Temperature scales and thermal expansion

Temperature measures the average kinetic energy of the thermal motion of particles, so measuring it requires a scale with fixed reference points. On the Celsius scale $0\,^\circ\text{C}$ is the melting point of ice and $100\,^\circ\text{C}$ is the boiling point of water at normal pressure, while the absolute Kelvin scale is shifted so that $T = t + 273{,}15$.

The Celsius and Kelvin degrees have the same size: $\Delta T = \Delta t$, so in thermal-expansion formulas a temperature difference may be inserted in either unit. The Fahrenheit scale is defined by $t_F = \frac{9}{5}t + 32$, and for temperature differences $\Delta t_F = \frac{9}{5}\Delta t$.

When a body is heated, the average distances between its particles grow, so all its linear dimensions increase. For a solid over a narrow temperature range, $L = L_0(1 + \alpha \Delta T)$, where $\alpha$ is the coefficient of linear expansion. The volume grows as $V = V_0(1 + \beta \Delta T)$, and for an isotropic solid $\beta \approx 3\alpha$.

Convert $25\,^\circ\text{C}$ to kelvin and to degrees Fahrenheit. What is the temperature difference $\Delta t = 10\,^\circ\text{C}$ in kelvin and in degrees Fahrenheit?

A steel rail at $20\,^\circ\text{C}$ has length $L_0 = 12{,}0$ m and a coefficient of linear expansion $\alpha = 1{,}2\cdot 10^{-5}\ \text{K}^{-1}$. Find $\Delta L$ and the length of the rail at $50\,^\circ\text{C}$.

An aluminium plate ($\alpha = 2{,}3\cdot 10^{-5}\ \text{K}^{-1}$) has a circular hole of diameter $d_0 = 5{,}000$ cm at $20\,^\circ\text{C}$. The plate is heated to $120\,^\circ\text{C}$. Will the hole grow or shrink? Find the new diameter and compare it with the expansion of a solid disc of the same diameter.

Thermal-expansion simulation (no need to heat real metal objects over an open flame).

1. Open an interactive thermal-expansion model (for example PhET or an equivalent) and choose a steel rod with $L_0 = 1{,}00$ m and $\alpha = 1{,}2\cdot 10^{-5}\ \text{K}^{-1}$.
2. Record the length $L$ at $\Delta T = 0,\ 10,\ 20,\ 30,\ 40,\ 50$ K and find $\Delta L$ for each value.
3. Plot $\Delta L$ against $\Delta T$ and check that you get a straight line through the origin with slope $\alpha L_0$.
4. Repeat for aluminium: its $\alpha$ is about twice as large, so the slope should be about twice as steep. Verify this numerically.
5. Switch on the bimetallic strip in the model and describe which way it bends when heated if the $\alpha$ values of the two layers differ.
6. Finish with a 2–3 sentence conclusion: how the slope of the graph is related to $\alpha$ and $L_0$, and why gaps in bridges and rails are not a builder's mistake but a consequence of expansion.

To show that the Celsius and Kelvin scales have the same degree size but different zeros, and that the length of a solid grows linearly with $\Delta T$; to link the slope of the graph to the product $\alpha L_0$ and to explain the bending of a bimetallic strip.

A three-part diagram.

• On the left — three vertical scales ($^\circ$C, K, $^\circ$F) with different zeros: the Celsius and Kelvin degrees are the same size ($\Delta T = \Delta t$), while the Fahrenheit degree is $5/9$ as large as the Celsius degree ($\Delta t_F = \frac{9}{5}\Delta t$); next to them the labels $T = t + 273{,}15$ and $t_F = \frac{9}{5}t + 32$.
• In the centre — a graph of $\Delta L$ against $\Delta T$: two straight lines through the origin, the first with slope $\alpha_{\text{steel}}L_0$ and the second with slope $\alpha_{\text{Al}}L_0$ (the aluminium one is steeper).
• On the right — a sketch of a bimetallic strip: straight at $\Delta T = 0$, and bending into an arc towards the layer with the smaller $\alpha$ when heated.

A steel bridge span of length $L_0 = 200\ \text{m}$ heats up to $+40\ ^\circ\text{C}$ in summer and cools down to $-20\ ^\circ\text{C}$ in winter. The linear expansion coefficient of steel is $\alpha \approx 1.2\times10^{-5}\ \text{K}^{-1}$.

The linear thermal expansion formula $\Delta L = \alpha L_0 \Delta T$ is at work. With $\Delta T = 40 - (-20) = 60\ \text{K}$ the span lengthens by $\Delta L = 200 \cdot 1.2\times10^{-5} \cdot 60 = 0.144\ \text{m}$, that is $14.4\ \text{cm}$; exactly that gap is left by the expansion joint, otherwise huge forces would build up in the junction. Note that the formula uses a temperature difference, and that difference is the same in degrees Celsius and in kelvins.

A metal (aluminium) lid of diameter $d_0 = 8.0\ \text{cm}$ is screwed tightly onto a glass jar. Hot water is poured over the lid, and its temperature rises from $20\ ^\circ\text{C}$ to $80\ ^\circ\text{C}$, that is by $\Delta T = 60\ \text{K}$.

For aluminium $\alpha \approx 2.3\times10^{-5}\ \text{K}^{-1}$, so the same linear expansion formula gives $\Delta d = \alpha d_0 \Delta T = 0.08 \cdot 2.3\times10^{-5} \cdot 60 \approx 1.1\times10^{-4}\ \text{m} = 0.11\ \text{mm}$. Glass expands about $2.5$ times less, so the lid stops gripping the neck — a direct application of the law of linear thermal expansion to solids.

On a frosty day a weather app on a phone shows $-15\ ^\circ\text{C}$, while the board at the airport you are flying to shows $5\ ^\circ\text{F}$. In the meteorological report the same temperature is given on the absolute scale: $T = 258.15\ \text{K}$.

This is the temperature-scale convention at work: $T_F = 1.8\,T_C + 32$ and $T = T_C + 273.15$. Check: $1.8 \cdot (-15) + 32 = 5\ ^\circ\text{F}$ and $-15 + 273.15 = 258.15\ \text{K}$. The reference points are melting ice ($0\ ^\circ\text{C} = 32\ ^\circ\text{F} = 273.15\ \text{K}$) and boiling water at normal pressure ($100\ ^\circ\text{C} = 212\ ^\circ\text{F}$); the Kelvin scale starts at absolute zero.