Question bank — Thermal expansion — linear, area, volumetric

T/F: A hole cut in a metal plate gets smaller when the plate is heated. ::: False. Every point of the plate moves radially outward by the same fraction, so the rim of the hole moves outward too — the hole grows with the very same $\beta$ as solid metal.

T/F: If you double the temperature rise $\Delta T$, you roughly double the stretch $\Delta L$. ::: True (to first order). Since $\Delta L=L_0\alpha \Delta T$ is linear in $\Delta T$, doubling $\Delta T$ doubles $\Delta L$ — as long as $\alpha \Delta T\ll 1$ so the exact $e^{\alpha \Delta T}$ is still close to $1+\alpha \Delta T$.

T/F: A temperature difference of $30{}^{\circ }C$ equals a difference of $30K$. ::: True. The two scales differ only by an offset of $273.15$, and offsets cancel in any difference — so $\Delta T(K)=\Delta T{(}^{\circ }C)$ exactly.

T/F: A solid block, a flat sheet, and a thin rod of the same metal, heated by the same amount, all grow the same fractional amount along any one edge. ::: True. Along a single direction each grows by factor $(1+\alpha \Delta T)$ regardless of shape — the difference is that the block stacks that factor in three directions (volume), the sheet in two (area), the rod in one (length).

T/F: $\beta =2\alpha$ is only an approximation; the exact area factor is $(1+\alpha \Delta T{)}^2$. ::: True. Exactly, $A=A_0(1+\alpha \Delta T{)}^2=A_0(1+2\alpha \Delta T+{\alpha }^2\Delta T^2)$. We drop ${\alpha }^2\Delta T^2\sim 10^{-10}$, leaving $\beta =2\alpha$.

T/F: Water always expands when heated. ::: False. Between $0$ and $4{}^{\circ }C$ water contracts on heating ($\gamma <0$) — the famous anomaly, see Anomalous expansion of water. Above $4{}^{\circ }C$ it behaves normally.

T/F: A gas has a single material constant $\gamma$ just like a solid. ::: False. For an ideal gas at constant pressure the fractional volume change per degree is $1/T$, which depends on the current absolute temperature — it is not a fixed material property. See Gas laws and Kinetic theory of gases.

T/F: Two rods of different metals, clamped end-to-end between fixed walls, will develop the same stress when heated. ::: False. Each has its own $\alpha$, so each wants to expand by a different amount; the thermal stress each develops depends on both its $\alpha$ and its stiffness (Young's modulus) — see Stress and strain.

"The bracket is $L=L_0{e}^{\alpha \Delta T}$, but since $e^x$ is curved and $\alpha \Delta T$ is tiny, we may as well use $L=L_0(1+\alpha \Delta T+\frac{1}{2}{\alpha }^2\Delta T^2)$ for accuracy." ::: The $\frac{1}{2}{\alpha }^2\Delta T^2$ term is of order $10^{-10}$ — utterly negligible and beyond any experimental precision. Keeping it is false precision; $L_0(1+\alpha \Delta T)$ is the honest working form.

"For area we use $\beta =2\alpha$, so for the hole in a plate we should use half of that, since a hole is empty." ::: Wrong. Emptiness has no expansion of its own; the hole tracks the metal around it. Its area uses the full $\beta =2\alpha$, exactly as if the hole were filled with the same metal.

"The rod expands, so its density stays the same — same atoms, same mass, same volume." ::: Mass stays the same, but volume grows, so density falls. Density $=m/V$ and $V$ increases with heating, so a hot object is slightly less dense than the cold one.

"A pendulum clock's brass rod lengthens in summer, so it swings faster and the clock gains time." ::: Backwards. Period $T=2\pi \sqrt{L/g}$ increases with $L$, so each swing takes longer and the clock runs slow. See Pendulum and simple harmonic motion.

"Converting to kelvin, $\Delta T=40+273-(10+273)=323-283=40K$, so I always add 273 before finding $\Delta T$." ::: The arithmetic happens to give the right number, but adding 273 to each temperature is wasted effort — the offsets cancel. Just take $40-10=30$ directly. (Adding 273 only matters for absolute temperatures, e.g. gas laws.)

"Thermal expansion happens because heating adds energy which creates extra material between the atoms." ::: No new material is created. Heating makes atoms vibrate harder, and because the interatomic potential well is asymmetric (steeper on the squeeze side), the average spacing drifts outward — same atoms, more room. See Interatomic potential energy curve.

Why is the deep cause of expansion the asymmetry of the potential well, not just "atoms move more"? ::: If the well were symmetric, harder vibration would swing the atom equally far in and out, leaving the average position unchanged — no expansion. Only because the well is steeper on the compression side does the time-averaged position shift outward.

Why does volume expand three times as fast as length ($\gamma =3\alpha$) rather than adding them up? ::: The three edges each grow by factor $(1+\alpha \Delta T)$, and volume is their product: $(1+\alpha \Delta T{)}^3\approx 1+3\alpha \Delta T$. The "3" is the exponent from three dimensions, not a sum.

Why do we work with the fractional change $\Delta L/L_0$ instead of the raw $\Delta L$? ::: The fraction is the intrinsic material property — the same whether the rod is measured in millimetres or kilometres. Raw $\Delta L$ scales with the object's size and so isn't a clean constant.

Why do bridges and railway tracks need expansion joints? ::: A long span accumulates a real, measurable $\Delta L=L_0\alpha \Delta T$ over the day-night or seasonal temperature swing; without a gap the ends would push into their supports and build up damaging thermal stress — see Stress and strain.

Why can a tight metal lid be loosened by running it under hot water? ::: The metal lid heats faster than the glass and expands more (higher $\alpha$), so its inner rim grows outward, loosening the grip. The hole in the lid gets bigger, not smaller.

Why do gases expand vastly more than solids for the same temperature rise? ::: In a solid, atoms are bound in a deep well and only edge apart slightly. In a gas the molecules are essentially free; raising $T$ raises their kinetic energy directly and, at fixed pressure, volume scales with absolute temperature — see Kinetic theory of gases.

What is $\Delta L$ when $\Delta T=0$? ::: Exactly zero. $\Delta L=L_0\alpha \cdot 0=0$ — no temperature change, no expansion. A useful sanity check on any expansion formula.

A ring is heated. Does the width of the metal (the annulus thickness) shrink to let the hole grow? ::: No. The metal itself expands in every direction, so the annulus gets thicker too. Both the hole and the metal enlarge together by the same fractional factor.

Can $\gamma$ ever be negative, and what does that mean physically? ::: Yes — water from $0$ to $4{}^{\circ }C$ has $\gamma <0$, meaning it shrinks on heating (and expands on cooling). This is why ice forms on top of ponds. See Anomalous expansion of water.

A thin ring is cut open (no longer a closed loop) and heated. Does the gap between the cut ends widen or close? ::: It widens. Every part of the ring, including the two cut faces, moves outward with the general expansion, so the ends separate — the ring behaves like a stretched-out version of itself.

If $\alpha \Delta T$ were not small (say a huge temperature swing), which formula must you use? ::: The exact one, $L=L_0{e}^{\alpha \Delta T}$, because the linear approximation $L_0(1+\alpha \Delta T)$ only holds while $\alpha \Delta T\ll 1$; for large exponents the neglected higher-order terms stop being negligible.

Two identical rods, one heated uniformly and one heated only at one end — same final average temperature. Same total expansion? ::: To first order, yes: total $\Delta L$ depends on the average temperature rise integrated along the rod, so equal average rise gives equal net expansion (ignoring second-order and stress effects). The non-uniform rod, however, carries internal thermal stress.