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 $2\, cm$ to $4 \,cm$ in $1$ hour, the atmospheric temperature will be (Thermal conductivity of ice, $K=4 \times 10^{-3}\, cal\, cm^{-1}\, s^{-1}\,{}^{\circ}C^{-1}$, density of ice $= 0.9\, g$ of $1$. Latent heat of fusion of ice $= 80\, cal\, g^{-1}$ 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,

$i_{H}=\frac{0-\left(-\theta\right)}{\left(y/KA\right)}=\frac{\theta\,KA}{y}=\frac{dQ}{dt}\quad\ldots\left(i\right)$
Also, rate of heat gain during fusion of ice,
$\frac{dQ}{dt}=L \frac{dm}{dt}=L\,\rho.A \frac{dy}{dt}\quad\ldots\left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{KA\theta}{y}=\rho AL \frac{dy}{dt}M$
$\int \limits^{4}_{2} ydy=\int \limits^{3600}_{0} \left(\frac{K\theta}{\rho L}\right)dt$
$\left[\frac{y^{2}}{2}\right]^{4}_{2}=\frac{K\theta}{\rho L}\left[t\right]^{3600}_{0}$
$\Rightarrow \frac{1}{2}\times\left[16-4\right]=\frac{4\times10^{-3} \times\theta \times\left(3600-0\right)}{0.9\times80}$
$\Rightarrow \theta=\frac{1}{2} \times \frac{12 \times 0.9 \times80}{4 \times3600 \times10^{-3}}$
$=30\,{}^{\circ}C$
$\therefore $ The atmospheric temperature $=-\theta=-30^{\circ}\,C$