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Concept 03 / 08Thermal Properties of Matter

Thermal Expansion of Solids: Linear, Areal and Volumetric Coefficients

Why solids expand at all (anharmonic bonds), the exact relations between alpha, beta and gamma, why beta equals 2 alpha only to first order, and why holes expand rather than shrink.

Figure 1
Almost everything in this section follows from one microscopic fact: the interatomic potential energy curve is not symmetric. The first principles origin of thermal expansion Model two neighbouring atoms in a solid by a potential U(r)U(r) with a minimum at the equilibrium separation r0r_0. Push the atoms closer than r0r_0 and the repulsion between the electron clouds rises extremely steeply; pull them apart and the attraction falls off gently. The well is therefore steep on the left and shallow on the right. At temperature TT an atom oscillates with total energy EE, exploring the range of rr where U(r)<EU(r) < E. Because the well is lopsided, the atom spends more of its excursion on the far side, and the time averaged separation ⟨r⟩\langle r \rangle is greater than r0r_0, and grows as EE, that is as TT, grows. Two consequences worth remembering. First, if the potential were exactly parabolic (a perfect harmonic oscillator, symmetric about r0r_0) the time averaged separation would be exactly r0r_0 at every temperature and solids would not expand at all. Thermal expansion is a purely anharmonic effect. Second, materials with deep, strong, high melting point bonds (tungsten, quartz, diamond) have small expansion coefficients, while soft, weakly bound solids (lead, zinc, most plastics) have large ones. Linear expansion Define the coefficient of linear expansion as a fractional rate of change, not as an absolute one: α=1LdLdT\alpha = \dfrac{1}{L}\dfrac{dL}{dT} at constant pressure, with SI unit K−1\text{K}^{-1} Integrating exactly gives L=L0exp⁡(∫T0Tα dT)L = L_0 \exp\left(\int_{T_0}^{T}\alpha \, dT\right), and if α\alpha is treated as constant this is L=L0eαΔTL = L_0 e^{\alpha \Delta T}. Expanding the exponential, L=L0(1+α ΔT+12α2ΔT2+⋯ )≈L0(1+α ΔT)L = L_0\left(1 + \alpha\,\Delta T + \tfrac{1}{2}\alpha^2 \Delta T^2 + \cdots\right) \approx L_0(1 + \alpha \, \Delta T) Because α\alpha is of order 10−5 K−110^{-5} \ \text{K}^{-1} for metals, αΔT\alpha \Delta T is about 10−310^{-3} even for a 100 K100 \ \text{K} swing, so the linear truncation is excellent. This also tells you that L=L0(1+αΔT)L = L_0(1+\alpha\Delta T) and L0=L(1−αΔT)L_0 = L(1 - \alpha \Delta T) are equivalent to the accuracy we work at, so it does not matter at which of the two temperatures you regard the length as the "original" one. Typical values: steel and iron 1.2×10−51.2 \times 10^{-5}, copper 1.7×10−51.7\times 10^{-5}, brass 1.9×10−51.9 \times 10^{-5}, aluminium 2.3×10−52.3\times 10^{-5}, ordinary glass 0.9×10−50.9\times 10^{-5}, fused quartz 0.5×10−60.5 \times 10^{-6}, invar 1.2×10−61.2\times 10^{-6}, all per kelvin. Areal expansion, and why the factor is 2 only to first order Take a square plate of side LL and area A=L2A = L^2. Heat it by ΔT\Delta T. Each side becomes L(1+αΔT)L(1+\alpha\Delta T), so A′=L2(1+αΔT)2=A(1+2αΔT+α2ΔT2)A' = L^2 (1+\alpha\Delta T)^2 = A\left(1 + 2\alpha\Delta T + \alpha^2 \Delta T^2\right) The diagram above shows exactly where these terms live. The new area is the old square, plus two rectangular strips each of area L⋅ΔLL \cdot \Delta L (these give the 2αΔT2\alpha\Delta T), plus one little corner square of area (ΔL)2(\Delta L)^2 (this gives the α2ΔT2\alpha^2\Delta T^2). The corner square is the second order term. So the true areal coefficient is β=2α+α2ΔT\beta = 2\alpha + \alpha^2 \Delta T and the familiar β=2α\beta = 2\alpha is the first order truncation. How good is it? The fractional error made by dropping the corner is α2ΔT22αΔT=αΔT2\dfrac{\alpha^2\Delta T^2}{2\alpha \Delta T} = \dfrac{\alpha \Delta T}{2}, which for α=2×10−5\alpha = 2\times 10^{-5} and ΔT=100 K\Delta T = 100 \ \text{K} is 10−310^{-3}, one tenth of one percent. That is why we never bother. Volume expansion follows identically: V′=V(1+αΔT)3=V(1+3αΔT+3α2ΔT2+α3ΔT3)V' = V(1+\alpha \Delta T)^3 = V(1 + 3\alpha\Delta T + 3\alpha^2\Delta T^2 + \alpha^3 \Delta T^3), so γ=3α\gamma = 3\alpha to first order, and exactly γ=3α+3α2ΔT+α3ΔT2\gamma = 3\alpha + 3\alpha^2\Delta T + \alpha^3\Delta T^2. Hence α:β:γ=1:2:3\alpha : \beta : \gamma = 1 : 2 : 3. Why this generalises beyond squares and cubes For an isotropic solid, heating multiplies every linear dimension, in every direction, by the same factor λ=1+αΔT\lambda = 1+\alpha\Delta T. The heated body is therefore geometrically similar to the original, a scaled copy. Any area scales as λ2\lambda^2 and any volume as λ3\lambda^3, whatever the shape. So β=2α\beta = 2\alpha and γ=3α\gamma = 3\alpha hold for a triangle, a circle, a gear wheel or a violin, not just a square. The same similarity argument settles the classic question about holes. Imagine a solid plate, and draw an imaginary closed curve on it. On heating, the material inside that curve expands by the same factor as everything else, so the curve itself grows by the factor λ\lambda. Now physically remove the material inside the curve. Removing it cannot change how the surrounding material moves, because in the uniform expansion no part of the plate was pushing or pulling on any other part; the whole body simply dilated. Therefore a cavity, hole or bore expands exactly as if it were filled with the same material. A hole in a heated plate gets bigger, never smaller. The inner diameter of a heated metal ring increases. Anisotropic solids Crystals that are not cubic expand differently along different axes. Then V′=V(1+αxΔT)(1+αyΔT)(1+αzΔT)V' = V(1+\alpha_x\Delta T)(1+\alpha_y \Delta T)(1+\alpha_z\Delta T), and to first order γ=αx+αy+αz\gamma = \alpha_x + \alpha_y + \alpha_z. Some crystals, for example calcite, actually contract along one axis while expanding along others. Useful derived scalings Density: ρ=mV\rho = \dfrac{m}{V}, so ρ=ρ01+γΔT≈ρ0(1−γΔT)\rho = \dfrac{\rho_0}{1+\gamma\Delta T} \approx \rho_0 (1 - \gamma \Delta T), a decrease. Moment of inertia: II has dimensions of mass times length squared, and mass is unchanged, so ΔII=2αΔT\dfrac{\Delta I}{I} = 2\alpha \Delta T. Time period of a pendulum: T=2πL/gT = 2\pi\sqrt{L/g} gives ΔTperiodTperiod=12ΔLL=12αΔθ\dfrac{\Delta T_{period}}{T_{period}} = \dfrac{1}{2}\dfrac{\Delta L}{L} = \dfrac{1}{2}\alpha\Delta\theta. Reading of a metal scale: if a scale calibrated at T0T_0 is used at a higher temperature TT, each of its divisions is now longer than it claims to be, so it under reports. The true length of an object equals the reading multiplied by (1+αscale(T−T0))(1+\alpha_{scale}(T - T_0)). Key Takeaways and Formulae: L=L0(1+αΔT)L = L_0(1+\alpha\Delta T), A=A0(1+βΔT)A = A_0(1+\beta \Delta T), V=V0(1+γΔT)V = V_0(1+\gamma\Delta T), with β=2α\beta = 2\alpha and γ=3α\gamma = 3\alpha for an isotropic solid, exact only to first order in αΔT\alpha\Delta T. Exact relations: β=2α+α2ΔT\beta = 2\alpha + \alpha^2\Delta T and γ=3α+3α2ΔT+α3ΔT2\gamma = 3\alpha + 3\alpha^2\Delta T + \alpha^3\Delta T^2; the corrections are of relative size αΔT∼10−3\alpha\Delta T \sim 10^{-3}. Anisotropic solid: γ=α1+α2+α3\gamma = \alpha_1+\alpha_2+\alpha_3. Cavities and holes expand with the same coefficient as the solid; a hole never shrinks on heating. Density falls: ρ≈ρ0(1−γΔT)\rho \approx \rho_0(1-\gamma\Delta T). Moment of inertia rises by 2αΔT2\alpha\Delta T. Pendulum period rises by 12αΔT\frac{1}{2}\alpha\Delta T. For two rods to keep a constant difference in length at all temperatures, you need L1α1=L2α2L_1\alpha_1 = L_2\alpha_2. Common mistakes: using β=3α\beta = 3\alpha or γ=2α\gamma = 2\alpha; thinking a hole contracts; forgetting that α\alpha is itself mildly temperature dependent so tabulated values are averages over a stated range; using ΔT\Delta T in Fahrenheit; applying α\alpha to a liquid (liquids have no fixed shape, so only γ\gamma is defined for them). Brain Teaser 1 A steel plate at 27∘C27^\circ\text{C} has a circular hole of diameter exactly 4.000 cm4.000 \ \text{cm}. A brass rivet has diameter 4.012 cm4.012 \ \text{cm} at 27∘C27^\circ\text{C} and is too fat to pass. To what temperature must the plate alone be heated so that the rivet just slips through? Take αsteel=1.2×10−5 K−1\alpha_{steel} = 1.2\times 10^{-5}\ \text{K}^{-1}. Solution. First the conceptual step: the hole expands, it does not close up, because the plate dilates as a whole. So heating the plate is the right strategy. We need the hole diameter to grow from 4.0004.000 to 4.012 cm4.012 \ \text{cm}, a fractional increase of ΔDD=0.0124.000=3.0×10−3\dfrac{\Delta D}{D} = \dfrac{0.012}{4.000} = 3.0\times 10^{-3} Setting this equal to αΔT\alpha \Delta T: ΔT=3.0×10−31.2×10−5=250 K\Delta T = \dfrac{3.0\times 10^{-3}}{1.2\times 10^{-5}} = 250 \ \text{K} So the plate must be raised to 27+250=277∘C27 + 250 = 277^\circ\text{C}. Two remarks an examiner likes. The rivet is assumed to stay at 27∘C27^\circ\text{C}; if you heated both together, the rivet would expand faster (brass has the larger α\alpha) and it would never fit. And in real riveting you do the opposite: cool the rivet in liquid nitrogen so it shrinks, insert it, and let it warm and lock tight. Brain Teaser 2 A student is worried that β=2α\beta = 2\alpha is "only an approximation" and wants to know if it could ever matter. A square aluminium plate (α=2.3×10−5 K−1\alpha = 2.3\times 10^{-5} \ \text{K}^{-1}) of area 1.0000 m21.0000 \ \text{m}^2 is heated through 200 K200 \ \text{K}. Compute the area increase exactly and with the first order formula, and find the percentage discrepancy. Solution. Exact: A′=A(1+αΔT)2A' = A(1+\alpha\Delta T)^2 with αΔT=2.3×10−5×200=4.6×10−3\alpha\Delta T = 2.3\times 10^{-5}\times 200 = 4.6\times 10^{-3}. A′=(1.0046)2=1.0092212A' = (1.0046)^2 = 1.0092212, so ΔAexact=9.2212×10−3 m2\Delta A_{exact} = 9.2212\times 10^{-3}\ \text{m}^2. First order: ΔA=2αΔT⋅A=9.2000×10−3 m2\Delta A = 2\alpha\Delta T \cdot A = 9.2000\times 10^{-3}\ \text{m}^2. Discrepancy =9.2212−9.20009.2212=2.3×10−3= \dfrac{9.2212 - 9.2000}{9.2212} = 2.3\times 10^{-3}, that is 0.230.23 percent, which is exactly the general estimate αΔT/2=2.3×10−3\alpha\Delta T / 2 = 2.3\times 10^{-3}, and is far smaller than the uncertainty in any tabulated α\alpha. Conclusion: the approximation is safe, and the reason to understand it is not numerical but conceptual, because it tells you β=2α\beta = 2\alpha is a statement about geometric similarity and not a coincidence. Brain Teaser 3 A pendulum clock with a steel pendulum keeps perfect time at 20∘C20^\circ\text{C}. It is moved to a room at 40∘C40^\circ\text{C}. Does it gain or lose, and by how much per day? αsteel=1.2×10−5 K−1\alpha_{steel} = 1.2\times 10^{-5}\ \text{K}^{-1}. Then: what temperature change would the same clock need if the pendulum were made of invar, α=1.2×10−6 K−1\alpha = 1.2\times 10^{-6}\ \text{K}^{-1}, to produce the same error? Solution. The period is Tp=2πL/gT_p = 2\pi\sqrt{L/g}, so ΔTpTp=12ΔLL=12αΔθ=12(1.2×10−5)(20)=1.2×10−4\dfrac{\Delta T_p}{T_p} = \dfrac{1}{2}\dfrac{\Delta L}{L} = \dfrac{1}{2}\alpha\Delta\theta = \dfrac{1}{2}(1.2\times 10^{-5})(20) = 1.2\times 10^{-4} The period becomes longer, so each swing takes more time, so the clock counts fewer seconds than it should: it runs slow and loses time. In one day of 8640086400 seconds, time lost=1.2×10−4×86400=10.4 s per day\text{time lost} = 1.2\times 10^{-4}\times 86400 = 10.4 \ \text{s per day} For invar, α\alpha is ten times smaller, so to lose the same 10.410.4 seconds per day you would need a temperature rise ten times larger, that is 200 K200 \ \text{K}. This is precisely why invar, an iron nickel alloy discovered by Guillaume (who won a Nobel Prize for it), is used in precision clocks and metrological standards. A common follow up: a clock taken to a colder room runs fast and gains, by the same formula with Δθ\Delta\theta negative.

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