Question

Water is densest at 4${}^{\circ }$C; this is the reason lakes do not completely freeze during extreme winters. Say, the atmosphere is at -${\theta }^{\circ }C(\theta >0)$,now the temperature of the lake's water starts to fall, and the denser water from the top sinks, getting the bottom layers up, and cooling them until the temperature reaches 4 ${}^{\circ }$C at the upper surface. Now, further reduction in temperature actually makes the water less dense! Hence the colder water stays on top, starts to freeze, until the top layer of ice reaches -${\theta }^{\circ }C$ and the bottom layer at 0 ${}^{\circ }$ C . And this layer of ice is the only area through which the water below loses heat (by conduction), slowly increasing the thickness of the ice (because ice is a bad conduction), and for a very long time the water at the bottom of the lake remains at 4 ${}^{\circ }$ C! Now, Let the thickness of the layer of ice be $y_1$, at an instant.How much time t, will it take to increase to thickness $y_2$?

$\rho \to$ density of the ice

K $\to$ thermal conductivity of ice.

L $\to$ Latent heat of fusion

$\theta \to$ temperature of surface (as mentioned earlier)

• A. $\frac{\rho L}{K\theta }(y_2-y_1)$
• B. $\frac{\rho L}{2K\theta }(y_2-y_1)$
• C. $\frac{\rho L}{K\theta }(y_2^2-y_1^2)$
• D. $\frac{\rho L}{2K\theta }(y_2^2-y_1^2)$

Answer 1

The correct option is D

$\frac{\rho L}{2K\theta }(y_2^2-y_1^2)$

Water in a lake starts freezing if the atmospheric temperature drops below 0${}^{\circ }$C. Let y be the thickness of ice layer in the lake at any instant t and atmospheric temperature is -$\theta$${}^{\circ }$C. The temperature of water in contact with the lower surface of ice will be zero. If A is the area of lake, heat escaping through ice in time dt is

$d{Q}_1=\frac{KA[0-(-\theta \left)\right]dt}{y}$

Now, suppose the thickness of ice layer increases by dy in time dt.due to escaping of above heat. Then $d{Q}_2=mL=\rho \left(dyA\right)L$

As $d{Q}_1=d{Q}_2$

hence, rate of growth of ice will be $\frac{dy}{dt}=\frac{K\theta }{\rho Ly}$

So, the time taken by ice to grow to a thickness y is $t=\frac{\rho L}{K\theta }{\int }_0^{y}ydy=\frac{\rho L}{2K\theta }{y}^2$

IF the thickness is increased from y1 to y2 then time taken $t=\frac{\rho L}{K\theta }{\int }_{y_1}^{y_2}ydy=\frac{\rho L}{2K\theta }(y_2^2-y_1^2).$