Physics / Linear Thermal Expansion of Copper
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EXPERIMENT 29 · PHYSICS LABORATORY

Linear Thermal Expansion of Copper theory

Understand linear thermal expansion of copper: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
ΔL = L_0 · α · ΔT ⇔ Y = α · X ⇔ α = S

Keep the physical quantities, graph variables and units consistent.

01Anharmonicity of Interatomic Potential and Origin of Thermal Dilatation

The physical origin of thermal expansion in crystalline solids is rooted in the asymmetric shape of the interatomic potential energy curve U(r). In a realistic crystal, the interatomic potential can be modeled by a Lennard-Jones (12-6) or Morse function: U(r) = U_0 [(r_0/r)¹² − 2(r_0/r)⁶], or expanded as a Taylor series about the equilibrium separation r_0: U(x) = c x² − g x³ − f x⁴, where x = r − r_0 represents atomic displacement, c is the harmonic restoring force constant, g is the cubic anharmonicity coefficient, and f is the quartic coefficient. If the potential were strictly harmonic (g = 0, f = 0), the Boltzmann distribution P(x) ∝ exp[−U(x)/(kT)] would be symmetric in ±x, meaning the expectation value ⟨x⟩ = ∫ x exp[−U(x)/(kT)] dx / ∫ exp[−U(x)/(kT)] dx would remain identically zero at all temperatures. However, because atomic electron shells strongly resist overlapping (producing a steep repulsive wall for x < 0) while attractive bonding forces decay gradually with distance (a shallow trough for x > 0), the cubic anharmonic term g is strictly positive. Evaluating the thermal expectation value using perturbation theory yields: ⟨x⟩ ≈ (3g / 4c²) · kT. Thus, as temperature T rises, the average interatomic separation increases linearly with T, giving rise to macroscopic thermal expansion. Without lattice anharmonicity, solids would not expand when heated.

U(x) = c x^2 - g x^3 - f x^4 \implies \langle x \rangle \approx \frac{3g}{4c^2} k_B T \implies \alpha = \frac{1}{r_0}\frac{d\langle x \rangle}{dT} = \frac{3g k_B}{4 c^2 r_0}
Why would a solid with a purely harmonic interatomic potential experience zero thermal expansion?

In a purely harmonic potential U(x) = cx², the energy well is perfectly symmetrical about the equilibrium position. When heated, atoms vibrate with larger amplitudes, but their time-averaged position ⟨x⟩ remains exactly at the center. Thermal expansion requires anharmonicity (the cubic term −gx³), which creates an asymmetric well that shifts the average atomic separation outward as energy increases.

02Linear, Areal, and Volumetric Expansion Relationships in Isotropic Solids

In isotropic materials where physical properties are identical in all spatial directions, thermal dilatation is described by a single scalar coefficient. Consider a rectangular block of dimensions L_x, L_y, L_z at baseline temperature T_0. When heated through ΔT, each linear dimension expands according to L_i(T) = L_{i,0}(1 + α ΔT). The resulting area of any face (e.g., A = L_x L_y) becomes A(T) = L_{x,0} L_{y,0}(1 + α ΔT)² = A_0(1 + 2α ΔT + α² ΔT²). Because typical expansion coefficients for metals are of order 10⁻⁵ K⁻¹, the second-order term α² ΔT² is negligible (e.g., (1.68 × 10⁻⁵ × 50)² ≈ 7 × 10⁻⁷), allowing the areal thermal expansion coefficient to be approximated as β ≈ 2α. Similarly, the expanded volume evaluates to V(T) = V_0(1 + α ΔT)³ = V_0(1 + 3α ΔT + 3α² ΔT² + α³ ΔT³). Neglecting higher-order terms yields V(T) ≈ V_0(1 + γ ΔT), establishing the fundamental thermodynamic relation γ ≈ 3α for isotropic solids. For anisotropic crystals such as calcite, zinc, or graphite, thermal expansion is a second-rank tensor α_ij, where expansion along different crystallographic axes can vary dramatically and may even exhibit negative thermal expansion along specific axes.

\beta = \frac{1}{A_0}\frac{dA}{dT} \approx 2\alpha \quad \text{and} \quad \gamma = \frac{1}{V_0}\frac{dV}{dT} \approx 3\alpha \quad (\text{for isotropic solids where } \alpha \Delta T \ll 1)
What is the mathematical relationship connecting the linear, areal, and volumetric thermal expansion coefficients in an isotropic solid?

For isotropic solids where αΔT is small compared to unity, the areal coefficient is β ≈ 2α and the volumetric coefficient is γ ≈ 3α. This derives directly from expanding (1 + αΔT)² and (1 + αΔT)³ and dropping the negligible higher-order terms in α² and α³.

03The Steam Dilatometer and Electrical Contact Sensing Principle

Measuring the thermal expansion of a 1.00-meter rod over a temperature elevation of 50 °C poses a delicate metrological challenge: the total elongation is only ΔL = 1.000 m × 16.80 × 10⁻⁶ K⁻¹ × 50 K = 0.840 mm (840 microns). If a mechanical micrometer screw is turned until it physically pinches the expanding rod, the compressive contact force can easily distort the frame, indent the soft copper rod tip, and compress the specimen longitudinally according to Hooke's law: δ_comp = F L_0 / (A Y). For a 6 mm diameter copper rod (A ≈ 2.83 × 10⁻⁵ m², Young's modulus Y ≈ 110 GPa), an anvil contact force of just 10 newtons causes a compressive strain of δ_comp ≈ 3.2 microns—introducing a 0.4% systematic measurement error. To eliminate contact force entirely, modern university steam dilatometers employ an electrical indicator circuit. The micrometer screw and specimen rod are placed in series with a 1.5 V DC source, a high-value current-limiting resistor, and a sensitive galvanometer or LED. As the micrometer screw is advanced slowly, the very first atomic contact between the polished spindle face and rod end closes the electrical circuit, instantly triggering a galvanometer deflection or lighting the indicator. The observer records the drum reading at this exact threshold of zero mechanical pressure, ensuring sub-micron accuracy.

\Delta L = L_0 \alpha \Delta T \quad \text{with} \quad \delta_{\text{compression}} = \frac{F L_0}{A Y} \to 0 \quad (\text{via electrical contact sensing})
Why is an electrical contact circuit used with the micrometer screw gauge in precision dilatometry?

If the micrometer screw is tightened manually against the rod until mechanical resistance is felt, the applied force flexes the support frame and compresses the copper rod elastically, introducing significant systematic error. The electrical circuit detects the instant of microscopic touch with zero contact pressure, eliminating mechanical deformation errors.

04Thermodynamic and Mechanical Stress in Constrained Thermal Expansion

When a solid bar is heated while its ends are rigidly constrained between unyielding walls, its natural tendency to expand is frustrated, generating immense internal compressive stress. The free thermal strain that would occur without constraint is ε_th = ΔL/L_0 = α ΔT. To maintain the bar at its original length (net strain ε_net = 0), the rigid supports must exert an opposing mechanical compressive strain equal in magnitude to the thermal strain: ε_mech = −σ / Y, where σ is the internal stress and Y is Young's modulus of the material. Equating strains yields the thermal stress equation: σ = −Y α ΔT. For a copper bar (Y = 110 GPa, α = 16.80 × 10⁻⁶ K⁻¹) heated through ΔT = 50 °C, the resulting compressive stress is σ = 110 × 10⁹ Pa × 16.80 × 10⁻⁶ K⁻¹ × 50 K = 92.4 MPa (924 atmospheres). If the bar has a cross-sectional area of 1.0 cm² (1.0 × 10⁻⁴ m²), the compressive thrust against the end walls reaches F = σ A = 9,240 newtons (~940 kg-force). This dramatic thermal stress explains why railway tracks require expansion gaps or prestressed continuous welding, why concrete bridges feature expansion joint combs, and why dilatometers must permit one end of the specimen to expand freely against the measuring instrument.

\sigma = -Y \alpha \Delta T \quad \text{and} \quad F = A Y \alpha \Delta T \quad (\text{for rigidly constrained ends})
What happens if a metal rod is heated while both of its ends are held completely rigid and immovable?

Immense internal compressive stress develops, equal to σ = Y α ΔT, where Y is Young’s modulus. For copper heated by 50 °C, this thermal stress exceeds 92 MPa, exerting thousands of newtons of thrust on the supports and risking severe buckling or mechanical failure.

05Temperature Non-Uniformity and Longitudinal Heat Loss Gradients

A principal source of systematic error in jacketed steam dilatometry is longitudinal temperature non-uniformity along the specimen rod. In an ideal dilatometer, saturated dry steam at 100 °C flows rapidly through the jacket, bathing the entire surface of the rod and establishing an isothermal boundary condition. However, at the two extremities where the rod emerges through rubber stoppers or glands to contact the fixed stop and micrometer spindle, significant conductive heat loss occurs along the exposed metal ends to the ambient room air. Furthermore, if condensed water pools along the bottom of the jacket, the lower surface of the rod may experience a different heat transfer coefficient than the upper surface exposed to turbulent steam vapor. The effective thermal elongation of the rod represents the integral of the local thermal strain across the full length: ΔL = ∫_0^{L_0} α [T(x) − T_0] dx = L_0 α (T_avg − T_0), where T_avg is the true length-averaged temperature. If the thermometer bulb is located near the central steam inlet, it may record 100.0 °C while the exposed rod ends are at only 85 °C to 90 °C, causing T_avg to be lower than the recorded temperature T_therm. Neglecting this end-loss gradient causes the calculated expansion coefficient α = ΔL / [L_0(T_therm − T_0)] to be systematically underestimated by 1% to 3%. To minimize this error, well-designed dilatometers utilize thermal insulating collars around the projecting tips and maintain high steam throughput.

\Delta L = \int_0^{L_0} \alpha [T(x) - T_0] dx = L_0 \alpha (\bar{T} - T_0) \quad \text{where } \bar{T} = \frac{1}{L_0} \int_0^{L_0} T(x) dx
Why might the measured expansion coefficient be systematically too low if the rod ends protrude outside the steam jacket?

The protruding metal ends conduct heat to the cool room air, causing them to stay cooler than the central steam chamber. If the thermometer only measures the hot center, the assumed temperature rise is higher than the true length-averaged temperature rise along the entire rod, causing the calculated expansion coefficient α = ΔL / (L₀ΔT) to be systematically underestimated.

06Instrumental Backlash, Zero Indexing, and Pitch Calibration of Micrometer Screws

The precision micrometer screw gauge is the primary displacement sensor in the dilatometer apparatus. A standard metric micrometer features a precision ground lead screw with a pitch of p = 0.500 mm, meaning one complete 360° revolution of the thimble advances the spindle by exactly half a millimeter. With 50 equal graduations etched along the circumference of the thimble, the least count (smallest directly readable division) evaluates to LC = pitch / divisions = 0.500 mm / 50 = 0.010 mm (10 microns). With careful interpolation using an optical loupe or vernier sleeve graduations, readings can be estimated to ±0.001 mm (1 micron). However, screw mechanisms are vulnerable to mechanical backlash: the slight clearance between the male screw threads and female nut threads required to permit low-friction rotation. If the screw is reversed during a measurement, the spindle does not move immediately until the thread clearance is traversed. To eliminate backlash error, the micrometer must always approach the contact point from the same direction (advancing forward toward the rod). Furthermore, the zero reading R_0 must be recorded under identical directional approach at baseline temperature before heating.

\text{Least Count (LC)} = \frac{\text{Pitch (mm)}}{\text{Thimble Divisions}} = \frac{0.500 \text{ mm}}{50} = 0.010 \text{ mm} \quad (\pm 0.001 \text{ mm estimated})
What is mechanical backlash in a micrometer screw, and how is it avoided during thermal expansion measurements?

Backlash is the mechanical slack or clearance between the internal mating screw threads, which causes lost motion when reversing rotational direction. It is eliminated by always rotating the micrometer thimble in the same forward direction when approaching the contact point for both cold and hot readings.

07Comparative Expansion Coefficients of Engineering Metals and Invar Alloys

The magnitude of the linear thermal expansion coefficient α varies widely across engineering materials, reflecting differences in crystal structure, bond strength, and melting points. Refractory metals with deep interatomic potential wells and high melting points generally exhibit small expansion coefficients: tungsten (T_m = 3,422 °C, α ≈ 4.5 × 10⁻⁶ K⁻¹), molybdenum (α ≈ 5.0 × 10⁻⁶ K⁻¹), and platinum (α ≈ 9.0 × 10⁻⁶ K⁻¹). Common engineering metals possess moderate expansion rates: iron and mild steel (α ≈ 11.5–12.0 × 10⁻⁶ K⁻¹), copper (α ≈ 16.80 × 10⁻⁶ K⁻¹), brass (α ≈ 19.0 × 10⁻⁶ K⁻¹), and aluminum (α ≈ 23.1 × 10⁻⁶ K⁻¹). The difference between copper and steel underlies the operation of bimetallic strips: when bonded together and heated, the copper expands more than the steel, forcing the strip to bend into an arc toward the steel side, widely exploited in mechanical thermostats, fire alarms, and thermal circuit breakers. In 1896, Charles Édouard Guillaume discovered Invar (Fe-36% Ni), an alloy possessing an extraordinarily low expansion coefficient: α_invar ≈ 1.2 × 10⁻⁶ K⁻¹ near room temperature. In Invar, an anomalous spontaneous volume magnetostriction arising from the alignment of ferromagnetic d-electron spins expands the lattice as temperature drops, almost exactly canceling the normal thermal contraction from lattice vibrations over the range −20 °C to +100 °C, earning Guillaume the 1920 Nobel Prize in Physics.

\alpha_{\text{Al}} (23.1) > \alpha_{\text{Brass}} (19.0) > \alpha_{\text{Cu}} (16.8) > \alpha_{\text{Steel}} (11.8) > \alpha_{\text{Invar}} (1.2) \quad [\times 10^{-6} \text{ K}^{-1}]
Why does a bimetallic strip made of copper and iron bend when heated, and in which direction does it bend?

Copper has a higher thermal expansion coefficient (16.8 × 10⁻⁶ K⁻¹) than iron (11.8 × 10⁻⁶ K⁻¹). When bonded together and heated, the copper side expands more than the iron side, forcing the strip to bend into an arc with the copper on the outside (convex side) and the iron on the inside (concave side).

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Connect the equations to a measurement.

The simulation calculates longitudinal elongation using the linear dilatation relationship ΔL = L_0 · α · ΔT, where L_0 = 1.0000 m is the baseline length, α = 16.80 × 10⁻⁶ K⁻¹ is the reference linear expansion coefficient for pure copper, and ΔT = T − T_0 is the temperature elevation above 20.00 °C. Elongation readings are rounded to 0.001 mm to match high-precision micrometer screw resolutions, and ordinary least-squares regression of ΔL against L_0 · ΔT across five temperature stations yields the reported linear expansion coefficient.