Exercises — Thermal expansion — linear, area, volumetric

Volume is a 3-dimensional quantity, so it uses the ==volume coefficient $\gamma$==. Why: a solid swells in all three directions, so the linear factor $(1+\alpha \Delta T)$ is cubed; the leading correction is $3\alpha \Delta T$. $\gamma =3\alpha .$

Length growth is $\alpha \Delta T=0.10\%=0.0010$. Area growth is $\beta \Delta T=2\alpha \Delta T$ because ==$\beta =2\alpha$== (two directions). $\frac{\Delta A}{A_0}=2\times 0.10\%=0.20\%.$ Why $\times 2$: each side grows by the same factor, and area is side $\times$ side, so the fractional growth doubles.

The hole gets larger. Why: imagine the hole filled with the same steel — that imaginary plug would expand outward by factor $(1+\alpha \Delta T)$. The real surrounding metal moves in the exact same way, so the boundary of the hole moves outward too. A hole scales up exactly like the solid, with area coefficient $\beta$.

$\Delta T=120-20=100K$ (a difference, identical in ${}^{\circ }C$ and $K$). $\Delta L=L_0\alpha \Delta T=2.00\times 1.7\times 10^{-5}\times 100=3.4\times 10^{-3}\text{m}=3.4\text{mm}.$ Why $\Delta T$ needs no conversion: we only convert to absolute kelvin when a formula needs $T$ itself (gas laws); a change is scale-independent.

Area problem → $\beta =2\alpha =1.8\times 10^{-5}{K}^{-1}$, and $\Delta T=30K$. $\Delta A=A_0\beta \Delta T=1.50\times 1.8\times 10^{-5}\times 30=8.1\times 10^{-4}{\text{m}}^2.$ That is $8.1{\text{cm}}^2$ — small, but real.

$\Delta V=V_0\gamma \Delta T=0.200\times 1.8\times 10^{-4}\times 50=1.8\times 10^{-3}{\text{cm}}^3.$ Why $\gamma$ is already given (not $\alpha$): liquids have no fixed shape, so only volume expansion is meaningful — tables list $\gamma$ directly.

The hole's diameter is a length, so it grows by $\alpha$ (linear), even though a hole "looks like" an area. Any single line across the hole is a length. We need the diameter to grow by $\Delta d=5.010-5.000=0.010\text{cm}$. $\Delta d=d\alpha \Delta T\Rightarrow \Delta T=\frac{\Delta d}{d\alpha }=\frac{0.010}{5.000\times 1.2\times 10^{-5}}.$ $\Delta T=\frac{0.010}{6.0\times 10^{-5}}=166.7K.$ Final temperature $=20+166.7=186.7^{\circ }\text{C}\approx 187^{\circ }\text{C}$.

Both the liquid and the container expand. The overflow is the difference: liquid wants more room than the growing can provides. Container volume coefficient: ${\gamma }_{\text{steel}}=3{\alpha }_{\text{steel}}=3.6\times 10^{-5}{K}^{-1}$. Effective (apparent) coefficient: ${\gamma }_{\text{app}}={\gamma }_{\text{liquid}}-{\gamma }_{\text{container}}=4.9\times 10^{-4}-3.6\times 10^{-5}=4.54\times 10^{-4}{K}^{-1}$. $\Delta V_{\text{overflow}}=V_0{\gamma }_{\text{app}}\Delta T=1000\times 4.54\times 10^{-4}\times 50=22.7{\text{cm}}^3.$ Why subtract: the can's cavity grows too, "absorbing" some liquid growth. Only the excess spills.

If the rod were free, it would stretch by strain $\frac{\Delta L}{L_0}=\alpha \Delta T$. The walls forbid this, so they squeeze the rod back by exactly that strain. Bridge to stress–strain: stress $=Y\times \text{strain}$. $\text{stress}=Y\alpha \Delta T=2.0\times 10^{11}\times 1.2\times 10^{-5}\times 50.$ $\text{stress}=1.2\times 10^8\text{Pa}=120\text{MPa (compressive)}.$ Why $L_0$ cancels: strain is fractional, so the length drops out — thermal stress depends only on material and $\Delta T$, not on how long the rod is.

Period $T=2\pi \sqrt{L/g}$. A small fractional length change gives a fractional period change half as big: $\frac{\Delta T_{\text{period}}}{T}=\frac{1}{2}\frac{\Delta L}{L}=\frac{1}{2}\alpha \Delta T_{\text{temp}}.$ Why the $\frac{1}{2}$: $T\propto \sqrt{L}$, and the square root halves any small fractional change (from $\sqrt{1+x}\approx 1+\frac{x}{2}$). $\frac{\Delta T_{\text{period}}}{T}=\frac{1}{2}\times 1.9\times 10^{-5}\times 15=1.425\times 10^{-4}.$ Longer period → clock runs slow. Seconds lost per day (86400 s): $86400\times 1.425\times 10^{-4}=12.3\text{s lost per day}.$

Here $\alpha \Delta T=4.5\times 10^{-6}\times 3000=0.0135$. Linear form: $L=1.000\times (1+0.0135)=1.013500\text{m}$. Exact form: $L=1.000\times e^{0.0135}=1.013591\text{m}$. Difference $\approx 9.1\times 10^{-5}\text{m}=0.091\text{mm}$, i.e. about $0.0090\%$ of the total expansion. Why so small even at extreme $\Delta T$: the dropped term is $\frac{1}{2}(\alpha \Delta T{)}^2=\frac{1}{2}(0.0135{)}^2\approx 9.1\times 10^{-5}$. Even at $3000K$, $\alpha \Delta T$ is still tiny, so the linear form is safe to $\sim 0.01\%$. The approximation only fails if $\alpha \Delta T$ approaches $1$ — which no solid survives.

With $\gamma <0$, the formula $\Delta V=V_0\gamma \Delta T$ has $\Delta T=1-2=-1K$ (cooling). $\Delta V=V_0\times (\text{negative})\times (\text{negative})=\text{positive}.$ So cooling from $2^{\circ }\text{C}$ to $1^{\circ }\text{C}$ makes the water expand — the opposite of normal. Why it matters: water is densest at $4^{\circ }\text{C}$. As a pond cools, water near $4^{\circ }\text{C}$ sinks; colder, lighter water (0–4°C) floats on top and freezes there. Ice forms on the surface, insulating the liquid below so fish survive. This is the only place in this chapter where a coefficient is negative.

Same $\Delta T$ for both, but brass has the larger $\alpha$, so the brass layer becomes longer than the steel layer. A longer strip on one side and a shorter strip on the other forces a curve: the longer (brass) side bows to the outside, the shorter (steel) side sits on the inside of the arc. Why: to keep both layers bonded while one is longer, the strip must wrap so the long layer traces the bigger radius. This is exactly how a thermostat switch trips a circuit.

Solid: expansion is a tiny fractional effect built on the material's fixed atomic spacing; the natural variable is the change $\Delta T$, and $\gamma =3\alpha \sim 10^{-5}{K}^{-1}$ — the same in $K$ or ${}^{\circ }C$ because it multiplies a difference. Gas: by Gas laws, at constant pressure $V\propto T$ (absolute). Here $V$ is directly proportional to $T$, so doubling absolute $T$ doubles $V$ — a huge effect (${\gamma }_{\text{gas}}=1/T\approx 3.4\times 10^{-3}{K}^{-1}$ at $293K$, ~100× a solid). Because $V$ hits $0$ at $T=0K$, you cannot shift the zero: you must use kelvin, not a difference. The bridge: solids barely change spacing (asymmetric potential, small effect); gas molecules are free-flying (Kinetic theory of gases), so their volume tracks absolute temperature outright.

Which coefficient for a liquid's volume? ::: $\gamma$ (only volume is defined for liquids). A hole's diameter grows with which coefficient? ::: $\alpha$ (a diameter is a length). Overflow when a filled container is heated uses which coefficient? ::: ${\gamma }_{\text{liquid}}-{\gamma }_{\text{container}}$. Thermal stress in a clamped rod depends on its length? ::: No — strain is fractional, so $L_0$ cancels; stress $=Y\alpha \Delta T$. Why does the pendulum error carry a factor $\frac{1}{2}$? ::: Because $T\propto \sqrt{L}$, so the fractional period change is half the fractional length change. When does $L_0(1+\alpha \Delta T)$ fail? ::: Only if $\alpha \Delta T$ nears $1$ — never for real solids.