Question

A lake surface is exposed to an atmosphere where the temperature is less than $0^{\circ }C$. If the thickness of the ice layer formed on the surface grows from $2cm$ to $4cm$ in $1$ hour, the atmospheric temperature will be (Thermal conductivity of ice, $K=4\times 10^{-3}calc{m}^{-1}{s}^{-1}{}^{\circ }{C}^{-1}$, density of ice $=0.9g$ of $1$. Latent heat of fusion of ice $=80cal{g}^{-1}$ Neglect the change of density during the state change. Assume that the water below the ice has $0^{\circ }C$ temperature every where.)

• A. $-20{}^{\circ }C$
• B. $0{}^{\circ }C$
• C. $-30{}^{\circ }C$
• D. $-15{}^{\circ }C$

Answer 1

Answer: (C)

The rate of heat flow through the layer of the ice,

$i_H=\frac{0-\left(-\theta \right)}{\left(y/KA\right)}=\frac{\theta KA}{y}=\frac{dQ}{dt}\dots \left(i\right)$
Also, rate of heat gain during fusion of ice,
$\frac{dQ}{dt}=L\frac{dm}{dt}=L\rho .A\frac{dy}{dt}\dots \left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{KA\theta }{y}=\rho AL\frac{dy}{dt}M$
$\underset{2}{\overset{4}{\int }}ydy=\underset{0}{\overset{3600}{\int }}\left(\frac{K\theta }{\rho L}\right)dt$
${\left[\frac{y^2}{2}\right]}_2^4=\frac{K\theta }{\rho L}{\left[t\right]}_0^{3600}$
$\Rightarrow \frac{1}{2}\times \left[16-4\right]=\frac{4\times 10^{-3}\times \theta \times \left(3600-0\right)}{0.9\times 80}$
$\Rightarrow \theta =\frac{1}{2}\times \frac{12\times 0.9\times 80}{4\times 3600\times 10^{-3}}$
$=30{}^{\circ }C$
$\therefore$ The atmospheric temperature $=-\theta =-30^{\circ }C$