Thermal Expansion

Heat a solid and its molecules vibrate through wider swings about their lattice positions, so their average separation grows and the body gets bigger.

That one sentence explains the gaps left between rails on the Janakpur railway, the rollers under a bridge girder, the sag designed into transmission lines, and the fact that a mercury thermometer works at all.

In this unit we put numbers on the effect: one coefficient for length, one for area, one for volume, and a special treatment for liquids, which cannot be heated without a container that expands too.

10.1 Linear expansion and its measurement

When a rod of length l_1 at temperature \theta_1 is heated to \theta_2, its length becomes l_2. Experiment shows the increase is proportional to the original length and to the temperature rise.

Strictly l_1 should be the length at 0\ ^{\circ}\text{C}, but because \alpha is of order 10^{-5} the difference between "length at 0\ ^{\circ}C" and "length at room temperature" changes the answer only in the fifth decimal place, so any convenient starting length is used.

Material	\alpha (\times 10^{-6} K⁻¹)	Material	\alpha (\times 10^{-6} K⁻¹)
Invar	1.2	Copper	17
Pyrex glass	3.2	Brass	19
Ordinary glass	9	Aluminium	23
Steel / iron / concrete	12	Lead	29

Invar (an iron–nickel alloy) expands so little that it is used for pendulum rods and survey tapes. Steel and concrete happen to have almost the same \alpha — which is the reason reinforced concrete does not tear itself apart in the Terai summer.

The rod under test stands on a fixed base inside a hollow steam jacket. Its upper end is just below the tip of a micrometer screw carried on a rigid frame.

The screw, the rod and a battery with an electric bell are connected in series, so the bell rings the instant the screw touches the rod.

1. With the rod at room temperature \theta_1 (read on the thermometer in the jacket), turn the screw down until the bell just rings; record the reading a_1. Measure the length l_1 of the rod with a metre rule.
2. Raise the screw well clear of the rod.
3. Pass steam through the jacket until the thermometer is steady at \theta_2. The rod has now expanded.
4. Turn the screw down again until the bell just rings; record a_2.

The expansion is \delta l = a_1 - a_2 (the screw has less distance to travel), so

10.2 Cubical expansion, superficial expansion and relation with linear expansion

10.3 Liquid Expansion: Absolute and apparent

A liquid has no shape of its own, so only its volume expansion is defined — there is no \alpha or \beta for a liquid. But a liquid must be held in a vessel, and when the vessel is heated it expands too.

The level in the vessel therefore rises by less than the liquid's true expansion. Two coefficients are needed.

In the weight-thermometer (specific-gravity bottle) method the bulb is filled with liquid of mass M at \theta_1, heated to \theta_2, and the mass m expelled is weighed. Since apparent expansion is measured by the liquid that leaves,

where M-m is the mass still filling the bulb at the higher temperature. Adding 3\alpha_g for the glass then gives the real expansivity.

Liquid	\gamma_r (K⁻¹)
Mercury	1.8\times10^{-4}
Water (near 20 °C)	2.1\times10^{-4}
Glycerine	5.3\times10^{-4}
Ethanol	1.1\times10^{-3}

Because \gamma_{\text{glass}} = 3\times(9\times10^{-6}) = 2.7\times10^{-5}\ \text{K}^{-1} is small compared with these, the apparent expansivity is usually only a little less than the real one — but the difference is easily measurable and must never be ignored in a calculation.

Water is the famous exception. From 0\ ^{\circ}C to 4\ ^{\circ}C it contracts on heating, and only above 4\ ^{\circ}C does it expand normally. Its density is therefore a maximum at 4\ ^{\circ}C.

This anomalous expansion is why ice floats, and why a lake such as Rara freezes from the surface downwards in winter, leaving water at 4\ ^{\circ}C at the bottom in which fish survive.

10.4 Dulong and Petit method of determining expansivity of liquid

Every method that uses a vessel gives the apparent expansivity first, and the real value only after the vessel's own expansion is known.

Dulong and Petit (1817) found a way to measure the real expansivity directly, using the fact that pressure, not volume, is what balances in a connected liquid.

Apparatus. The liquid fills a U-shaped glass tube whose two vertical limbs are joined by a narrow horizontal tube at the bottom.

One limb is surrounded by a jacket of melting ice at 0\ ^{\circ}\text{C}; the other is surrounded by a steam jacket at \theta\ ^{\circ}\text{C}.

The heights of the two columns above the common horizontal base are measured with a travelling microscope, so nothing touches the liquid.

(a) A bimetallic strip: two metals that expand by different amounts

(b) Linear expansion of a 30 m rail for a 40 K rise

Chapter summary

• Linear expansivity \alpha = \dfrac{l_2-l_1}{l_1\Delta\theta}, so l_2 = l_1(1+\alpha\Delta\theta); unit K⁻¹.
• Pullinger's apparatus measures \alpha: rod in a steam jacket, expansion read on a micrometer screw that completes a bell circuit; \alpha = \delta l / (l_1\Delta\theta).
• A_2 = A_1(1+\beta\Delta\theta) and V_2 = V_1(1+\gamma\Delta\theta), with \beta = 2\alpha, \gamma = 3\alpha and \alpha:\beta:\gamma = 1:2:3.
• A cavity or hole expands exactly as if it were filled with the surrounding material.
• A liquid has two expansivities: real \gamma_r and apparent \gamma_a, with \gamma_r = \gamma_a + \gamma_g where \gamma_g = 3\alpha_g for the vessel. By the weight thermometer, \gamma_a = m/[(M-m)\Delta\theta].
• Water expands anomalously: it contracts from 0 to 4\ ^{\circ}C and has maximum density at 4\ ^{\circ}C.
• Dulong and Petit: balancing columns give h_\theta = h_0(1+\gamma\theta), so \gamma_r = (h_\theta-h_0)/(h_0\theta), with no correction for the vessel.

Practice questions

Group A — Multiple choice (1 mark each)

1. For an isotropic solid the ratio \alpha : \beta : \gamma is [1]a1:2:3b3:2:1c1:3:2d2:3:1
2. When a metal plate with a hole in it is heated, the hole [1]agets smallerbgets largercstays the samedfirst shrinks, then grows
3. The SI unit of the coefficient of linear expansion is [1]ambK⁻¹cm K⁻¹dit has no unit
4. The real expansivity of a liquid equals its apparent expansivity plus the [1]alinear expansivity of the vesselbcubical expansivity of the vesselcsuperficial expansivity of the vesseldnothing
5. Dulong and Petit's method directly determines the [1]aapparent expansivitybreal expansivityclinear expansivitydsuperficial expansivity

Answers to Group A

1. (a) — from (1+\alpha\Delta\theta)^n for n = 1,2,3.

2. (b) — a cavity expands like the material that would fill it.

3. (b) — it is a fractional length change per kelvin, so only K⁻¹ is left.

4. (b) — \gamma_r = \gamma_a + \gamma_g, where \gamma_g = 3\alpha_g.

5. (b) — it balances pressures, so the vessel's expansion cancels out.

Group B — Short answer (5 marks each)

1. Define the coefficient of linear expansion and show that \gamma = 3\alpha for an isotropic solid. [5]
2. With a labelled diagram, describe how the linear expansivity of a metal rod is determined using Pullinger's apparatus. [5]
3. Distinguish between the real and apparent expansivity of a liquid and show that \gamma_r = \gamma_a + \gamma_g. [5]
4. A glass flask of volume 500\ \text{cm}^{3} is completely filled with mercury at 0\ ^{\circ}\text{C}. On heating to 80\ ^{\circ}\text{C}, 6.0\ \text{cm}^{3} of mercury overflows. If \gamma_{\text{Hg}} = 1.8\times10^{-4}\ \text{K}^{-1}, find the coefficient of linear expansion of the glass. [5]
5. A brass scale is correct at 15\ ^{\circ}\text{C}. It is used at 35\ ^{\circ}\text{C} and reads a length of 50.00\ \text{cm}. What is the true length? Take \alpha_{\text{brass}} = 1.9\times10^{-5}\ \text{K}^{-1}. [5]

Answers to Group B

4. \gamma_a = \dfrac{6.0}{500\times80} = 1.5\times10^{-4}\ \text{K}^{-1}. Then \gamma_g = \gamma_r - \gamma_a = 1.8\times10^{-4} - 1.5\times10^{-4} = 3.0\times10^{-5}\ \text{K}^{-1}, so \alpha_g = \gamma_g/3 = 1.0\times10^{-5}\ \text{K}^{-1}.

5. At 35\ ^{\circ}C each scale division has grown, so every division now measures more than it claims. True length = 50.00\left(1 + 1.9\times10^{-5}\times20\right) = 50.00(1+3.8\times10^{-4}) = 50.019\ \text{cm}.

Group C — Long answer (8 marks each)

1. (a) Derive the relations \beta = 2\alpha and \gamma = 3\alpha, stating clearly the approximation used. [4](b) Describe Dulong and Petit's method for the absolute expansivity of a liquid and derive \gamma = \dfrac{h_\theta - h_0}{h_0\theta}. Explain why no correction for the expansion of the glass is needed. [4]
2. (a) An iron rim of diameter 79.80\ \text{cm} at 20\ ^{\circ}\text{C} is to be fitted on a wooden cart wheel of diameter 80.00\ \text{cm}. To what temperature must the rim be heated? Take \alpha_{\text{iron}} = 1.2\times10^{-5}\ \text{K}^{-1}. [4](b) In a Dulong and Petit experiment on a liquid, the column at 0\ ^{\circ}\text{C} is 60.00\ \text{cm} high and the column at 100\ ^{\circ}\text{C} is 61.20\ \text{cm} high. Find the real expansivity of the liquid, and its apparent expansivity when it is kept in a glass vessel with \alpha = 9\times10^{-6}\ \text{K}^{-1}. [4]

Answers to Group C

2. (a) The rim must expand from 79.80 to 80.00\ \text{cm}, so \Delta d = 0.20\ \text{cm} and

\Delta\theta = \frac{\Delta d}{d_1\alpha} = \frac{0.20}{79.80\times1.2\times10^{-5}} = 208.9\ \text{K}

The rim must be heated to 20 + 208.9 \approx 229\ ^{\circ}\text{C}. On cooling it grips the wheel tightly — the standard way a wheelwright fits a tyre.

(b) \gamma_r = \dfrac{61.20-60.00}{60.00\times100} = \dfrac{1.20}{6000} = 2.0\times10^{-4}\ \text{K}^{-1}.

\gamma_g = 3\times9\times10^{-6} = 2.7\times10^{-5}\ \text{K}^{-1}, so \gamma_a = 2.0\times10^{-4} - 0.27\times10^{-4} = 1.73\times10^{-4}\ \text{K}^{-1}.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. Define the coefficients of linear, area and volume expansion of solids and prove that \alpha : \beta : \gamma = 1 : 2 : 3. Give a numerical and eight applications. [10]
2. Distinguish between real and apparent expansion of a liquid, derive the relation between them, and explain the anomalous expansion of water. [10]

Model answer D1: expansion of solids (10 marks)

Introduction. Most solids expand on heating because the atoms vibrate more strongly and the average spacing between them grows. The change can be in length, area or volume.

Coefficients.

• Linear expansivity \alpha: the increase in length per unit original length per degree rise, \alpha = \dfrac{\Delta L}{L_{0}\,\Delta\theta} (unit K⁻¹).
• Areal expansivity \beta = \dfrac{\Delta A}{A_{0}\,\Delta\theta}.
• Volume expansivity \gamma = \dfrac{\Delta V}{V_{0}\,\Delta\theta}.

Proof that \alpha : \beta : \gamma = 1 : 2 : 3. Take a cube of side L_{0} heated through \Delta\theta.

1. New side L = L_{0}(1 + \alpha\Delta\theta).
2. Area of a face: A = L^{2} = L_{0}^{2}(1 + \alpha\Delta\theta)^{2} = A_{0}(1 + 2\alpha\Delta\theta + \alpha^{2}\Delta\theta^{2}). As \alpha is of the order 10^{-5}, the square term is negligible: A = A_{0}(1 + 2\alpha\Delta\theta), so \beta = 2\alpha.
3. Volume: V = L^{3} = V_{0}(1 + \alpha\Delta\theta)^{3} \approx V_{0}(1 + 3\alpha\Delta\theta), so \gamma = 3\alpha.
4. Hence \alpha : \beta : \gamma = 1 : 2 : 3.

Material	\alpha (10^{-6}\ \text{K}^{-1})
Aluminium	23
Copper	17
Steel / iron	12
Glass (ordinary)	9
Pyrex	3
Invar	1

Numerical. A steel railway rail 30\ \text{m} long is laid at 20^{\circ}C. In summer the temperature rises to 50^{\circ}C (\alpha = 1.2\times10^{-5}\ \text{K}^{-1}).

Then \Delta L = 30\times1.2\times10^{-5}\times30 = 1.08\times10^{-2}\ \text{m} = 10.8\ \text{mm}.

If there were no gap, the rail would exert an enormous thermal stress: Y\alpha\Delta\theta = 2\times10^{11}\times1.2\times10^{-5}\times30 = 7.2\times10^{7}\ \text{Pa}.

Eight applications (and precautions).

1. Expansion gaps between rails, so that tracks do not buckle in the summer sun.
2. Expansion joints in bridges and concrete roads, like the metal combs on the Kathmandu flyovers.
3. Bimetallic strip (brass and iron): brass expands more, so the strip bends; it is used in thermostats in irons and heaters and in fire alarms.
4. Shrink fitting: a steel tyre is heated, slipped over a wheel and cooled so that it grips it tightly.
5. Riveting: hot rivets contract on cooling and pull the plates tight.
6. Pyrex and Invar: laboratory glassware and clock pendulums are made of low-expansion materials to avoid cracking and keep time.
7. Sagging of power lines in summer: the Nepal Electricity Authority leaves slack in the transmission lines.
8. Pendulum and balance wheels of watches are compensated for the temperature.

Conclusion. The three coefficients are in the ratio 1 : 2 : 3. Expansion is useful in thermostats and fittings, but its effects must be allowed for in engineering.

Model answer D2: expansion of liquids and anomalous expansion of water (10 marks)

Introduction. Liquids expand much more than solids (\gamma about 10 times larger) and have no shape of their own, so only volume expansion is meaningful. A liquid is always kept in a container which also expands, and that complicates the measurement.

Real and apparent expansion.

• Real (absolute) expansion \gamma_{r}: the actual increase in the volume of the liquid, the expansion of the container being taken into account.
• Apparent expansion \gamma_{a}: the expansion as it appears, when the expansion of the container is ignored; it is the visible rise in the liquid level.

Relation. Let a vessel of volume V_{0} be full of liquid at 0^{\circ}C and heated through \Delta\theta.

1. The vessel's volume increases by V_{0}\gamma_{g}\Delta\theta.
2. The liquid's true volume increases by V_{0}\gamma_{r}\Delta\theta.
3. The liquid that overflows (the apparent increase) is the difference: V_{0}\gamma_{a}\Delta\theta = V_{0}\gamma_{r}\Delta\theta - V_{0}\gamma_{g}\Delta\theta.
4. Therefore

Numerical. For mercury, \gamma_{r} = 1.8\times10^{-4}\ \text{K}^{-1}, and for the glass \gamma_{g} = 3\alpha = 3\times9\times10^{-6} = 2.7\times10^{-5}\ \text{K}^{-1}.

The apparent expansion is \gamma_{a} = 1.8\times10^{-4} - 0.27\times10^{-4} = 1.53\times10^{-4}\ \text{K}^{-1}. That is why mercury rises visibly in a glass thermometer.

Anomalous expansion of water. Most liquids expand regularly, but water contracts when heated from 0^{\circ}C to 4^{\circ}C, and expands only above 4^{\circ}C.

• The volume is a minimum and the density is a maximum (1000 kg m⁻³) at 4^{\circ}C.
• Reason: below 4^{\circ}C, there are ice-like clusters of water molecules held by hydrogen bonds in an open structure. As the temperature rises the clusters break and the molecules pack closer, which outweighs the normal expansion. Above 4^{\circ}C the normal expansion dominates.

Temperature (^{\circ}C)	Density of water (\text{kg m}^{-3})
0	999.84
4	999.97 (maximum)
10	999.70
100	958.4

Consequences. In winter, the surface water of a lake cools to 4^{\circ}C, becomes denser and sinks. The colder water (below 4^{\circ}C) is lighter and stays at the top and freezes.

Ice floats, so the lake freezes from the top downward, and the water at the bottom stays at 4^{\circ}C, saving the fish and plants, as in the lakes of Gosaikunda and Rara.

Conclusion. Real expansion is the sum of apparent expansion and the expansion of the vessel. The anomalous expansion of water, with its maximum density at 4^{\circ}C, protects aquatic life in the cold.