Spread the love

Growth of Ice on Lake

Growth of Ice on Lake :- The growth of ice on a lake is a fascinating natural process that occurs during cold weather conditions. When the air temperature drops below freezing (0°C or 32°F), the heat from the lake’s surface is transferred to the colder air, causing the water at the surface to lose heat and freeze.

Consider a lake of surface area A. Let at any time t thickness of ice on the surface of the lake = y and air temperature = –θ° C. The temperature of water in contact with the lower surface of ice = 0°C

Rate of heat loss from the surface of lake to atmosphere, through ice slab of thickness y is given by :-

$\displaystyle \frac{dQ}{dt}=\frac{kA\left[ 0-(-\theta ) \right]}{y}$

$\displaystyle \Rightarrow dQ=\frac{kA\theta }{y}dt$ …..(1)

This dQ is the heat loss of extra thickness(dx) of ice.

Mass of this extra thickness of ice, dm = ρ_iceV = ρ_iceA dy

$\displaystyle \Rightarrow dQ=\left( dm \right){{L}_{f}}=A{{\rho }_{ice}}{{L}_{f}}dy$ …..(2)

From equations (1) and (2),

$\displaystyle \frac{kA\theta }{y}dt=A{{\rho }_{ice}}{{L}_{f}}dy$

$\displaystyle \Rightarrow \int\limits_{0}^{t}{dt}=\frac{{{\rho }_{ice}}{{L}_{f}}}{k\theta }\int\limits_{0}^{y}{ydy}$

$\displaystyle \Rightarrow \left[ t \right]_{0}^{t}=\frac{{{\rho }_{ice}}{{L}_{f}}}{k\theta }\left[ \frac{{{y}^{2}}}{2} \right]_{0}^{y}$

So time taken by ice to grow a thickness y is

$\displaystyle \Rightarrow t=\left[ \frac{{{\rho }_{ice}}{{L}_{f}}}{2k\theta } \right]{{y}^{2}}$

Note :-

(1). Time taken by ice to grow from thickness y₁ to thickness y₂ is

$\displaystyle t={{t}_{2}}-{{t}_{1}}=\frac{{{\rho }_{ice}}{{L}_{f}}}{2k\theta }\left( y_{2}^{2}-y_{1}^{2} \right)$

Or $\displaystyle t\propto \left( y_{2}^{2}-y_{1}^{2} \right)$

(2). If at t = 0, y = 0, then the ratio of time taken to double and triple the thickness :

$\displaystyle {{t}_{1}}:{{t}_{2}}:{{t}_{3}}:.....{{t}_{n}}={{1}^{2}}:{{2}^{2}}:{{3}^{2}}:.....{{n}^{2}}$

$\displaystyle \Delta {{t}_{1}}:\Delta {{t}_{2}}:\Delta {{t}_{3}}:.....={{1}^{2}}-{{0}^{2}}:{{2}^{2}}-{{1}^{2}}:{{3}^{2}}-{{2}^{2}}:.....$

$\displaystyle \Delta {{t}_{1}}:\Delta {{t}_{2}}:\Delta {{t}_{3}}:.....=1:3:5:7:.....$

Example 1.

Ice starts forming in a lake with water at 0°C when the atmospheric temperature is -10°C. If the time taken for the first 1 cm of ice to be formed is 7 hours, then the time taken for the thickness of ice to change from 1 cm to 2 cm is:
(A) 7 hours
(B) 14 hours
(C) 21 hours
(D) 3.5 hours

Solution:

Using

$\displaystyle \Delta {{t}_{1}}:\Delta {{t}_{2}}={{1}^{2}}-{{0}^{2}}:{{2}^{2}}-{{1}^{2}}$

We get

$\displaystyle \frac{\Delta {{t}_{1}}}{\Delta {{t}_{2}}}=\frac{1}{3}$

$\displaystyle \Rightarrow \Delta {{t}_{2}}=3\times \Delta {{t}_{1}}=3\times 7=21hours$

Example 2.

If time taken for the formation of 1 cm of ice(thickness) on a lake is 4 minutes, then calculate time taken in the formation of 4^th cm of ice.

Solution:

Using

$\displaystyle \Delta {{t}_{1}}:\Delta {{t}_{4}}={{1}^{2}}-{{0}^{2}}:{{4}^{2}}-{{3}^{2}}$

$\displaystyle \Rightarrow \frac{\Delta {{t}_{1}}}{\Delta {{t}_{4}}}=\frac{1}{7}$

$\displaystyle \Rightarrow \Delta {{t}_{4}}=7\times \Delta {{t}_{1}}=7\times 4=28\min .$

Example 3.

In a $10 m$ deep lake, the bottom is at a constant temperature of ${4 °C}$ . The air temperature is constant at -4 ${°C}$ . The thermal conductivity of ice is $3$ times that of water. Neglecting the expansion of water on freezing, what is the maximum thickness (in $m$ ) of ice that can be formed?