Question

Show that the thickness of the ice on a lake increases with the square root of the time in cold weather, making the following simplifying assumptions. Let the water temperature be a constant$\mathbf{10}\mathbf{°}\mathit{C}$, the air temperature a constant$\mathbf{-}\mathbf{10}\mathbf{°}\mathit{C}$, and assume that at any given time the ice forms a slab of uniform thickness x. The rate of formation of ice is proportional to the rate at which heat is transferred from the water to the air. Let t=0when x=0.

Short answer

The thickness of the ice on a lake increases with the square root of the time in cold weather when air and water are constant.

Step-by-step solution

Definition of Heat

Heat is a form of energy that is transferred between two materials of various temperatures.This transfer of energy occurs due to differences in the average translational kinetic energy per molecule within the two materials.

Given Parameters

Given that water temperature is constant at$10°C$and the air temperature is constant at$-10°C$and the ice forms a slab of uniform thickness x.

When temperature and air are constant

At time t the thickness of the ice is x, and at time $t+dt$the thickness of the ice is $x+dx.$

Given that the rate of formation of ice is proportional to the rate at which heat transformed from water to air is$\frac{dx}{dt}=\alpha \frac{dQ}{dt}...\left(1\right)$

From the previous exercise we have

$\frac{dQ}{dt}=kA\frac{dT}{dr}...\left(1.1\right)$

Substitute this value in equation (1), we get

$\frac{dx}{dt}=\alpha kA\frac{dT}{dr}$

When surface at x=0  and take integration

Now, let the surface be at x=0 when the distance dx goes upward, then the change in temperature with position is

$$\begin{gathered} \frac{dT}{dx}=\frac{T_W-T_A}{dx} \\ =\frac{20}{dx} \end{gathered}$$Therefore,

$$\begin{gathered} \frac{dx}{dt}=20\alpha kA\frac{1}{dx} \\ dxdx=20\alpha kAdt \end{gathered}$$

Integrating both sides

$$\begin{gathered} \left(\int 1dx\right)dx=20\alpha kA\int 1dt \\ xdx=\left(20\alpha kA\right)t+\beta \end{gathered}$$

Using the boundary condition, we get $\beta =0$.

Now it is helpful to use our assumptions to substitute the value dx. After a very short time, the thickness of the ice is$x=0+dx$ .

Therefore, it can write as$x^2=\left(20\alpha kA\right)t$ or,$x=\gamma \sqrt{t}$

Where $\gamma =\sqrt{20\alpha kA}$is constant.

Hence, the thickness of the ice on a lake increases with the square root of the time in cold weather when air and water are constant.