Question

A lake surface is exposed to an atmosphere where the temperature is less than $0$. If the thickness of the ice layer formed on the surface grown from 2 cm to 4 cm in 1 hour, the atmospheric temperature will be (Thermal conductivity office, $K=4\times {10}^{-3}cal{cm}^{-1}{s}^{-1}{}^{-1}$, density of ice $0.9g{cc}^{-1}$. 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 $0$ temperature every where.)

• A. $-20$
• B. $0$
• C. $-30$
• D. $-15$

Answer 1

Answer: (C)

$-30$

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

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

Also, rate of heat gain during fusion of ice,

$\frac{dQ}{dt}=L\frac{dm}{dt}=L\rho .A\frac{dy}{dt}$ $\dots$ (ii)

From equations (i) and (ii), we get

$\frac{KA\theta }{y}=\rho AL\frac{dy}{dt}$

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

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

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

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

$\therefore$ The atmospheric temperature $=-\theta =-30$