\theta }{\rho l}t$

$\Rightarrow t=\frac{\rho l{h}^2}{2k\theta }$