# How to calculate thermal expansion

Thermal expansion is a physical property of a substance (gas, liquid or solid) to modify its shape (length, area or volume) function of temperature. Thermal expansion relates with the expansion and contraction of particles in a substance function of temperature.

Thermal expansion can also be regarded as a fractional change in size of a material/substance caused by a change of temperature.

Image: Particles expansion and contraction

Thermal expansion has effect on gases, liquids and solids. From the mathematical point of view, thermal expansion can be described as:

• linear (one direction, 1-D)
• areal (two directions, 2-D)
• volumetric (three directions, 3-D)

The linear and areal (also called superficial) thermal expansion applies only to solids. The volumetric (also called cubical) thermal expansion applies to both solids and liquids. For gases, thermal expansion is described by the ideal gas law and it is treated differently.

### Linear thermal expansion

Image: Linear thermal expansion

Linear thermal expansion applies mostly to solids. Knowing the initial length L0 [m] of a given solid (e.g. metal rod), the temperature difference ΔT [ºC] and the coefficient of linear expansion of the solid α [1/ºC], the change in length ΔT [m] of the solid can be calculated as:

$\Delta L = \alpha \cdot L_0 \cdot \Delta T \tag{1}$

The change in length is directly proportional with the change in temperature. The higher the temperature difference the higher the increase in length of the material (e.g. metal rod).

The length difference ΔL is equal with the subtraction of the initial length L0 from the final length L:

$\Delta L = L – L_0 \tag{2}$

By replacing (2) in (1), we can calculate the final length (after thermal expansion) function of the initial length, temperature difference and linear thermal expansion coefficient.

$\bbox[#FFFF9D]{L = L_0 \cdot (1+ \alpha \cdot \Delta T)} \tag{3}$

The coefficient of linear thermal expansion is not constant but varies slightly with temperature. Therefore, the mathematical expression can be applied only to small temperature variations.

### Areal thermal expansion

Image: Areal thermal expansion

Thermal expansion also applies to surfaces. Imagine a sheet of metal with a defined area. If heated, the same sheet of metal will have a slightly bigger area.

Knowing the initial area A0 [m2] of a given solid (e.g. metal sheet), the temperature difference ΔT [ºC] and the coefficient of linear expansion of the solid α [1/ºC], the change in area ΔA [m2] of the solid can be calculated as:

$\Delta A = 2 \cdot \alpha \cdot A_0 \cdot \Delta T \tag{4}$

The change in area is directly proportional with the change in temperature. The higher the temperature difference the higher the increase in surface of the material (e.g. metal sheet).

The area difference ΔA is equal with the subtraction of the initial area A0 from the final area A:

$\Delta A = A – A_0 \tag{5}$

By replacing (5) in (4), we can calculate the final area (after thermal expansion) function of the initial area, temperature difference and linear thermal expansion coefficient.

$\bbox[#FFFF9D]{A = A_0 \cdot (1+ 2 \cdot \alpha \cdot \Delta T)} \tag{6}$

To demonstrate the mathematical expression (6), let’s assume that the area is the square of the length:

$A = L^2 \tag{7}$

Replacing (3) in (7), gives:

$A = L_{0}^2 \cdot \left ( 1 + 2 \cdot \alpha \cdot \Delta T + \alpha^2 \cdot \Delta T^2 \right ) \tag{8}$

Since the coefficient of thermal expansion is very small (e.g. for steel 12·10-6 1/ºC), the quadratic term of the equation (8) can be neglected. Assuming that the initial area is equal with the square of the initial length:

$A_0 = L_{0}^2 \tag{9}$

equation (8) becomes (6).

The same principle applies to areal thermal expansions. The coefficient of linear thermal expansion is not constant but varies slightly with temperature. Therefore, the mathematical expression can be applied only to small temperature variations.

### Volumetric thermal expansion

Image: Thermal expansion (volumetric)

Thermal expansion causes variations in volume for solids and liquids function of temperature.

Knowing the initial volume V0 [m3] of a given solid, the temperature difference ΔT [ºC] and the coefficient of linear expansion of the solid α [1/ºC], the change in volume ΔV [m3] of the solid can be calculated as:

$\Delta V = 3 \cdot \alpha \cdot V_0 \cdot \Delta T \tag{10}$

The change in volume is directly proportional with the change in temperature. The higher the temperature difference the higher the increase in volume of the material.

The volume difference ΔV is equal with the subtraction of the initial volume V0 from the final volume V:

$\Delta V = V – V_0 \tag{11}$

By replacing (11) in (10), we can calculate the final volume (after thermal expansion) function of the initial volume, temperature difference and linear thermal expansion coefficient.

$\bbox[#FFFF9D]{V = V_0 \cdot (1+ 3 \cdot \alpha \cdot \Delta T)} \tag{12}$

To demonstrate the mathematical expression (12), let’s assume that the volume is the cube of the length:

$V = L^3 \tag{12}$

Replacing (3) in (12), gives:

$V = L_{0}^3 \cdot \left ( 1 + 3 \cdot \alpha \cdot \Delta T + 3 \cdot \alpha^2 \cdot \Delta T^2 + \alpha^3 \cdot \Delta T^3 \right ) \tag{14}$

Since the coefficient of thermal expansion is very small, the cubic and quadratic terms of the equation (14) can be neglected. Assuming that the initial volume is equal with the cube of the initial lenght:

$V_0 = L_{0}^3 \tag{15}$

equation (14) becomes (12).

For volumetric thermal expansion calculations we can use the coefficient of volumetric thermal expansion β instead of the coefficient of linear thermal expansion α.

$\beta \approx 3 \cdot \alpha \tag{16}$

which gives the equation for the change in volume:

$\bbox[#FFFF9D]{\Delta V = \beta \cdot V_0 \cdot \Delta T} \tag{17}$

The same principle applies to volumetric thermal expansions. The coefficient of volumetric thermal expansion is not constant but varies slightly with temperature. Therefore, the mathematical expression can be applied only to small temperature variations.

The coefficient of thermal expansion are obtained from experimental data. In the table below you can find the values of the thermal expansion coefficient for common substances.

 Material Coefficient of linear expansion Coefficient of volume expansion Solids Aluminum 25·10-6 75·10-6 Brass 19·10-6 56·10-6 Copper 17·10-6 51·10-6 Gold 14·10-6 42·10-6 Iron 12·10-6 35·10-6 Invar 0.9·10-6 2.7·10-6 Lead 29·10-6 87·10-6 Silver 18·10-6 54·10-6 Glass 9·10-6 27·10-6 Glass 3·10-6 9·10-6 Quartz 0.4·10-6 1·10-6 Concrete 12·10-6 36·10-6 Marble 7·10-6 21·10-6 Liquids Ether 1650·10-6 Ethyl 1100·10-6 Petrol 950·10-6 Glycerin 500·10-6 Mercury 180·10-6 Water 210·10-6 Gases Air and most other gases at atmospheric pressure 3400·10-6

Source:
College Physics, openstax, Rice University
Wikipedia

### Thermal expansion examples

Example 1 (linear thermal expansion). Assuming a bridge, made of steel, with an initial total length of 1500 m at -20 ºC, calculate the length difference at 40 ºC and its total length.

Image: Suspension bridge

Step 1. Write down the known parameters of the problem:

$\begin{split} L_0 &= 1500 \text{ m}\\ T_0 &= -20 \text{ }^\circ \text{C}\\ T &= 40 \text{ }^\circ \text{C}\\ \alpha &= 12 \cdot 10^{-6} \text{ 1/}^\circ \text{C} \end{split}$

Step 2. Calculate the temperature difference

$\Delta T = T – T_0 = 40 – (-20) = 40+20 = 60 \text{ }^\circ \text{C}$

Step 3. Calculate the length difference

$\Delta L = \alpha \cdot L_0 \cdot \Delta T = 12 \cdot 10^{-6} \cdot 1500 \cdot 60 = 1.08 \text{ m}$

Step 4. Calculate the total final length

$L = L_0 + \Delta L = 1500 + 1.08 = 1501.08 \text{ m}$

The change in length is very small compared with the initial length of the bridge. However it’s noticeable and can cause structural problems if not taken into account in the design phase. Because of the thermal expansion, metal bridges are build up from several sections which have air gaps between them, in order to allow thermal expansion function of temperature change.

Thermal expansion has also a big impact on railway tracks. A 10 km railway track is not made up from a single piece of steel but divided into several pieces with air gaps (expansion spaces) between them. In winter the air gaps are bigger becasue the rails have smaller length and in summer the air gaps are bearly noticeable becasue the rails have increase length due to thermal expansion.

Example 2 (area thermal expansion). Assuming a football pitch is made of aluminium and has an initial total area of 7140 m2 at -10 ºC, calculate the area difference at 30 ºC and its total area.

Image: Football pitch (field)

Step 1. Write down the known parameters of the problem:

$\begin{split} A_0 &= 7140 \text{ m}^2\\ T_0 &= -10 \text{ }^\circ \text{C}\\ T &= 30 \text{ }^\circ \text{C}\\ \alpha &= 25 \cdot 10^{-6} \text{ 1/}^\circ \text{C} \end{split}$

Step 2. Calculate the temperature difference

$\Delta T = T – T_0 = 30 – (-10) = 30 + 10 = 40 \text{ }^\circ \text{C}$

Step 3. Calculate the area difference

$\Delta A = 2 \cdot \alpha \cdot A_0 \cdot \Delta T = 2 \cdot 25 \cdot 10^{-6} \cdot 7140 \cdot 40 = 14.28 \text{ m}^2$

Step 4. Calculate the total final area

$A = A_0 + \Delta A = 7140 + 14.28 = 7154.28 \text{ m}^2$

Example 3 (volumetric thermal expansion). For this example we are going to assume that we have a 50 L steel tank full with gasoline at -20 ºC. What is going to be the volume difference for both tank and fuel at 40 ºC? Is the fuel going to fit in the fuel tank?

Image: Fuel tank

Step 1. Write down the known parameters of the problem:

$\begin{split} V_{0t} &= 50 \text{ L}\\ V_{0f} &= 50 \text{ L}\\ T_0 &= -20 \text{ }^\circ \text{C}\\ T &= 40 \text{ }^\circ \text{C}\\ \alpha &= 12 \cdot 10^{-6} \text{ 1/}^\circ \text{C}\\ \beta &= 950 \cdot 10^{-6} \text{ 1/}^\circ \text{C}\\ \end{split}$

Step 2. Calculate the temperature difference

$\Delta T = T – T_0 = 40 – (-20) = 40 + 20 = 60 \text{ }^\circ \text{C}$

Step 3. Calculate the volume difference of the fuel tank (assuming it’s solid steel)

$\Delta V_t = 3 \cdot \alpha \cdot V_{0t} \cdot \Delta T = 3 \cdot 12 \cdot 10^{-6} \cdot 50 \cdot 60 = 0.108 \text{ L}$

Step 4. Calculate the volume difference of the fuel (gasoline)

$\Delta V_f = \beta \cdot V_{0f} \cdot \Delta T = 950 \cdot 10^{-6} \cdot 50 \cdot 60 = 2.85 \text{ L}$

Step 5. Calculate the excess volume of fuel

$V_{ex} = \Delta V_f – \Delta V_t = 2.85 – 0.108 = 2.742 \text{ L}$

We can see that there is more fuel than the full capacity of the tank, which means that the excess fuel will spill.

You can also check your results using the calculator below.

### Thermal Expansion Calculator

 L0 [m] α [1/°C] T0 [°C] T [°C] Linear expasion L [m] = ΔL [m] = A0 [m2] α [1/°C] T0 [°C] T [°C] Areal expasion A [m2] = ΔA [m2] = V0 [m3] α [1/°C] T0 [°C] T [°C] Volumetric solid expasion V [m3] = ΔV [m3] = V0 [m3] β [1/°C] T0 [°C] T [°C] Volumetric gas expasion V [m3] = ΔV [m3] =

### Bimetallic Strips

A bimetallic strip is made up from two metals, bond together, with different thermal expansion coefficient.

Image: Bimetallic strip

The two strips of metal are bond together at a reference temperature (e.g. 20 °C), having equal lengths. When the temperature change, because they have different thermal expansion coefficient, the length change (ΔL) of each strip will be different. Being bond together, the strip will bend function of temperature change.

Bimetallic are used as switches in electric circuits to open/close electric contacts function of external temperature or current through the circuit.

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