Thermal Expansion

Coefficient of linear expansion: the fractional increase in length of a solid per unit rise in temperature,

$\alpha=\frac{L_2-L_1}{L_1(\theta_2-\theta_1)}=\frac{\Delta L}{L_1\Delta\theta}$

Its unit is $\text{K}^{-1}$ (or $^\circ\text{C}^{-1}$).

Dependence on length: No. Because the change in length is divided by the original length, $\alpha$ is independent of the length (and size or shape) of the conductor. It depends only on the material, and slightly on temperature.

A seconds pendulum has a period of 2 s. On heating, the length of the brass rod increases, so the period increases and the clock runs slow, i.e. it loses time.

$T=2\pi\sqrt{\frac lg},\qquad l'=l(1+\alpha\Delta\theta)\;\Rightarrow\;\frac{T'}{T}=\sqrt{1+\alpha\Delta\theta}\approx1+\tfrac12\alpha\Delta\theta$

With $\Delta\theta=35-10=25$ K and $\alpha=2\times10^{-5}\ \text{K}^{-1}$:

$\frac{\Delta T}{T}=\tfrac12\times2\times10^{-5}\times25=2.5\times10^{-4}$

Time lost per day $=86400\times2.5\times10^{-4}=21.6\ \text{s}$.

The clock loses about $21.6$ s per day.

The mercury that overflows is the difference between the expansion of the mercury and the expansion of the flask's volume:

$V_{\text{overflow}}=V_0(\gamma_{Hg}-\gamma_g)\,\Delta\theta$

$=400\times(18\times10^{-5}-1.2\times10^{-5})\times 80=400\times 16.8\times10^{-5}\times 80\approx 5.38\ \text{cm}^3$

About $5.4\ \text{cm}^3$ of mercury overflows.

It means that when the temperature of mercury rises by $1^\circ$C, each unit volume of mercury expands by $1.8\times10^{-5}$ of its original volume (e.g. $1\ \text{m}^3$ increases by $1.8\times10^{-5}\ \text{m}^3$).

At $10^\circ$C the tape contracts, so its markings are closer than their nominal spacing and it needs more graduations to cover the same real distance. The reading is therefore larger than the true length.

A true length of $1000\ \text{m}$ is covered by $\dfrac{1000}{1-\alpha\Delta\theta}$ nominal metres, with $\alpha\Delta\theta=12\times10^{-6}\times10=1.2\times10^{-4}$:

$\text{reading}=\frac{1000}{1-1.2\times10^{-4}}\approx1000(1+1.2\times10^{-4})=1000.12\ \text{m}$

The tape shows about $1000.12\ \text{m}$ (an error of $0.12\ \text{m}$).

A metal scale is calibrated to be correct at one particular temperature only. At a higher temperature it expands, so each division becomes longer than its marked value and the scale reads a length smaller than the true one. At a lower temperature it contracts, so each division is shorter and the reading is larger than the true length. The error is $l\,\alpha\,\Delta\theta$, so the scale is accurate only at its calibration temperature.

(c) the nature of the material

The coefficient of linear expansion ($\alpha$) is a characteristic property of the substance itself. It does not depend on the original length, the temperature rise or the area of cross-section — these affect only the change in length, $\Delta L = L_0 \alpha \,\Delta T$, not the coefficient itself.

Derivation of final length

Let a metal rod have length $L_0$ at temperature $\theta_1$. It is heated to temperature $\theta_2$, so the rise in temperature is

$\Delta\theta = \theta_2 - \theta_1$

Experiment shows that for small temperature changes, the increase in length $\Delta L$ is directly proportional to:

• the original length $L_0$, and
• the rise in temperature $\Delta\theta$.

So,

$\Delta L \propto L_0 \,\Delta\theta$

$\Delta L = \alpha L_0 \,\Delta\theta$

where $\alpha$ is the coefficient of linear expansion — the fractional increase in length per unit rise in temperature. Its SI unit is $\text{K}^{-1}$ (or $^\circ\text{C}^{-1}$).

The final length is the original length plus the increase:

$L = L_0 + \Delta L$

$L = L_0 + \alpha L_0 \,\Delta\theta$

$\boxed{L = L_0\left[1 + \alpha(\theta_2 - \theta_1)\right]}$

This is the required expression for the final length of the rod.

Why railway tracks are laid with small gaps

Railway tracks are made of iron or steel, which has a appreciable coefficient of linear expansion. In summer, the rails get heated and tend to expand; in winter they contract. If the rails were fixed end to end with no gap, the expansion in hot weather would make them buckle and bend sideways, while contraction in cold weather would set up large thermal stresses that could crack the rails or damage the joints. A small gap left between successive rails gives the metal room to expand freely in summer without buckling, and the gap closes up in winter when the rails contract. This keeps the track safe and undistorted throughout the year.

Real expansion: the actual increase in volume of a liquid on heating, i.e. the expansion when the expansion of the containing vessel is also taken into account.

Apparent expansion: the increase in volume of the liquid that appears to us when it is heated in a vessel, without allowing for the expansion of the vessel.

Coefficients. For a liquid of volume $V_0$ at $0^\circ\text{C}$ heated through $\Delta\theta$:

• Coefficient of real expansion: $\gamma_r = \dfrac{\text{real increase in volume}}{V_0\,\Delta\theta}$
• Coefficient of apparent expansion: $\gamma_a = \dfrac{\text{apparent increase in volume}}{V_0\,\Delta\theta}$

Relation between $\gamma_r$, $\gamma_a$ and $\gamma_g$

Let a vessel contain liquid of volume $V_0$ up to a mark at the initial temperature. On heating through $\Delta\theta$:

• The vessel's internal volume increases by $V_0\gamma_g\Delta\theta$.
• The liquid's volume actually increases by $V_0\gamma_r\Delta\theta$ (real expansion).
• The liquid appears to rise above the mark by the apparent increase, which is the real expansion of the liquid minus the expansion of the vessel:

$V_0\gamma_a\Delta\theta = V_0\gamma_r\Delta\theta - V_0\gamma_g\Delta\theta$

Cancelling $V_0\Delta\theta$,

$\gamma_a = \gamma_r - \gamma_g \quad\Longrightarrow\quad \boxed{\gamma_r = \gamma_a + \gamma_g}$

Thus the coefficient of real expansion of a liquid equals the sum of its coefficient of apparent expansion and the coefficient of volume expansion of the vessel.

Coefficients of expansion of a solid

1. Coefficient of linear expansion ($\alpha$): It is the fractional increase in length per unit rise in temperature. $\alpha = \frac{L - L_0}{L_0\,\Delta\theta} = \frac{\Delta L}{L_0\,\Delta\theta}$
2. Coefficient of superficial (areal) expansion ($\beta$): It is the fractional increase in area per unit rise in temperature. $\beta = \frac{A - A_0}{A_0\,\Delta\theta} = \frac{\Delta A}{A_0\,\Delta\theta}$
3. Coefficient of volume (cubical) expansion ($\gamma$): It is the fractional increase in volume per unit rise in temperature. $\gamma = \frac{V - V_0}{V_0\,\Delta\theta} = \frac{\Delta V}{V_0\,\Delta\theta}$

Relation between $\beta$ and $\alpha$

Consider a thin rectangular sheet of length $L$ and breadth $b$ at temperature $\theta_0$. Let the temperature rise by $\Delta\theta$.

Initial area, $A_0 = L\,b$

Final length $= L(1+\alpha\,\Delta\theta)$, final breadth $= b(1+\alpha\,\Delta\theta)$

Final area, $A = L(1+\alpha\,\Delta\theta)\cdot b(1+\alpha\,\Delta\theta)$ $A = A_0\,(1+\alpha\,\Delta\theta)^2 = A_0\,(1+2\alpha\,\Delta\theta+\alpha^2\Delta\theta^2)$

Since $\alpha$ is very small ($\sim 10^{-5}$), the term $\alpha^2\Delta\theta^2$ is negligible. So $A = A_0\,(1+2\alpha\,\Delta\theta)$ $A - A_0 = 2\alpha\,A_0\,\Delta\theta$ $\frac{A-A_0}{A_0\,\Delta\theta} = 2\alpha$ $\boxed{\beta = 2\alpha}$

Relation between $\gamma$ and $\alpha$

Consider a cube of side $L$ at temperature $\theta_0$. Let the temperature rise by $\Delta\theta$.

Initial volume, $V_0 = L^3$

Final side $= L(1+\alpha\,\Delta\theta)$

Final volume, $V = L^3(1+\alpha\,\Delta\theta)^3 = V_0\,(1+3\alpha\,\Delta\theta+3\alpha^2\Delta\theta^2+\alpha^3\Delta\theta^3)$

Neglecting the very small terms $\alpha^2\Delta\theta^2$ and $\alpha^3\Delta\theta^3$, $V = V_0\,(1+3\alpha\,\Delta\theta)$ $V - V_0 = 3\alpha\,V_0\,\Delta\theta$ $\frac{V-V_0}{V_0\,\Delta\theta} = 3\alpha$ $\boxed{\gamma = 3\alpha}$

Change in volume of the cube and decrease in density

Let the cube of side $a$ at temperature $\theta_0$ be heated to temperature $\theta$, so that $\Delta\theta = \theta - \theta_0$.

Initial volume, $V_0 = a^3$

Final side, $a' = a(1+\alpha\,\Delta\theta)$

Final volume, $V = a'^3 = a^3(1+\alpha\,\Delta\theta)^3$ $V = V_0\,(1+3\alpha\,\Delta\theta) \quad \text{(neglecting small terms)}$

Change in volume: $\Delta V = V - V_0 = V_0\,(1+3\alpha\,\Delta\theta) - V_0$ $\boxed{\Delta V = 3\alpha\,V_0\,\Delta\theta = 3\alpha\,a^3\,\Delta\theta}$

Decrease in density:

Let $m$ be the mass of the cube. Mass does not change on heating, so

Initial density, $\rho_0 = \dfrac{m}{V_0}$

Final density, $\rho = \dfrac{m}{V} = \dfrac{m}{V_0(1+3\alpha\,\Delta\theta)} = \dfrac{\rho_0}{1+3\alpha\,\Delta\theta}$

Expanding and neglecting the small term $3\alpha\,\Delta\theta$ in the denominator (binomial approximation), $\rho = \rho_0\,(1+3\alpha\,\Delta\theta)^{-1} \approx \rho_0\,(1-3\alpha\,\Delta\theta)$

Since $\Delta\theta > 0$ and $\alpha > 0$, the term $3\alpha\,\Delta\theta$ is positive, so $\rho < \rho_0$

Thus the density of the solid decreases on heating. The fractional decrease in density is $\frac{\rho_0-\rho}{\rho_0} = 3\alpha\,\Delta\theta = \gamma\,\Delta\theta$

This shows that the coefficient of volume expansion $\gamma$ also measures the fractional decrease in the density of a solid per unit rise in temperature.

The ends are fixed, so the wire cannot contract. The contraction it would have had is prevented, and the wire is stretched by the same amount, which produces the extra tension.

Thermal strain: $\dfrac{\Delta L}{L}=\alpha\,\Delta T=12\times10^{-6}\times10=1.2\times10^{-4}$

Young's modulus: $Y=\dfrac{F/A}{\Delta L/L}$, so $F=YA\,\alpha\,\Delta T$

Area: $A=\pi\left(\dfrac{4\times10^{-3}}{2}\right)^2=\pi(2\times10^{-3})^2=1.257\times10^{-5}\ \text{m}^2$

$F=2\times10^{11}\times1.257\times10^{-5}\times1.2\times10^{-4}\approx3.0\times10^{2}\ \text{N}$

The tension increases by about $300\ \text{N}$. (The length 8 m cancels out.)

In winter the metal parts contract. Where an outer part fits around an inner one (such as a bearing on a shaft), the clearance between them shrinks or the outer part grips the inner part, so the parts bind and become jammed. In addition, lubricating oil becomes more viscous (thicker) in the cold and no longer reduces friction well.