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 \(10^{\circ} \mathrm{C}\), the air temperature a constant \(-10^{\circ} \mathrm{C}_{7}\) 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 \(t=0\) when \(x=0\).

Short answer

The thickness x of the ice increases as the square root of time t: \( x = \sqrt{2c k A \Delta T \, t} \).

Step-by-step solution

- Understand the Problem

You're asked to show that the thickness of the ice on a lake increases with the square root of the time in cold weather. Given: Water temperature is constant at 10°C, air temperature is constant at -10°C, ice forms a uniform slab of thickness x at time t, and the rate of formation of ice is proportional to the rate of heat transfer from water to air.

- Express Heat Transfer

Use the formula for the rate of heat conduction through a slab: \[ Q = -kA \frac{dT}{dx} \] where - Q is the rate of heat transfer - k is the thermal conductivity - A is the area - \frac{dT}{dx} is the temperature gradient

- Define Temperature Gradient

Given that temperature difference between water and air is: \[ \Delta T = 10 - (-10) = 20 \text{°C} \] and the thickness of the ice is x, the temperature gradient can be expressed as: \[ \frac{dT}{dx} = \frac{20}{x} \]

- Set Up the Proportionality

The rate of ice formation is proportional to the rate of heat transfer. Let the proportional constant be c. Then: \[ \frac{dx}{dt} = c \frac{Q}{L} = c \frac{kA \Delta T}{x} \] where L is the latent heat of fusion (constant)

- Integrate the Equation

Rewrite the equation: \[ y = \frac{dx}{dt} = c \frac{kA \Delta T}{x} \] Integrate both sides with respect to time t: \[ \int x\, dx = \int c k A \Delta T \, dt \] This yields: \[ \frac{x^2}{2} = c k A \Delta T \, t \]

- Solve for x

Multiply both sides by 2: \[ x^2 = 2c k A \Delta T \, t \] Take the square root of both sides: \[ x = \sqrt{2c k A \Delta T \, t} \] Therefore, the thickness of the ice increases with the square root of time.

Key concepts

Rate of Heat Conduction

In thermodynamics, the rate of heat conduction describes how heat transfers through a material. Heat always flows from regions of higher temperature to regions of lower temperature. This process is governed by Fourier's Law of Heat Conduction.
In this exercise, we use the formula:
\[ Q = -kA \frac{dT}{dx} \]
where:

• Q is the rate of heat transfer
• k is the thermal conductivity of the material
• A is the cross-sectional area perpendicular to the heat flow
• \( \frac{dT}{dx} \) is the temperature gradient across the material

The negative sign indicates that heat flows from higher to lower temperature zones.
By understanding how heat conducts through the material, we can determine the rate at which ice forms on the lake surface when it is cold.

Temperature Gradient

The temperature gradient represents the rate of change of temperature with distance. It plays a crucial role in determining heat transfer. In the exercise, the temperature difference between the water and air is given as 20°C. This gradient creates a driving force for heat to move from the warmer water to the colder air through the ice layer.
The temperature gradient can be expressed mathematically as:
\[ \frac{dT}{dx} = \frac{\Delta T}{x} \]
In this case:

• \(\Delta T\) is the temperature difference (20°C)
• x is the thickness of the ice slab

This simplifies to \( \frac{20}{x} \).
Understanding this concept allows us to set up the proportionality needed later to relate ice thickness to heat transfer.

Integration in Calculus

To find a relationship between the thickness of the ice and time, we need to use integration. Integration helps sum up small changes over time to give us a cumulative result.
Given the rate of ice formation is proportional to the rate of heat transfer, our equation is:
\[ \frac{dx}{dt} = c \frac{kA \Delta T}{x} \]
To solve for x, we need to integrate both sides:
\[ \begin{aligned} & \int x \, dx = \int c k A \Delta T \, dt \] Performing the integration yields:
\[ \frac{x^2}{2} = c k A \Delta T t \]
Solving for x:
\[ x = \sqrt{2c k A \Delta T t} \]Hence, the thickness of the ice grows with the square root of time. This calculus concept allows us to comprehensively understand the dynamics of ice formation.