2.10 Thermal Conductivity

7 min read•december 31, 2022

Krish Gupta

Daniella Garcia-Loos

AP Physics 2 🧲

61 resources

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Thermal Conductivity

Thermal conductivity refers to how well an object can transfer thermal energy. Objects that have high thermal conductivity are good conductors of heat and vice versa. Thermal conductivity of a substance is given by the letter k.

Here are some key points about thermal conductivity:

Thermal conductivity is a property of a material that describes how easily heat can be conducted through it.
Materials with high thermal conductivity are good at conducting heat, while materials with low thermal conductivity are poor at conducting heat.
The thermal conductivity of a material is determined by its atomic structure, the arrangement and motion of its atoms and molecules, and the strength of its bonds.
Thermal conductivity is usually measured in watts per meter per kelvin (W/m-K).
The thermal conductivity of a material depends on its temperature, with most materials having higher thermal conductivity at higher temperatures.
The thermal conductivity of a material also depends on the presence of impurities or defects in the material, which can disrupt the flow of heat.
Thermal conductivity is an important property for materials used in applications that involve the transfer of heat, such as insulation, heat exchangers, and heat sinks.

Metals

Metals tend to have a much higher value of thermal conductivity than organic solids, liquids and gases. This is partly due to the sea of electrons and metallic bonding in metals as opposed to the covalent bonding found in many non-metals.

Let’s look at an example. Let’s say you are out on a sunny day and you decide to go fishing. You have a metal fishing pole and a wooden fishing pole. Let’s say you have been fishing for a while now. You’d expect both the fishing poles to be the same temperature as the environment because heat would have flown until thermal equilibrium was established.

You would be right. If you measured the temperatures of either fishing poles using a thermometer you’d get the same number. BUT if you were to touch each fishing pole which one do you think would feel hotter? The metal one right? So why is it that the metal feels hotter than wood even though they are at the same temperature? This is because a metal has a much higher thermal conductivity than wood so it transfers heat to your hand much faster.

The same would be true on a winter day. If you left a plastic water bottle and a metal water bottle out in the month of December you’d feel that the metal one is much colder even though truly they are at the same temperature. This is because metal is great at transferring heat so it sucks the heat out of your hand much faster than wood.

Here are some key points about the thermal conductivity of metals:

Metals are generally good at conducting heat, due to their metallic bonding and the ability of their valence electrons to move freely through the lattice.

The thermal conductivity of a metal is typically higher than the thermal conductivity of non-metallic materials, such as ceramics, plastics, and insulation.

The thermal conductivity of a metal depends on its atomic structure, the arrangement and motion of its atoms and molecules, and the strength of its bonds.

The thermal conductivity of a metal also depends on its temperature, with most metals having higher thermal conductivity at higher temperatures.

The thermal conductivity of a metal also depends on the presence of impurities or defects in the material, which can disrupt the flow of heat.

Examples of metals with high thermal conductivity include copper, silver, and gold, while examples of metals with low thermal conductivity include aluminum, lead, and zinc.

Fourier's Conduction Law

Now that we know that objects with greater thermal conductivity allow for faster heat flow, we need an equation to find how much faster? 10 times? 100 times?

To answer that question we have Fourier’s Conduction Law: Q/Δt=kAΔT/L

This law states that the rate at which heat flows depends on the thermal conductivity of teh object, the cross sectional area and the temperature gradient. The temperature gradient is defined as the ratio of the temperature difference to the length of the object 🧮 Since this is not a calculus course, we can assume that the rate of heat flow is the same across the object or even across a junction of two objects.

Here are some key points about Fourier's conduction law:

Fourier's conduction law is a mathematical relationship that describes how heat is conducted through a solid material.
According to Fourier's conduction law, the heat flow (Q) through a material is proportional to the temperature gradient (dT/dx) and the cross-sectional area (A) of the material, and is inversely proportional to the thickness (L) of the material.
Mathematically, Fourier's conduction law can be written as Q = -kA(dT/dx)/L, where k is the thermal conductivity of the material.
The negative sign in this equation indicates that heat flows from hot to cold, and the thermal conductivity (k) is a measure of how easily heat can flow through the material.
Fourier's conduction law is based on the idea that heat is conducted through a material by the movement of atoms and molecules, which transfer energy from one location to another through collisions and interactions.
Fourier's conduction law is used to predict the heat flow through a material under different conditions, such as different temperature gradients, cross-sectional areas, and thicknesses. It is also used to design materials and structures that are good at conducting or insulating heat.

Example Problem:

You are asked to design an experiment to measure the thermal conductivity of two different metals, aluminum and copper. The aluminum sample has a thickness of 0.5 cm, and the copper sample has a thickness of 0.75 cm.

(a) Describe the experimental setup you would use to measure the thermal conductivity of the aluminum and copper samples.

(b) Explain how you would measure the temperature gradient across the samples, and what factors you would need to control in order to obtain accurate results.
(c) Explain how you would calculate the thermal conductivity of each metal from the data you collect.
(d) Based on your calculations, which metal do you expect to have a higher thermal conductivity? Why?
(e) Discuss any sources of error that could affect the accuracy of your measurements, and how you would mitigate or correct for these errors."

To solve this problem, you could follow these steps:

(a) Design an experimental setup to measure the thermal conductivity of the aluminum and copper samples. One way to do this would be to use a hot plate to heat one end of each sample, and a thermocouple to measure the temperature at the hot and cold ends of the sample. You could place the samples on a thermally insulating surface to minimize heat loss to the surroundings.

(b) Measure the temperature gradient across the samples by placing the thermocouple at the hot and cold ends of the sample and recording the temperature readings. To obtain accurate results, you would need to control for factors such as the temperature of the hot plate, the distance between the hot and cold ends of the sample, and the time elapsed during the measurement.

(c) Calculate the thermal conductivity of each metal from the data you collect using the formula Q = -kA(dT/dx)/L, where Q is the heat flow through the sample, k is the thermal conductivity of the material, A is the cross-sectional area of the sample, dT/dx is the temperature gradient across the sample, and L is the thickness of the sample. To calculate the heat flow (Q), you could measure the power of the hot plate and the time elapsed during the measurement, and use the formula Q = P*t. To calculate the cross-sectional area (A) of the sample, you could measure the width and height of the sample.

(d) Based on your calculations, you would expect copper to have a higher thermal conductivity than aluminum. This is because copper has a higher atomic weight and a more ordered crystal structure than aluminum, which allows it to conduct heat more efficiently.

(e) Some sources of error that could affect the accuracy of your measurements include:

Uncertainty in the temperature readings from the thermocouple. To mitigate this error, you could use a calibrated thermocouple and take multiple readings to average out any random error.
Uncertainty in the dimensions of the sample. To mitigate this error, you could use a micrometer or other precise measuring tool to measure the thickness and cross-sectional area of the sample.
Heat loss to the surroundings. To mitigate this error, you could use a thermally insulating surface to minimize heat loss and ensure that most of the heat flows through the sample.
Changes in the temperature of the hot plate over time. To mitigate this error, you could use a thermostatically controlled hot plate and measure the temperature of the hot plate at regular intervals during the measurement.

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