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Thermodynamics Learning Material
Tutorial ID	Title	Tutorial	Video Tutorial	Revision Notes	Revision Questions
13.2	Thermal Expansion

In this Physics tutorial, you will learn:

What is the phenomenon of thermal expansion (contraction)?
How does it differ from the other types of matter's volume change?
How many types of thermal expansion contraction are there?
What are the factors affecting thermal expansion (contraction)?
Where is the phenomenon of thermal expansion (contraction) encountered in daily life?
How is the phenomenon of thermal expansion (contraction) applied in technology?
What are the main types of thermometers?
What property of matter do most thermometers use to operate?

Introduction

You may have noticed that when a balloon is left for a long time on a cold surface such as on a tiled floor, it shrinks. On the other hand, when you forgot it near a heat source, it may explode. Why does this occur?

Do you know why the railway tracks have a certain distance from each other?

Why electric wires are not completely stretched?

What happens to the volume of objects when they are heated up? Cooled down?

These and many other questions will get answers during the explanation of thermal expansion phenomenon, provided in this tutorial.

What is Thermal Expansion and Thermal Contraction?

Volume of objects can change in three ways:

By adding or removing material from it. This is a mechanical process, during which we change the mass of the object as well. However, the density of material does not change.
By stretching or compressing an object manually, without adding to or removing material from it, i.e. the mass of object does not change. This is a mechanical process too, and any change in one dimension is accompanied by another (reverse) change in the other dimension. For example if you pull a chewing gum, it becomes longer but also thinner at the same time. The same phenomenon can be also observed when we pull an elastic thread. Anyway, the density of material still does not change considerably.
By supplying or removing some heat energy to the object. In this process, mass does not change. However, there is a change in volume because when the object becomes hotter its molecules vibrate more than before and thus, the volume increases as the particles in the outer layers of the object move farther from each other.

In this tutorial, we are concerned about the third method of objects' change in dimensions, i.e. in thermal expansion (contraction) as this process occurs without any mechanical work done on the object. We need only a heat source to make an object expand, or we can just move it away from the heat source to make it contract. It is obvious that the temperature of objects increase when they expand and decreases when they contract. By definition,

Thermal expansion is the general increase in the volume of a material as its temperature is increased. Also, thermal contraction, is the general decrease in the volume of a material as its temperature is decreased.

Thermal expansion (contraction) has no unit; it is represented through a fractional change in length, area or volume of material in respect to the original dimensions. Let's see this quantity more closely in the next paragraph.

Linear Thermal Expansion and Contraction

In reality, all objects expand or contract thermally in all dimensions (in 3 D). However, when one dimension is much greater than the other two (such as a long and thin bars for example), only the dimension corresponding to the greater value (usually the length) is considered. Given this, we can think about the factors affecting the amount of thermal expansion or contraction of a long bar of original length L₀. They are:

Type of material. Obviously, different types of materials experience different amounts of thermal expansion (contraction) for the same conditions. Usually denser materials expand or contract less than the other less dense materials as the particles of dense materials are more strongly bound to each other and it is more difficult to make them move away from their equilibrium positions. There is a quantity known as the coefficient of linear expansion, α (measured in Kelvin or Celsius degree at power -1), which represents numerically the type of material in the formula of thermal expansion (contraction).
Length of material. Obviously, longer the material, more it expands for the same conditions. For example, a long iron bar expands more millimeters than an iron nail if they experience the same process of heating, despite they are made of the same material. Original length of material is represented in formula by the letter L₀, as stated earlier.
Change in temperature. When the change in any object's temperature is slight, it does not experience any considerable change in length. On the other hand, if there is a considerable change in object's temperature, its dimensions change a lot. We can use either Celsius or Kelvin scale but not Fahrenheit to represents the change in temperature of an expanding (contracting) object. Thus, for temperatures measured in Celsius scale, we can write Δt = t - t⁰ for the change in temperature where t is the final and t⁰ the initial temperature of the object, while for temperatures measured in Kelvin scale we can write ΔT = T - T₀ for the same thing.

Putting all the above factors together, we obtain the formula of linear thermal expansion (contraction):

∆L = L₀ × α × ∆T

where ΔL = L - L₀ is the amount of linear extension (compression) a bar with original length experiences when the temperature changes by ΔT = T - T₀ degrees, and L is the final length of the bar.

If we are interested to find the final length of the bar, the above formula becomes:

L - L₀ = L₀ × α × (T - T₀)

L = L₀ + L₀ × α × (T - T₀)

When factoring L₀ we obtain for the final length L of the bar after experiencing thermal expansion or contraction:

L = L₀ × (1 + α × ∆T)

Thus, when the bar heats up, ΔT is positive as T > T₀. Therefore, the final length of the bar will be greater than its its initial length. On the other hand, when the bar cools down, ΔT is negative as T > T₀. This means the bar will be shorter than before at the end of process.

Remark!

You can also use temperatures in Celsius scale in the formula as the change in temperature in both Kelvin and Celsius scale is the same. Or maybe you can switch from Kelvin to Celsius an vice-versa during the solution of a problem if needed.

Example 1

The linear thermal expansion coefficient of iron at 20°C is 11.8 × 10^-6 K-1. What is the length of a 12 m long railway track during a hot summer day where the temperature of the track becomes 42°C?

Solution 1

Let's write the clues first. We have:

α = 11.8 × 10^-6 K-1
t⁰ = 20°C
t = 42°C
L₀ = 12 m
L = ?

Applying the linear thermal expansion formula

L = L₀ [1 + α × (t - t₀]

we obtain for the change in the bar's length after substituting the known values:

L=12 m × [1 + 11.8 × 10^-6 K^-1 × (42-20)°C]
= 12 m × [1 + 11.8 × 10^-6 1/K × 22K]
= 12 m × [1 + 259.6 × 10^-6 ]
= 12 m × [1 + 0.0002596]
= 12 m × 1.0002596
= 12.0031152 m
≈12.003 m

This means in this example, the track elongates by about 3 mm because of weather conditions.

Area Thermal Expansion and Contraction

When an object is foil-like, it has two relevant dimensions, length and width. Height is very small to take into consideration. Therefore, another coefficient, known as area thermal expansion coefficient, β is introduced. Mathematically, we ca write β = 2α for the same material, because each dimension experiences the same degree of thermal expansion or contraction. For example, since the linear expansion coefficient of iron is 11.8 × 10^-6 K-1 (as discussed in the previous example), the area thermal expansion coefficient of iron is 2 × 11.8 × 10^-6 K-1 = 23.6 × 10^-6 K-1. The formula is similar to that used for linear thermal expansion, i.e.

A = A₀ × (1 + β × ∆T)

where A is the final area and A₀ is the original area, and ΔT = T - T₀ is the change in temperature.

Example 2

A tin sheet must cover the upper part of a rectangular storage room of floor dimensions 5 m × 4 m as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Thermal Expansion

If the linear expansion coefficient of tin (in 200 C is 20 × 10^-6 K - 1, calculate the minimum area of the tin cover, so that the materials inside the storage room be protected from weather conditions. Take the minimum temperature in winter in the specific region equal to -30°C. Suppose that the setup is made at normal weather conditions, i.e. at 20°C.

Solution 2

First, we calculate the area coefficient of thermal expansion, β. Thus, given that the linear coefficient of thermal expansion for tin is α = 20 × 10^-6 K-1, we obtain for the corresponding area coefficient of thermal expansion β:

β = 2 × α
= 2 × 20 × 10^-6 K^-1
= 40 × 10^-6 K^-1

The area to be covered represents the original area A₀ in the formula. Thus, we have

A₀ = 5m × 4m
= 20 m²

Also, the change in temperature is calculated by taking the difference between final and initial temperature, i.e.

∆t = t - t₀
= -30°C - 20°C
= -50°C

Therefore, the area of the tin sheet in winter after experiencing thermal contraction is:

A = A₀ × (1 + β × ∆T)
= 20 × [1 + 40 × 10^-6 × (-50)]
= 20 × (1 - 2000 × 10^-6 )
= 20 × (1 - 2 × 10^-3 )
= 20 × (1 - 0.002)
= 20 × 0.998 = 19.96 m²

This means at least other 0.04 m₂ of tin are needed at the setup moment to make possible the entire area coverage of the room. Therefore, at least 20 m₂ + 0.04 m₂ = 20.04 m₂ of tin are needed to cover the entire room in order to prevent issues in winter.

Volume Thermal Expansion and Contraction

As stated earlier, when an object expands or contract due to the change in temperature, it experiences a volume thermal expansion or contraction. Since objects extend in three dimensions in space, they experience a linear expansion or contraction for each dimension. This mean the coefficient of volume expansion or contraction γ is triple the corresponding linear expansion coefficient α, i.e.

γ = 3 × α

For example, given that the linear thermal expansion of mercury is α = 61 × 10^-6 K-1, its corresponding coefficient of volume thermal expansion is

γ = 3 × α
= 3 × 61 × 10^-6 K^-1
= 183 × 10^-6 K^-1

In this way, if an object has a volume V⁰ at temperature T₀, its volume V at another temperature T becomes

V = V₀ × (1 + γ × ∆T)

Example 3

A plastic object of volume 40 cm³ is brought near a heat source in which the air temperature is 80°C. Given that the linear expansion coefficient of plastics at 20°C is 100 × 10^-6 K-1, calculate the final volume of the object after experiencing thermal expansion.

Solution 3

We have the following clues:

V⁰ = 40 cm³ = 40 × 10^-6 m³ = 4 × 10^-5 m³
t⁰ = 20°C
t = 80°C
α = 100 × 10^-6 K^-1 = 10^-4 K^-1
V = ?

We have:

V = V₀ × (1 + γ × ∆T)
= V₀ × [1 + 3α × (t - t₀)]
= 4 × 10^-5 × [1 + 3 × 10^-4 × (80 - 20)]
= 4 × 10^-5 × (1 + 3 × 10^-4 × 60)
= 4 × 10^-5 × (1 + 180 × 10^-4 )
= 4 × 10^-5 × (1 + 0.0180)
= 4 × 10^-5 × 1.018
= 4.072 × 10^-5 m²
= 40.72 cm³

This means the object's volume increases by 0.72 cm³ during the given process of thermal expansion.

Thermal Expansion and Contraction in Daily Life

Thermal expansion and contraction are encountered very often in daily life, and many engineering applications of this phenomenon are of a very wide range. Some of them include:

Railway tracks are placed at a certain distance from each other to avoid issues caused by thermal expansion during hot days in summer.
Electricity or telephone cables bend down in summer and strain in winter due to thermal expansion and contraction respectively.
Metal rings are heated prior to being fixed to a wooden barrel. The rings contract when they cool down and thus, they hold the barrel tightly as shown in the figure.

Bimetallic Strip. Thermostat

A bimetallic strip is a system composed by two different metal strips, which are placed side by side and welded together. This process is carried out in normal temperature, so initially the system looks like this:

Let's suppose that the coefficient of linear thermal expansion of metal 1 is greater than that of metal 2. This means the metal 1 can expand or contract more than the metal 2. As a result, when the bimetallic strip is heated up, it bends around the material with the smallest coefficient of linear thermal expansion (metal 2) as it remains shorter (metal 1 expands more than metal 2). On the other hand, when the bimetallic strip is cooled down, the metal 1 contracts more than the metal 2 because it has a greater coefficient of thermal expansion and contraction as well. As a result, the system bends around the metal 1 as shown in the figure.

This property of different amount of thermal expansion or contraction two metals experience when they are exposed to the same change in temperature, is used to construct thermostats, which are equipment that keep heaters at constant temperature by controlling the current flow in an electric circuit through bimetallic strips. Thus, if the heater is at the desired temperature, the bimetallic strip is at straight position, as shown in the figure.

When the heater is delivering more heat than needed, the bimetallic strip bends around the metal with the smallest coefficient of linear expansion. As a result, the contact is open, so the current flow in the circuit stops as shown in the figure below. This state lasts until the bimetallic strip cools down to the desired temperature and therefore it straightens again. Then the current starts flowing again through the circuit as the contact closes, and so on.

This is basically the operating principle of a thermostat.

Example 4

Two bimetallic strips made of different materials at room temperature (200

are shown in the figure below.

The coefficients for linear thermal expansion of the four materials are:

aluminum => 24 × 10^-6 K^-1
iron => 12 × 10^-6 K^-1
bronze => 18 K^-1
steel => 11 K^-1

What is the shape of the two bimetallic strips when they are heated up at a few hundred Celsius degrees?

Solution 4

When a bimetallic strip is heated, it bends around the material with the least coefficient of thermal expansion, as it expands less than the material with the smallest coefficient. Therefore, the first bimetallic strip will bend around the iron strip when heated, as the coefficient of thermal expansion of iron is smaller than that of aluminum, while the second bimetallic strip will bend around steel when heated, as steel has a smaller coefficient of thermal expansion than bronze. However, the bending in the second strip will not be of the same degree as in the first strip; the first bimetallic strip will bend more than the second one, as the difference in the coefficients of linear thermal expansion in the first bimetallic strip is much greater than in the second.

The two bimetallic strips will look as in the figure below when heated.

Types of Thermometers

All thermometers except digital (electronic) thermometers use the property of thermal expansion (contraction) to measure the temperature. Below, we will briefly explain the construction and the operating principle of all types of thermometers available.

a) Clinical thermometer.

It is used to measure the temperature of human body. A clinical thermometer uses the expanding property of mercury, whose level rises or drops in a narrow column according the body temperature. Therefore, mercury here acts as a capillary liquid.

The range of most clinical thermometers varies from 35°C to 42°C as the body temperature of a healthy person is typically about 37°C.

b) Household (home) thermometer.

Since the range of air temperatures are wider than those of the human body this thermometer uses (coloured) alcohol instead of mercury as capillary liquid. (Colour is used to make the reading more visible as alcohol itself is colourless). Alcohol thermometers are used instead of mercury thermometers in very cold regions because alcohol has a lower freezing point than mercury. If mercury freezes, it won't move in the tube, so a liquid that has a freezing point that is lower than the temperature it is measuring must be used for this purpose.

The range of most room thermometers varies from -30°C to + 50°C.

c) Lab thermometer.

This kind of thermometer must have a much wider range than room thermometers as it is used in various thermal and chemical processes, which involve many substances that are not commonly used in daily life.

Below, a lab thermometer that uses colored alcohol as capillary liquid is shown.

Lab thermometers contain a variety of capillary liquids and ranges but the most typical capillary liquid is alcohol and their range usually varies between -50°C to 250°C.

d) Metallic thermometer.

This kind of thermometer is used to measure the temperature of melting metals, so its range must be very wide. We cannot use glass as cover or mercury and alcohol as capillary liquids because they cannot resist to such high temperatures. (For example, iron melts at about 1600°C). Therefore, a completely different technique must be used for this purpose.

Most metallic thermometers constructors use the properties of bimetallic strip to measure the temperatures of very hot objects. A round bimetallic strip with a pointer at its end is placed on a metal end, which is placed in contact with the hot object. Since heat travels through metal, it reaches the bimetallic strip making it bend more than when it is cold. Then, a temperature scaling is made accordingly, producing a new type of thermometer, which uses a completely different technique compared to the other types of thermometers explained earlier.

A simple scheme of a metallic thermometer is shown below.

The range of metal thermometers typically varies from -200°C to +1600°C.

e) Digital thermometer.

This kind of thermometer uses the digital technology to convert the heat energy into electric energy. There are two main types of digital thermometers: contact thermometers, which are operated in a similar way to clinical ones, and infra-red (contactless) thermometers which are operated remotely, in a similar way to remote controls.

Digital thermometers use the fact that hotter the object, more infrared radiation it emits. This radiation is converted by the thermometer's software into digital values accordingly.

Example 5

A technician wants to measure the temperature of a liquid substance that is hotter than the temperature of boiling water (100°C). Which thermometer is more suitable for this purpose?

Solution 5

Thermometers such as home, clinical and digital thermometers are excluded as options because they have narrow ranges of temperature measurements. Therefore, only metallic and lab thermometers do remain as possible options. However, metallic thermometers are not so finely graded since they include a wide range of temperatures. Therefore, lab thermometers are more suitable for this purpose.

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