Thermal Properties of Matter · Lesson 2 of 8
Thermal Expansion and Thermal Stress
“Derive linear, area and volume expansion, explain water’s anomaly, and solve fitting, measurement and constrained-expansion problems.”
• Define expansivities and derive the small-change relations αA ≈ 2αl and αV ≈ 3αl. • Explain expansion of holes, temperature-dependent coefficients and the density maximum of water. • Calculate fitting temperatures, tape corrections and density changes with stated approximations. • Distinguish free thermal strain from mechanical strain and derive constrained thermal stress.
Heating can change dimensions
A warm metal lid may loosen on a bottle, the liquid level in a thermometer rises as it warms, and a partly inflated balloon grows in warm surroundings. These observations concern changes of dimensions, not just temperature readings. Thermal expansion is the change in a body’s dimensions associated with a temperature change. Most familiar substances expand on heating over ordinary ranges, but the sign and size of the change depend on the material and temperature.
A useful microscopic picture is that the average separation between neighbouring particles can change as their thermal motion changes. The particles themselves do not need to inflate like tiny balloons. For engineering calculations we summarize the measured dimensional response by an expansion coefficient. A coefficient used as constant is an approximation over a specified temperature interval, not a claim of exact temperature independence.
The coefficient αl has units K⁻¹ because fractional expansion is dimensionless. Temperature differences in kelvins and Celsius degrees have the same magnitude. Thus a 20 °C rise can be inserted as 20 K in this equation, but a final temperature of 20 °C is not a change of 293.15 K. On cooling, ΔT is negative, so the usual positive αl gives a contraction.
| Material | Representative αl / (10⁻⁵ K⁻¹) |
|---|---|
| Aluminium | 2.5 |
| Brass | 1.8 |
| Iron | 1.2 |
| Copper | 1.7 |
| Silver | 1.9 |
| Gold | 1.4 |
| Pyrex glass | 0.32 |
| Lead | About 2.9 |
Copper’s listed expansivity is about 1.7/0.32 ≈ 5.3 times that of Pyrex. Low-expansion glass is useful where dimensions should change little. The representative lead value is about 2.9 × 10⁻⁵ K⁻¹. Representative values may differ with composition and range, so a problem’s explicitly supplied coefficient takes priority over a generic table.
Area and volume: derive rather than memorize
Problem
A rectangular isotropic sheet has original sides a and b. Show that its area-expansion coefficient is twice its linear coefficient to first order.
- 1.Each side changes by the same fraction: Δa = αlaΔT and Δb = αlbΔT.
- 2.Expand the new area: (a + Δa)(b + Δb) = ab + aΔb + bΔa + ΔaΔb.
- 3.Divide the area increase by A0 = ab: ΔA/A0 = 2αlΔT + (αlΔT)².
- 4.For a small fractional length change, the squared term is much smaller than the first-order term. Hence ΔA/A0 ≈ 2αlΔT and αA ≈ 2αl.
- 5.The conclusion assumes the same coefficient in both in-plane directions. It need not hold unchanged for an anisotropic sheet.
| Substance | Representative αV / K⁻¹ |
|---|---|
| Aluminium | 7 × 10⁻⁵ |
| Brass | 6 × 10⁻⁵ |
| Iron | 3.55 × 10⁻⁵ |
| Paraffin | 58.8 × 10⁻⁵ |
| Ordinary glass | 2.5 × 10⁻⁵ |
| Pyrex glass | 1 × 10⁻⁵ |
| Hard rubber | 2.4 × 10⁻⁴ |
| Invar | 2 × 10⁻⁶ |
| Mercury | 18.2 × 10⁻⁵ |
| Water | 20.7 × 10⁻⁵, away from its anomalous region |
| Ethanol | 110 × 10⁻⁵ |
These quoted values indicate scale and material differences; they are not exact constants across all temperatures. Invar, an iron–nickel alloy, and Pyrex have especially low expansivities. Ethanol expands more strongly than mercury under a comparable rise. Independently rounded table entries need not satisfy αV = 3αl to every digit. Liquid expansion is measured relative to a vessel, whose own expansion also matters in a precise measurement. Most importantly, the positive representative water entry cannot describe the anomalous 0–4 °C interval.
At 0 °C, 1/T = 1/273.15 ≈ 3.66 × 10⁻³ K⁻¹, much larger than typical solid expansivities. Near 300 K it is about 3.33 × 10⁻³ K⁻¹. Unlike the approximately constant coefficients often used for solids over a short range, the ideal-gas coefficient decreases as absolute temperature increases. The graph of copper’s coefficient approaching a flatter range therefore cannot be generalized to every substance or every regime.
Water contracts when heated from 0 °C to about 4 °C
Cool liquid water from room temperature and it initially contracts. Near 4 °C its density is greatest. Further cooling toward 0 °C makes it expand, so its density falls. Equivalently, heating water from 0 °C to 4 °C decreases its volume. In that interval a positive volume-expansion coefficient would give the wrong sign. At the density maximum the small-change volume coefficient passes through zero.
This behaviour controls the order in which a lake cools. Surface water cooling toward 4 °C becomes denser and sinks, while less dense water rises. Once the surface cools below about 4 °C it becomes less dense and tends to stay on top, where it can freeze. Ice also floats and forms an insulating surface layer. Liquid water can consequently survive beneath it. This is a qualitative environmental consequence, not a prediction that every lake must have the same temperature at every depth.
An opening expands too: fitting rings and shafts
Imagine filling a circular hole with a matching piece of the sheet’s material. On uniform heating, the filled sheet would expand smoothly, including that imagined piece. Removing the piece again leaves a larger hole, not a smaller one. Diameters and radii therefore follow the same linear-expansion factor as other lengths in a uniformly heated free isotropic sheet.
Problem
At 27 °C an iron ring has diameter 5.231 m and a wooden wheel rim has diameter 5.243 m. Find the ring temperature needed for a just-fit diameter using αl = 1.20 × 10⁻⁵ K⁻¹.
- 1.The required fractional diameter increase is (5.243 − 5.231)/5.231.
- 2.Set this equal to αl(T − 27 °C), giving a required rise about 191.17 °C.
- 3.The final ring temperature is about 218.17 °C, or approximately 218 °C with the stated precision.
- 4.Only the ring is heated in this model. When it cools after fitting, its tendency to contract helps it grip the rim.
Problem
A steel shaft diameter is 8.70 cm and a stationary wheel hole diameter 8.69 cm, both initially at 27 °C. Separately, a copper-sheet hole is 4.24 cm across at 27 °C and is heated to 227 °C. Use αsteel = 1.20 × 10⁻⁵ K⁻¹ and αcopper = 1.70 × 10⁻⁵ K⁻¹.
- 1.For the shaft, set 8.69 = 8.70[1 + αsteel(T − 27)]. The result is T ≈ −68.8 °C, about −69 °C.
- 2.This assumes the wheel remains at its initial temperature and the shaft’s coefficient is approximately constant over the interval.
- 3.For the copper hole, Δd = d0αcopperΔT = 4.24 × 1.70 × 10⁻⁵ × 200 = 0.014416 cm.
- 4.The diameter increases by about 1.44 × 10⁻² cm. Treating the hole as contracting would reverse the physical result.
A heated measuring tape changes the size of its divisions
Problem
A steel tape is calibrated at 27.0 °C and reads 63.0 cm for a steel rod at 45.0 °C. Both have α = 1.20 × 10⁻⁵ K⁻¹. Find the rod’s actual hot length and its length at the calibration temperature.
- 1.At 45 °C every physical tape division is larger by factor 1 + α(45 − 27) = 1.000216.
- 2.The hot rod’s actual length is 63.0 × 1.000216 = 63.013608 cm. The calculated correction is 0.0136 cm.
- 3.Cooling the rod to 27 °C divides its length by the same factor, giving 63.0 cm. Equal expansivities explain why the tape reading stays the same when both are at the same changed temperature.
- 4.The extra displayed digits show the modelled correction; they do not create extra measurement precision. At the original three-significant-figure precision, the total length is still reported as 63.0 cm.
Free expansion does not automatically create thermal stress
If a rod can expand freely and uniformly, it changes its length without requiring an axial support force. If rigid supports prevent the same expansion, they supply forces that elastically compress the rod relative to the length it would freely adopt at the new temperature. The actual total length change can be zero even though the thermal tendency and mechanical strain are both nonzero. Add their contributions rather than calling the actual constrained length change αlLΔT.
Problem
A 5 m steel rail of area 40 cm² is fully constrained while heated by 10 °C. Use α = 1.2 × 10⁻⁵ K⁻¹ and Y = 2 × 10¹¹ Pa.
- 1.Its free thermal strain would be αΔT = 1.2 × 10⁻⁴. The supports impose an equal and opposite mechanical strain, so actual total strain is zero.
- 2.The stress is compressive with magnitude YαΔT = 2.4 × 10⁷ Pa.
- 3.Area 40 cm² = 4.0 × 10⁻³ m², so the reaction magnitude is 9.6 × 10⁴ N, of order 10⁵ N.
- 4.The large compressive force can promote buckling. Actual design must check stability and material limits, not just compute a shortening.
Problem
A 1.8 m brass wire of diameter 2.0 mm is fixed at both ends and cools from 27 °C to −39 °C. Use α = 2.0 × 10⁻⁵ K⁻¹ and Y = 0.91 × 10¹¹ Pa. Separately, free end-to-end brass and steel rods are each 50 cm long at 40 °C and heat to 250 °C, with αbrass = 2.0 × 10⁻⁵ and αsteel = 1.2 × 10⁻⁵ K⁻¹.
- 1.For the wire, ΔT = −66 °C. Its tendency to contract is prevented, so tensile stress is Yα(66) = 1.2012 × 10⁸ Pa.
- 2.Multiply by area π(1.0 × 10⁻³)² to obtain tension 377 N, about 3.8 × 10² N. Neglect the stated small initial tension.
- 3.For the free rods, ΔT = 210 °C. Brass extends 50 × 2.0 × 10⁻⁵ × 210 = 0.210 cm; steel extends 0.126 cm.
- 4.The combined extension is 0.336 cm, about 0.34 cm. No axial thermal stress is forced merely by joining these rods end to end when their external ends can move freely.
Mass conservation links expansion to density
Problem
Glycerine has quoted volume expansivity 49 × 10⁻⁵ K⁻¹. Estimate its fractional density change for a 30 °C rise.
- 1.The fractional volume increase is x = αVΔT = 0.0147.
- 2.To first order, Δρ/ρ0 ≈ −0.0147, a density decrease of 1.47%.
- 3.Retaining the denominator in the same volume model gives Δρ/ρ0 = 1/(1 + 0.0147) − 1 = −0.014487, a 1.45% decrease.
- 4.These are first-order and unexpanded-denominator versions of the stated approximate model, not opposite physical answers. Both predict decreasing density.
Quick Quiz
Quiz
A freely heated uniform sheet has a circular hole. What happens to the hole for ordinary positive expansion?
An imagined matching insert expands with the sheet, so the opening grows as well.
Why is αV ≈ 3αl for an isotropic solid?
Expand (1 + αlΔT)³ and retain only the first-order contribution.
Liquid water is heated from 1 °C to 3 °C at ordinary pressure. What is its usual volume response?
This lies in the anomalous range approaching the volume minimum near 4 °C.
A uniformly heated rod is perfectly prevented from expanding. What is zero in the simple constrained model?
A compressive mechanical strain cancels the free thermal expansion tendency.
For a fixed mass with small positive volume expansion x, its fractional density change is approximately
Density is inversely proportional to volume, so the first-order change has the opposite sign.
Practice Problems
- A freely expanding aluminium rod is 1.20 m long at 20 °C. Use αl = 2.5 × 10⁻⁵ K⁻¹ to find its length at 70 °C. Then estimate the fractional area and volume changes for a small isotropic aluminium block over the same interval.
- A circular hole in a copper plate is 12.0 mm across at 25 °C. Find its diameter increase at 125 °C using αl = 1.7 × 10⁻⁵ K⁻¹. Explain why the opening expands instead of closing.
- A steel ruler is calibrated at 20 °C and used at 50 °C. It reads 80.0 cm for an object at that temperature. Calculate the modelled actual length using α = 1.2 × 10⁻⁵ K⁻¹ and identify the direction of the correction.
- A brass rod is fixed between rigid supports and heated by 25 °C. Use Y = 90 GPa and α = 2.0 × 10⁻⁵ K⁻¹ to find the sign and magnitude of the stress. Explain how its total strain can be zero while stress is not.
- A liquid has volume expansivity 8.0 × 10⁻⁴ K⁻¹. Estimate the fractional density change for a 20 °C rise using both the first-order expression and the unexpanded denominator of the linear-volume model. State the fixed-mass assumption.
Key Takeaways
• Expansion coefficients describe a material and temperature interval; distinguish a reading from a temperature change. • Area and volume factors 2 and 3 require isotropic small expansion; holes expand with the surrounding sheet. • Water has a density maximum near 4 °C and does not follow one positive expansion coefficient throughout 0–4 °C. • Support constraints, rather than temperature change alone, generate the simple axial thermal stress.