Question

A lake surface is exposed to an atmosphere where the temperature is < $0^{\circ }$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, $T_a$ will be (Thermal conductivity of ice $K=4m,\times {10}^{-3}cal/cm/s{/}^{\circ }C$; density of ice = 0.9 gm/cc. Latent heat of fusion of ice = 80 cal/gm. Neglect the change of density during the state change. Assume that the water below the ice has $0^{\circ }$ temperature every where)

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

Answer 1

The correct option is C $-{30}^{\circ }$C

This problem is based on growth of ice on ponds

When temperature of atmosphere falls below 0°C, the water in lake or ponds starts freezing.

Let the time t, thickness of ice is y cm and atmospheric pressure is -T°C.

The temperature of water in control with the lower surface of ice be0°C

$\frac{d\theta }{dt}=L_f\left(\frac{dm}{dt}\right)$ ($L_F$= latent heat of fusion)

$\frac{\Delta T}{R}=L_f\frac{d}{dt}\left(A\rho y\right)$ (Ay=Volume, $\rho$=density)

$\frac{0-(-T)}{\frac{y}{KA}}=L_{f}A\rho \frac{dy}{dt}$

And hence time taken by ice to grow a thickness y

$t=\frac{\rho L_f}{KT}{\int }_0^{y}ydy$ (1)

$t=\frac{1}{2}\frac{\rho L_f}{KT}y$

Time does not depend on area of pond.

Now come to question, using the formula.

Now in (1)

$t=\frac{\rho L_f}{KT}{\int }_{y_1}^{y_2}ydy$

$t-\frac{1}{2}\frac{\rho L_f}{KT}[y_2^2-y_1^2]$

$3600=\frac{0.9\times 80}{2\times 4\times {10}^{-3}}\times T[16-4]$

$300=\frac{72}{8\times {10}^{-3}}\times T$

$T=\frac{72\times 1000}{300\times 8}\times T$

$T=-30°C$