# Thermal expansion of liquids and solids

References: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.7 – 1.8

The volume thermal expansion coefficient of a substance as its temperature is increased at constant pressure is defined as the fractional change in volume per degree kelvin, that is

$\displaystyle \beta\equiv\frac{\Delta V/V}{\Delta T} \ \ \ \ \ (1)$

Example 1 For mercury, ${\beta=1.80\times10^{-4}\mbox{ K}^{-1}}$ so if we have a typical mercury thermometer with a cylindrical bulb ${h=1\mbox{ cm}}$ long and with a radius of ${r=0.2\mbox{ cm}}$, and the scale on the thermometer is 1 mm per degree, then we can work out the inside diameter ${2\rho}$ of the tube. We get

 $\displaystyle V$ $\displaystyle =$ $\displaystyle \pi r^{2}h=1.26\times10^{-7}\mbox{ m}^{3}\ \ \ \ \ (2)$ $\displaystyle \Delta V$ $\displaystyle =$ $\displaystyle \beta V\Delta T\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1.80\times10^{-4}\right)\left(1.26\times10^{-7}\right)\left(1\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.27\times10^{-11}\mbox{ m}^{3}\ \ \ \ \ (5)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\Delta V}{\pi\left(10^{-3}\mbox{ m}\right)}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8.5\times10^{-5}\mbox{ m} \ \ \ \ \ (7)$

The diameter is therefore about 0.2 mm.

Example 2 For water, ${\beta}$ varies a lot in the liquid region. At ${100^{\circ}\mbox{C}}$, it is ${7.5\times10^{-4}\mbox{ K}^{-1}}$ and decreases to zero at ${4^{\circ}\mbox{ C}}$. Between the freezing point at ${0^{\circ}\mbox{C}}$ and ${4^{\circ}\mbox{ C}}$, ${\beta}$ is actually negative, with its largest negative value of ${\beta=-0.68\times10^{-4}\mbox{ K}^{-1}}$ at the freezing point. That is, melting ice actually contracts (becomes denser) as its temperature increases to ${4^{\circ}\mbox{ C}}$ which is the reason that ice floats. If ${\beta}$ were positive over the entire liquid range of water, a cooling lake would start to freeze from the bottom up rather than from the top down as it does in nature.

Incidentally, you might think that because ${\beta>0}$ for temperatures between ${4^{\circ}\mbox{ C}}$ and ${100^{\circ}\mbox{ C}}$, ice might sink in hot water (before it melts, of course). However, at standard pressure, the density of boiling water is ${0.9584\mbox{ g cm}^{-3}}$ while the density of ice at ${0^{\circ}\mbox{ C}}$ is ${0.9167\mbox{ g cm}^{-3}}$ so ice floats even in boiling water.

For solids, we can define a linear thermal expansion coefficient as the fractional change of length per degree of increase in temperature:

$\displaystyle \alpha\equiv\frac{\Delta L/L}{\Delta T} \ \ \ \ \ (8)$

Example 3 For steel, ${\alpha=1.1\times10^{-5}\mbox{ K}^{-1}}$. Assuming this value is constant over the range of outdoor air temperatures, we can estimate the change in length of a 1 km steel bridge between winter and summer. In Dundee, the temperature doesn’t vary as much as in more continental locations, but we’ll take a cold day in Dundee to be ${0^{\circ}\mbox{ C}}$ and a hot day to be ${25^{\circ}\mbox{ C}}$, so ${\Delta T=25}$. The change in length is therefore

$\displaystyle \Delta L=\left(1.1\times10^{-5}\right)\left(25\right)\left(10^{3}\right)=0.275\mbox{ m} \ \ \ \ \ (9)$

Thus the change in length is far from negligible, which is the reason why long bridges are built in sections with expansion joints in between.

Example 4 One type of thermometer consists of a spiral consisting of two different metal strips (with different values of ${\alpha}$) bonded together. Since the metals expand at different rates, the spiral winds and unwinds as the temperature changes. A dial can be attached to the end of the spiral to measure its position and thus give a measure of temperature.

Example 5 If a solid is not isotropic, it has different values of ${\alpha}$ in each direction, so in rectangular coordinates we have ${\alpha_{x}}$, ${\alpha_{y}}$ and ${\alpha_{z}}$ defined as ${\Delta x/\left(x\Delta T\right)}$ and so on. For a rectangular solid we have

 $\displaystyle V$ $\displaystyle =$ $\displaystyle xyz\ \ \ \ \ (10)$ $\displaystyle \Delta V$ $\displaystyle =$ $\displaystyle \left(x+\Delta x\right)\left(y+\Delta y\right)\left(z+\Delta z\right)-xyz\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle yz\Delta x+xz\Delta y+xy\Delta z+\mathcal{O}\left(\Delta x^{2}\right)\ \ \ \ \ (12)$ $\displaystyle \frac{\Delta V}{V}$ $\displaystyle =$ $\displaystyle \frac{\Delta x}{x}+\frac{\Delta y}{y}+\frac{\Delta z}{z}+\mathcal{O}\left(\Delta x^{2}\right)\ \ \ \ \ (13)$ $\displaystyle \frac{\Delta V}{V\Delta T}$ $\displaystyle =$ $\displaystyle \alpha_{x}+\alpha_{y}+\alpha_{z}+\mathcal{O}\left(\Delta x^{2}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \beta \ \ \ \ \ (15)$

Thus to first order in changes in length

$\displaystyle \beta=\alpha_{x}+\alpha_{y}+\alpha_{z} \ \ \ \ \ (16)$

## 4 thoughts on “Thermal expansion of liquids and solids”

1. gwrowe Post author

That’s right. ${\mathcal{O}}$ stands for ‘order’ and ${\mathcal{O}\left(x^{2}\right)}$ means terms of order ${x^{2}}$ (and higher) in an expansion such as a Taylor series, where ${x}$ is small.