Intermediate · 12 min
Thermal expansion of solids
Calculate small length changes caused by a temperature change.
Length can change with temperature
Many solids expand when warmed and contract when cooled. For a small temperature interval where the material's coefficient can be treated as constant, linear expansion is ΔL = α × L₀ × ΔT. Here L₀ is original length and α is the coefficient of linear expansion, usually stated per kelvin or per degree Celsius. A temperature change of 1 K has the same size as a change of 1 °C.
Worked example
For α = 10 × 10⁻⁶ per °C, L₀ = 2 m, and ΔT = 20 °C: ΔL = 10 × 10⁻⁶ × 2 × 20 = 0.0004 m = 0.4 mm
Separate the change from the final length
The expansion formula gives a length change, not the final length. Add it to the original length for warming with positive α. A negative temperature change gives contraction. The model assumes the object can expand freely and its temperature is reasonably uniform. If expansion is constrained, internal stresses can develop instead. Expansion joints let structures accommodate length changes.
Worked example
A 1 m rod expands by 0.0003 m. Final length = 1 + 0.0003 = 1.0003 m The increase is 0.3 mm, not 0.3 m.
Plan for a length change
- Calculate the expansion of a freely expanding 5 m model rail with α = 10 × 10⁻⁶ per °C during a 40 °C increase. Express the change in millimetres.
- Explain why a joint that allows movement can matter even when the percentage length change is small.