# Coefficient of thermal expansion

Material Properties
Specific heat $c=\frac{T}{N}\left(\frac{\partial S}{\partial T}\right)$
Compressibility $\beta=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)$
Thermal expansion $\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)$

When the temperature of a substance changes, the energy that is stored in the intermolecular bonds between atoms changes. When the stored energy increases, so does the length of the molecular bonds. As a result, solids typically expand in response to heating and contract on cooling; this dimensional response to temperature change is expressed by its coefficient of thermal expansion (CTE).

Different coefficients of thermal expansion can be defined for a substance depending on whether the expansion is measured by:

• linear thermal expansion (CLTE)
• area thermal expansion
• volumetric thermal expansion

These characteristics are closely related. The volumetric thermal expansion coefficient can be defined for both liquids and solids. The linear thermal expansion can only be defined for solids, and is common in engineering applications.

Some substances expand when cooled, such as freezing water, so they have negative thermal expansion coefficients.

## Thermal expansion coefficient

The thermal expansion coefficient is a thermodynamic property of a substance.

It relates the change in temperature to the change in a material's linear dimensions. It is the fractional change in length per degree of temperature change.

$\alpha={1\over L_0}{\partial L \over \partial T}$

dL = L0 x ( alpha x dT )

where $L_0\$ is the original length, $L\$ the new length, and $T\$ the temperature.

## Linear thermal expansion

${\Delta L \over L_0} = \alpha_L \Delta T$

or equivalently:

L1 = L0(1 + αL(T2T1))

The linear thermal expansion is the one-dimensional length change with temperature. This equation works reasonably well as long as αL < < 1 and ΔT are relatively small.

Consider a rod of length L = 1m with α = 0.1, if the change in temperature ΔT = 2 (say from 0 to 2 degree C) ΔL will be 0.1 x 1 x 2 = 0.2

On the other hand if we consider the same temperature change in two steps that is:
1) ΔT = 1 (from 0 to 1 degree C), with L0 = 1m at 0 degree C

       ΔL will be 0.1 x 1 x 1 = 0.1 m


and
2) ΔT = 1 (from 1 to 2 degree C), with L0 = 1+0.1 = 1.1 m at 1 degree C

       ΔL will be 0.1 x 1.1 x 1 = 0.11m


adding the two we get a total change in length of 0.21 m

and hence a contradiction. This occurred because our example αL was very large, compared to typical values of the order 10 - 5. If αL is much smaller, this discrepancy will be very small.

### Proof that the discrepancy will be small

Consider a change of temperature from T0 to T1 to T2, resulting in a change in length from L0 to L1 to L2:

L1 = L0(1 + αL(T1T0))
L2 = L1(1 + αL(T2T1))

Combining these equations

\begin{alignat}{2} L_2 & = L_0(1 + \alpha_L (T_1 - T_0))(1 + \alpha_L (T_2 - T_1))\\ & = L_0(1 + \alpha_L (T_2-T_0)) + \alpha_L^2 L_0 (T_1 - T_0) (T_2 - T_1)\\ \end{alignat}

The second term becomes negligible in the case αL < < 1 and that the temperature changes aren't too large, resulting in:

L2 = L0(1 + αL(T2T0))

As expected.

## Area thermal expansion

The change in area with temperature can be written:

${\Delta A \over A_0} = \alpha_A \Delta T$

For exactly isotropic materials, the area thermal expansion coefficient is very closely approximated as twice the linear coefficient.

$\alpha_A\cong 2\alpha_L$
${\Delta A \over A_0} = 2 \alpha_L\Delta T$

## Volumetric thermal expansion

The change in volume with temperature can be written[1]:

${\Delta V \over V_0} = \alpha_V \Delta T$

The volumetric thermal expansion coefficient can be written

$\alpha_V =\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p=-{1\over\rho} \left(\frac{\partial \rho}{\partial T}\right)_{p}$

where $T\$ is the temperature, $V\$ is the volume, $\rho\$ is the density, derivatives are taken at constant pressure $p\$; $\beta\$ measures the fractional change in density as temperature increases at constant pressure.

For exactly isotropic materials, the volumetric thermal expansion coefficient is very closely approximated as three times the linear coefficient.

$\alpha_V\cong 3\alpha_L$
${\Delta V \over V_0} = 3 \alpha \Delta T$

Proof:

$\alpha_V = \frac{1}{V} \frac{\partial V}{\partial T} = \frac{1}{L^3} \frac{\partial L^3}{\partial T} = \frac{1}{L^3}\left(\frac{\partial L^3}{\partial L} \cdot \frac{\partial L}{\partial T}\right) = \frac{1}{L^3}\left(3L^2 \frac{\partial L}{\partial T}\right) = 3 \cdot \frac{1}{L}\frac{\partial L}{\partial T} = 3\alpha_L$

This ratio arises because volume is composed of three mutually orthogonal directions. Thus, in an isotropic material, one-third of the volumetric expansion is in a single axis (a very close approximation for small differential changes). Note that the partial derivative of volume with respect to length as shown in the above equation is exact, however, in practice it is important to note that the differential change in volume is only valid for small changes in volume (i.e., the expression is not linear). As the change in temperature increases, and as the value for the linear coefficient of thermal expansion increases, the error in this formula also increases. For non-negligible changes in volume:

$({L + }{\Delta L})^3 = {L^3 + 3L^2}{\Delta L} + {3L}{\Delta L}^2 + {\Delta L}^3 \,$

Note that this equation contains the main term, $3L^2\$, but also shows a secondary term that scales as $3L{\Delta L}^2 = {3L^3}{\alpha}^2{\Delta T}^2\,$, which shows that a large change in temperature can overshadow a small value for the linear coefficient of thermal expansion. Although the coefficient of linear thermal expansion can be quite small, when combined with a large change in temperature the differential change in length can become large enough that this factor needs to be considered. The last term, ${\Delta L}^3\$ is vanishingly small, and is almost universally ignored.

## Anisotropy

In anisotropic materials the total volumetric expansion is distributed unequally among the three axes and if the symmetry is monoclinic or triclinic even the angles between these axes are subject to thermal changes. In such cases it is necessary to treat thermal expansion as a tensor that has up to six independent elements. A good way to determine the elements of the tensor is to study the expansion by powder diffraction.

## Thermal expansion coefficients for some common materials

The expansion and contraction of material must be considered when designing large structures when using tape or chain to measure distances for land surveys, when designing molds for casting hot material, and in other engineering applications when large changes in dimension due to temperature are expected. The range for α is from 10-7 for hard solids to 10-3 for organic liquids. α varies with the temperature and some materials have a very high variation. Some values for common materials, given in parts per million per Celsius degree: (NOTE: This can also be in kelvins as the changes in temperature are a 1:1 ratio)

NOTE: Theoretically, the coefficient of linear expansion can be found from the coefficient of volumetric expansion (β=3α). However, for liquids, α is calculated through the experimental determination of β, hence it is more accurate to state β here than α. (The formula β=3α is usually used for solids)[2]

coefficient of linear thermal expansion α coefficient of volumetric thermal expansion β
material α in 10-6/K at 20 °C β(=3α) in 10-6/K at 20 °C
Gasoline ~317 950[2]
Ethanol ~250 750[3]
Water ~69 207[4]
Mercury ~61 182[4]
Rubber 77 231
PVC 52 156
Benzocyclobutene 42 126
Magnesium 26 78
Aluminium 23 69
Brass 19 57
Silver 18[5] 54
Stainless steel 17.3 51.9
Copper 17 51
Gold 14 42
Nickel 13 39
Concrete 12 36
Steel, depends on composition 11.0 ~ 13.0 33.0 ~ 39.0
Iron 11.1 33.3
Carbon steel 10.8 32.4
Platinum 9 27
Glass 8.5 25.5
Gallium(III) arsenide 5.8 17.4
Indium phosphide 4.6 13.8
Tungsten 4.5 13.5
Glass, borosilicate 3.3 9.9
Silicon 3 9
Invar 1.2 3.6
Diamond 1 3
Quartz (fused) 0.59 1.77
Oak (perpendicular to the grain) 54 [6] 162
Pine (perpendicular to the grain) 34 102

## Applications

For applications using the thermal expansion property, see bi-metal and mercury thermometer.

Thermal expansion is also used in mechanical applications to fit parts over one another, e.g. a bushing can be fitted over a shaft by making its inner diameter slightly smaller than the diameter of the shaft, then heating it until it fits over the shaft, and allowing it to cool after it has been pushed over the shaft, thus achieving a 'shrink fit'. Induction shrink fitting is a common industrial method to pre-heat metal components between 150˚C and 300˚C thereby causing them to expand and allow for the insertion or removal of another component.

There exist some alloys with a very small CTE, used in applications that demand very small changes in physical dimension over a range of temperatures. One of these is Invar 36, with a coefficient in the 0.6x10-6 range. These alloys are useful in aerospace applications where wide temperature swings may occur.

## References

1. ^ Turcotte, Donald L.; Schubert, Gerald (2002). Geodynamics (2nd Edition ed.). Cambridge. ISBN 0-521-66624-4.
2. ^ a b Thermal Expansion
3. ^ Textbook: Young and Geller College Physics, 8e
4. ^ a b Properties of Common Liquid Materials
5. ^ Thermal Expansion Coefficients
6. ^ WDSC 340. Class Notes on Thermal Properties of Wood