Question

Ice has formed on a shallow pond and a steady state has been reached with the air above the ice at \(-5.20^{\circ} \mathrm{C}\) and the bottom of the pond at \(3.98^{\circ} \mathrm{C}\). If the total depth of ice \(+\) water is \(1.42 \mathrm{~m}\), how thick is the ice? (Assume that the thermal conductivities of ice and water are \(1.67\) and \(0.502 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively.)

Short answer

To find the thickness of the ice, apply the heat conduction equation, equate the heat conducted through the ice and water and solve the equation.

Step-by-step solution

Understand the problem

We have a shallow pond with ice and water, the pond is \1.42m deep. The temperatures at different layers of the pond are given along with the thermal conductivities of ice and water. We need to find the thickness of the ice.

Apply heat conduction equation

Heat conduction is expressed by the equation Q=\(kA{\Delta}T / d\), where: Q is heat conduction, k is the thermal conductivity, A is the cross sectional area, \({\Delta}T\) is temperature difference and d is the thickness of the material.

Equate heat conducted through ice and water

Since the problem reached the steady state, the heat conducted through the ice must be equal to the heat conducted through the water. Hence, \(k_{ice}A{\Delta}T_{ice} / d_{ice}\) = \(k_{water}A{\Delta}T_{water} / d_{water}\). Here, ice and water share the same cross-sectional area which cancels out so we get \(k_{ice}{\Delta}T_{ice} / d_{ice}\) = \(k_{water}{\Delta}T_{water} / d_{water}\).

Substitute with given values and solve for \(d_{ice}\)

Plug in the given values: \(1.67({-5.20 - 3.98)}/d_{ice} = 0.502(3.98)/ (1.42 - d_{ice}) \). After solving this equation, we can find the value of \(d_{ice}\), the thickness of the ice.

Key concepts

Heat Conduction Equation

When we talk about heat transfer, the heat conduction equation is a fundamental concept to consider. It's expressed mathematically as:

\( Q = kA\frac{\Delta T}{d} \).

This equation represents the rate at which heat (\( Q \)) is transferred through a material. Here, \( k \) stands for the material's thermal conductivity, which is a measure of its ability to conduct heat. The variable \( A \) signifies the cross-sectional area through which heat is being transferred, and \( \Delta T \) describes the temperature difference across the material. Finally, \( d \) represents the thickness of the material.

When applying the heat conduction equation, it's crucial to pay attention to the units and ensure they are consistent. This equation is particularly useful for solving problems involving steady state heat transfer, like estimating the thickness of ice on a pond, given the thermal conductivity of ice and the temperature difference across it.

Steady State Heat Transfer

Moving on from the general heat conduction equation, we arrive at the concept of steady state heat transfer. This occurs when the temperature gradient within an object remains constant over time, meaning that the rate at which heat enters a section of the object is equal to the rate at which it exits. In the scenario of the ice-covered pond, the steady state is reached when the heat conducted through the ice is the same as the heat conducted through the water, given the constant temperatures on either side of the ice and water layers.

The beauty of steady state heat transfer is that it simplifies our calculations. We don't have to worry about time-dependent changes in temperature or thermal energies. The equation for heat conduction, which we covered in the previous section, becomes particularly convenient to use since we can set up an equalization based on the fact that the transfer rates through both materials are the same at the steady state.

Thermal Conductivity

Finally, let's dive deeper into the concept of thermal conductivity (\( k \)). It's one of the key parameters in the heat conduction equation and varies between different materials. Thermal conductivity is defined as the ability of a material to conduct heat and is measured in watts per meter per degree Kelvin or Celsius (\( W/m\cdot K \) or \( W/m\cdot ^\circ C \)).

Materials with high thermal conductivity, like metals, are excellent heat conductors, which is why they feel cold to the touch; they effectively transfer heat away from your body. In contrast, materials with low thermal conductivity, such as rubber or wood, act as insulators.

In our ice pond problem, understanding the different conductivities of ice and water is critical. Ice, with a higher thermal conductivity, will transfer heat more efficiently than water, affecting the distribution of temperature and consequently, the thickness of the ice layer. By comparing the known conductivities and temperature differences of the ice and water, we can solve for the unknown thickness of the ice.