13.2 Thermal Expansion of Solids and Liquids

Temperature changes cause materials to expand or contract, affecting their dimensions and properties. This phenomenon, known as thermal expansion, plays a major role in engineering and everyday life, from bridge design to the way thermometers work.

At the atomic level, higher temperature means greater kinetic energy for particles. They vibrate more vigorously, which increases the average distance between them. That increased spacing adds up across billions of atoms, producing a measurable change in size. Different materials expand at different rates, which is why material choice matters so much in design.

Thermal Expansion

Thermal Expansion Effects

When you heat a solid or liquid, it expands. When you cool it, it contracts. This applies to all three spatial dimensions:

• Linear expansion affects length (think of a metal rod getting slightly longer).
• Area expansion affects surface area (a metal sheet grows in both length and width).
• Volume expansion affects the total volume (water in a heated tank takes up more space).

Because mass stays constant while volume changes, thermal expansion also changes an object's density. A heated object becomes slightly less dense.

Not all materials expand equally. Each material has its own coefficient of thermal expansion, which quantifies how much it expands per degree of temperature change. Aluminum, for instance, expands roughly twice as much as glass for the same temperature increase.

Linear Expansion Calculations

For a solid object changing length along one dimension, use:

$\Delta L = \alpha L_0 \Delta T$

where:

• $\Delta L$ = change in length
• $\alpha$ = coefficient of linear expansion (units: $\text{K}^{-1}$ or $^\circ\text{C}^{-1}$)
• $L_0$ = original length
• $\Delta T$ = change in temperature

The coefficient $\alpha$ is a material property that tells you how much one unit of length expands per degree of temperature change. For steel, $\alpha \approx 1.2 \times 10^{-5} \text{ K}^{-1}$. For aluminum, $\alpha \approx 2.5 \times 10^{-5} \text{ K}^{-1}$.

To find the final length after heating or cooling:

$L_f = L_0 + \Delta L$

Example: A 1.000 m steel rod is heated by 10 K.

1. Find $\Delta L$: $\Delta L = (1.2 \times 10^{-5} \text{ K}^{-1})(1.000 \text{ m})(10 \text{ K}) = 1.2 \times 10^{-4} \text{ m}$
2. Find $L_f$: $L_f = 1.000 \text{ m} + 0.00012 \text{ m} = 1.00012 \text{ m}$

The change is tiny for a single rod, but across a 500 m bridge span, these small expansions add up to centimeters.

Volume Expansion Determination

For three-dimensional expansion (especially useful for liquids), use:

$\Delta V = \beta V_0 \Delta T$

where:

• $\Delta V$ = change in volume
• $\beta$ = coefficient of volume expansion
• $V_0$ = original volume
• $\Delta T$ = change in temperature

For most solids, the volume coefficient is approximately three times the linear coefficient:

$\beta \approx 3\alpha$

This relationship comes from the fact that volume depends on three linear dimensions. If each dimension expands by a factor involving $\alpha$, the combined volume change scales as roughly $3\alpha$ (the higher-order terms are negligibly small).

Example: For steel with $\alpha \approx 1.2 \times 10^{-5} \text{ K}^{-1}$:

$\beta \approx 3(1.2 \times 10^{-5}) = 3.6 \times 10^{-5} \text{ K}^{-1}$

A 1.000 $\text{m}^3$ steel block heated by 10 K expands by:

$\Delta V = (3.6 \times 10^{-5} \text{ K}^{-1})(1.000 \text{ m}^3)(10 \text{ K}) = 3.6 \times 10^{-4} \text{ m}^3$

For liquids, $\beta$ is typically much larger than for solids. This is why liquid-in-glass thermometers work: the liquid expands much more than the glass tube, so the liquid level rises visibly.

Thermal Stress Analysis

Thermal stress arises when an object is constrained so it can't freely expand or contract. If a steel rail is bolted rigidly at both ends and the temperature rises, the rail tries to expand but can't. That blocked expansion creates internal stress, which can cause buckling, deformation, or fracture.

The thermal stress in a constrained object is:

$\sigma = E \alpha \Delta T$

where:

• $\sigma$ = thermal stress (in Pascals)
• $E$ = elastic modulus (Young's modulus, a measure of the material's stiffness)
• $\alpha$ = coefficient of linear expansion
• $\Delta T$ = change in temperature

This equation assumes the object is completely prevented from expanding. In practice, partial constraint produces proportionally less stress.

Engineers reduce thermal stress in two main ways:

1. Expansion joints or gaps allow materials room to expand. Bridges have visible gaps between sections for exactly this reason.
2. Matching materials with similar $\alpha$ values prevents differential expansion. A bimetallic strip deliberately uses mismatched coefficients so the strip bends with temperature change, which is useful in thermostats.

A common everyday example: pouring boiling water into a thick glass can shatter it because the inner surface expands rapidly while the outer surface stays cool. The resulting thermal stress exceeds the glass's strength. Thin glass or tempered glass handles this better.

Thermal Equilibrium and Energy Transfer

When two objects at different temperatures are in contact, energy transfers from the hotter object to the cooler one. This continues until both reach the same temperature, a state called thermal equilibrium. During this process, the hotter object contracts slightly as it cools, and the cooler object expands slightly as it warms. Both objects are adjusting their dimensions as they settle toward a shared final temperature.