Question

A lake surface is exposed to an atmosphere where the temperature is less than $0^{o}C$. If the thickness of the ice layer formed on the surface grows from $2cm$ to $4cm$ in $1$ hour, the atmosphere temperature will be (Thermal conductivity of ice,$K=4\times {10}^{-3}calc{m}^{-1}{s}^{-1}{}^o{C}^{-1}$, density of ice $=0.9gc{c}^{-1}$. 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 $0^{o}C$ temperature every where.)

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

Answer 1

The correct option is C $-{30}^{o}C$

The rate of heat flow through the layer of the ice formed ice $-\theta <0^{o}C$

$i_H=\frac{0-(-\theta )}{\left(y/KA\right)}=\frac{\theta KA}{y}=\frac{dQ}{dt}$ .........(i)

Also, rate of heat gain during fusion of ice,
$\frac{dQ}{dt}=L\frac{dm}{dt}=L\rho .A\frac{dy}{dt}$ ........(ii)

From equations (i) and (ii), we get
$\frac{KA\theta }{y}=\rho AL\frac{dy}{dt}$

${\int }_2^{4}ydy={\int }_0^{3600}\left(\frac{K\theta }{\rho L}\right)dt$

${\left[\frac{y^2}{2}\right]}_2^4=\frac{K\theta }{\rho L}[t{]}_0^{3600}$

$\Rightarrow \frac{1}{2}\times [16-4]=\frac{4\times {10}^{-3}\times \theta \times (3600-0)}{0.9\times 80}$

$\Rightarrow \theta =\frac{1}{2}\times \frac{12\times 0.9\times 80}{4\times 3600\times 10-3}={30}^{o}C$

$\therefore$ the atmospheric temperature $=-\theta =-{30}^{o}C$