Chapter 10 · Thermal Properties of Matter

10.1 Temperature, the Ideal-Gas Law and Thermal Expansion

Temperature measures how hot a body is; heat is energy that flows because of a temperature difference. Gases lead us to absolute zero and $pV=nRT$, and almost everything expands when heated: $\Delta L=\alpha L\,\Delta T$.

A doctor reads $38.5\ ^\circ\text{C}$ on a thermometer and knows you have a fever; a weather report from New York says $95\ ^\circ\text{F}$ and means the same kind of heatwave as $35\ ^\circ\text{C}$ in Chennai. Both rely on something so familiar that we rarely question it: a column of liquid in a thin tube that grows longer when it gets hotter. The same expansion that makes thermometers work also buckles railway lines, cracks glass dishes and pops the lids off tight jars when you hold them under hot water.

This section sorts out the basic ideas. We separate temperature from heat, build temperature scales from two fixed points, and see how the pressure of a gas at fixed volume points to a lowest possible temperature — absolute zero — and to the kelvin scale used in every formula of this chapter. Along the way we meet the ideal-gas equation $pV=nRT$. Then we measure thermal expansion: how much a rod lengthens, a plate spreads and a liquid swells for each kelvin, why a bimetallic strip curls, and why water — densest at $4\ ^\circ\text{C}$ — behaves so strangely.

Heat, temperature and temperature scales

Touch a metal railing and a wooden bench on a cold morning: the metal feels colder, although both have been outside all night and are at the same temperature. Your skin is a poor thermometer — it senses how fast energy leaves your hand, not how hot the object is. Physics needs a sharper language.

📘 Definition 10.1 — Temperature and heat

Temperature is a measure of how hot or cold a body is. When two bodies are placed in contact, energy flows from the one at higher temperature to the one at lower temperature until their temperatures become equal; they are then in thermal equilibrium (you will meet this idea formally in Chapter 11).

Heat $Q$ is the energy transferred between a system and its surroundings because of a temperature difference between them. Heat is energy in transit: its SI unit is the joule, $\text{J}$. The older unit, the calorie, is the heat that raises the temperature of $1\ \text{g}$ of water by $1\ ^\circ\text{C}$ (from $14.5$ to $15.5\ ^\circ\text{C}$): $1\ \text{cal}=4.186\ \text{J}$.

A body does not "contain heat", just as a bank account does not contain "deposits": heat describes energy on its way in or out. A spoonful of boiling water is hotter than a bathtub of warm water, yet the bathtub can melt far more ice — temperature and the amount of energy a body can give up are different things.

Thermometers. Any property that changes steadily with temperature can be used to measure it: the length of a mercury or alcohol column, the pressure of a gas held at fixed volume, the electrical resistance of a platinum wire. To turn the property into numbers we choose two reproducible fixed points and divide the interval between them into equal steps. The traditional pair, both at standard atmospheric pressure, are the ice point (pure ice melting) and the steam point (pure water boiling).

📘 Definition 10.2 — Celsius and Fahrenheit scales

On the Celsius scale the ice point is $0\ ^\circ\text{C}$ and the steam point $100\ ^\circ\text{C}$: $100$ equal divisions. On the Fahrenheit scale the ice point is $32\ ^\circ\text{F}$ and the steam point $212\ ^\circ\text{F}$: $180$ equal divisions.

📐 Result 10.1 — Converting between temperature scales

For any two scales that are linear in the same thermometric property, the fraction of the way from the ice point to the steam point is the same:

$$\frac{X-X_\text{ice}}{X_\text{steam}-X_\text{ice}}=\text{same on every scale}.$$

For Celsius ($t_C$) and Fahrenheit ($t_F$):

$$\frac{t_F-32}{180}=\frac{t_C}{100},\qquad t_F=\tfrac95\,t_C+32 .$$

A temperature difference converts without the offset: $\Delta t_F=\tfrac95\,\Delta t_C$, so a change of $1\ ^\circ\text{C}$ equals a change of $1.8\ ^\circ\text{F}$.

✏️ Example 1 — Converting temperatures
(a) Normal human body temperature is often quoted as $98.6\ ^\circ\text{F}$. Express it in degrees Celsius and in kelvin (using $T=t_C+273.15$, explained below). (b) At what temperature do the Celsius and Fahrenheit thermometers show the same number? (c) The temperature in a desert falls by $25\ ^\circ\text{C}$ between noon and midnight. What is this fall in $^\circ\text{F}$ and in kelvin?
(a) $t_C=\tfrac59(t_F-32)=\tfrac59(98.6-32)=\tfrac59(66.6)=37.0\ ^\circ\text{C}$, and $T=37.0+273.15=310.2\ \text{K}$.
(b) Put $t_F=t_C=x$: $x=\tfrac95x+32$, so $-\tfrac45x=32$ and $x=-40$. Both scales read $-40$ at the same temperature.
(c) A difference: $\Delta t_F=1.8\times25=45\ ^\circ\text{F}$. The kelvin and the Celsius degree are the same size, so the fall is $25\ \text{K}$.
Check $t_F=1.8(-40)+32=-72+32=-40$ ✓. Notice that in (c) adding $32$ would be wrong: the offset belongs to temperatures, not to differences.
✏️ Example 2 — A badly calibrated thermometer
A thermometer has a uniform bore, but its scale was printed carelessly: it reads $-1\ ^\circ$ in melting ice and $103\ ^\circ$ in steam at standard pressure. (a) What is the true Celsius temperature when it reads $51\ ^\circ$? (b) At what temperature does it give the correct reading?
Idea A uniform bore means the reading $X$ is linear in the true temperature. Between the fixed points it spans $103-(-1)=104$ of its own divisions for $100$ true degrees, so $\dfrac{X-(-1)}{104}=\dfrac{t_C}{100}$.
(a) $t_C=\dfrac{100\,(51+1)}{104}=\dfrac{5200}{104}=50.0\ ^\circ\text{C}$.
(b) Put $X=t_C=x$: $100(x+1)=104x$, so $4x=100$ and $x=25\ ^\circ\text{C}$.
Check At $25\ ^\circ\text{C}$ the faulty reading is $-1+1.04\times25=25.0$ ✓. Below $25\ ^\circ\text{C}$ it reads low, above it reads high. Set $25\ ^\circ\text{C}$ in Interactive 1 and watch the fourth thermometer agree with the Celsius one.
Interactive 1 — Four Thermometers, One Temperature

Four identical liquid-in-glass thermometers sit in the same bath, so their columns are always the same height; only the printed scales differ — Celsius, Fahrenheit, kelvin and the faulty scale of Example 2. Drag the bath temperature or pick a preset; the dashed lines mark the ice and steam points on every scale. Press Play to sweep the temperature up and down.

37.0 °C
Celsius—
Fahrenheit—
kelvin—
faulty scale—

The ideal-gas equation and absolute temperature

A mercury thermometer and an alcohol thermometer, both calibrated at $0$ and $100\ ^\circ\text{C}$, disagree slightly at $50\ ^\circ\text{C}$, because the two liquids do not expand in exactly the same way. Gases are far better behaved. At low density, every gas obeys the same simple laws, found experimentally in the seventeenth and eighteenth centuries.

📐 Law 10.2 — Boyle's law and Charles' law

For a fixed amount of a gas at low density:

  • Boyle's law: at constant temperature, $pV=\text{constant}$ — halve the volume and the pressure doubles.
  • Charles' law: at constant pressure, the volume increases linearly with the temperature; every gas, extrapolated, would reach zero volume at the same temperature, about $-273\ ^\circ\text{C}$.

Similarly, at constant volume the pressure increases linearly with temperature, and every gas extrapolates to zero pressure at the same temperature.

The constant-volume gas thermometer. Trap a little gas in a bulb, keep its volume fixed with a mercury column, and record its pressure at several temperatures. The points lie on a straight line. Do it again with a different gas, or with less gas in the bulb: you get different straight lines, but at low pressures all of them extrapolate to $p=0$ at the same temperature, $-273.15\ ^\circ\text{C}$. Nothing can be colder than the temperature at which a gas would exert no pressure at all. This natural zero is the starting point of the absolute scale.

📘 Definition 10.3 — Absolute zero and the kelvin scale

Absolute zero is the lowest possible temperature, $-273.15\ ^\circ\text{C}$. The absolute (kelvin) temperature $T$ is measured from it in steps the same size as the Celsius degree:

$$T=t_C+273.15 ,$$

with the SI unit the kelvin, $\text{K}$ (no degree sign). The scale is fixed by assigning the triple point of water — the single temperature and pressure at which ice, liquid water and water vapour coexist — the value $273.16\ \text{K}$ exactly ($0.01\ ^\circ\text{C}$). A constant-volume gas thermometer then measures

$$T=273.16\ \text{K}\times\frac{p}{p_\text{tp}},$$

where $p_\text{tp}$ is the gas pressure at the triple point (in the limit of low gas density).

Absolute temperature is never negative, and a temperature difference is the same number in kelvin and in Celsius degrees. Inside any formula that contains $T$ itself — a ratio of temperatures, $pV=nRT$, $T^4$ in Section 10.3 — always use kelvin.

Combining Boyle's and Charles' laws, $\dfrac{pV}{T}$ is constant for a fixed amount of gas, and experiment shows that the constant is proportional to the number of moles. This gives one equation for every gas at low density.

📐 Result 10.3 — The ideal-gas equation

For $n$ moles of a gas at low density, with pressure $p$, volume $V$ and absolute temperature $T$,

$$pV=nRT ,$$

where $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant and $n=\dfrac{m}{M}$ (mass divided by molar mass). A gas that obeys this equation exactly is an ideal gas; real gases come close at low pressure and high temperature, far from liquefying. For a fixed amount of gas, $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$.

Chapter 12 explains why gases obey this law, from the motion of their molecules. For now it is a well-tested experimental fact — and a very useful one.

✏️ Example 3 — Reading a gas thermometer
A constant-volume gas thermometer registers a pressure of $6.80\times10^4\ \text{Pa}$ when its bulb is at the triple point of water, and $9.20\times10^4\ \text{Pa}$ when the bulb is in a beaker of hot oil. Find the temperature of the oil in kelvin and in degrees Celsius.
Idea At constant volume $p\propto T$, and the scale is anchored at $273.16\ \text{K}$.
Kelvin $T=273.16\times\dfrac{9.20\times10^4}{6.80\times10^4}=273.16\times1.353=369.6\ \text{K}$.
Celsius $t_C=369.6-273.15=96.4\ ^\circ\text{C}$.
Check The pressure is about $35\%$ higher, so the absolute temperature is about $35\%$ higher: $1.35\times273\approx370\ \text{K}$ ✓. Using $^\circ\text{C}$ in the ratio would be nonsense — the triple point is $0.01\ ^\circ\text{C}$, and the ratio would explode.
✏️ Example 4 — A gas cylinder in the sun
A rigid steel cylinder of volume $0.0500\ \text{m}^3$ holds gas at $27\ ^\circ\text{C}$ and $2.00\times10^5\ \text{Pa}$. (a) How many moles of gas does it contain? (b) Left in the sun, the cylinder warms to $87\ ^\circ\text{C}$. Find the new pressure. (c) Its safety valve opens whenever the pressure exceeds $2.20\times10^5\ \text{Pa}$. What fraction of the gas has escaped once the cylinder has settled at $87\ ^\circ\text{C}$?
(a) $T_1=300\ \text{K}$. $\ n=\dfrac{pV}{RT}=\dfrac{2.00\times10^5\times0.0500}{8.31\times300}=4.01\ \text{mol}$.
(b) Fixed $n$ and $V$: $\dfrac{p_2}{p_1}=\dfrac{T_2}{T_1}=\dfrac{360}{300}$, so $p_2=2.40\times10^5\ \text{Pa}$.
(c) The valve holds the pressure at $2.20\times10^5\ \text{Pa}$; at fixed $V$ and $T$, $n\propto p$, so the gas left is $\dfrac{2.20}{2.40}=0.917$ of what would have been there. Fraction escaped $=1-0.917=0.083$, about $8.3\%$.
Check Moles left $=\dfrac{2.20\times10^5\times0.0500}{8.31\times360}=3.68\ \text{mol}$, and $\dfrac{4.01-3.68}{4.01}=0.083$ ✓.

Thermal expansion of solids

In a solid the atoms vibrate about their positions in a lattice. The force between neighbours is not symmetric: it is much harder to push two atoms together than to pull them apart. As the temperature rises the vibrations grow, the atoms spend more time on the "far" side, and the average spacing increases. Multiplied over billions of atoms in a row, the tiny increase in spacing becomes a measurable growth in length — proportional, for moderate changes, to the original length and to the rise in temperature.

📘 Definition 10.4 — Coefficients of linear, area and volume expansion

When the temperature of a body changes by a small amount $\Delta T$, its length $L$, area $A$ and volume $V$ change by

$$\Delta L=\alpha L\,\Delta T,$$ $$\Delta A=\beta A\,\Delta T,$$ $$\Delta V=\gamma V\,\Delta T,$$

which define the coefficients of linear expansion $\alpha=\dfrac{\Delta L}{L\,\Delta T}$, area expansion $\beta$ and volume expansion $\gamma$. Each has unit $\text{K}^{-1}$ (the same as $(^\circ\text{C})^{-1}$, since only a temperature difference appears). They depend a little on temperature; the values below are for room temperature.

Solid$\alpha$ ($10^{-5}\ \text{K}^{-1}$)Solid$\alpha$ ($10^{-5}\ \text{K}^{-1}$)
aluminium2.3steel1.2
brass1.9soda glass0.90
copper1.7invar (Fe–Ni alloy)0.12

Liquids have no fixed shape, so only $\gamma$ matters for them, and it is larger: about $2.1\times10^{-4}\ \text{K}^{-1}$ for water at $20\ ^\circ\text{C}$, $1.8\times10^{-4}\ \text{K}^{-1}$ for mercury and $1.1\times10^{-3}\ \text{K}^{-1}$ for ethanol.

📐 Result 10.4 — Area and volume coefficients of an isotropic solid

For a solid that expands equally in all directions (isotropic), and for $\alpha\,\Delta T\ll1$,

$$\beta=2\alpha,\qquad \gamma=3\alpha .$$

Every length of the body — including the diameter of a hole cut in it — scales by the same factor $1+\alpha\,\Delta T$.

Why. A cube of side $L$ heated by $\Delta T$ has side $L(1+\alpha\Delta T)$ and volume $L^3(1+\alpha\Delta T)^3$, which multiplies out to $L^3\big(1+3\alpha\Delta T$ $+\,3\alpha^2\Delta T^2+\alpha^3\Delta T^3\big)$. With $\alpha\Delta T\sim10^{-3}$ the squared and cubed terms are a thousand times smaller and are dropped, leaving $\Delta V=3\alpha V\Delta T$. The same argument for a square gives $\beta=2\alpha$. A hole expands exactly as the material that would fill it — a heated washer gets a bigger hole, which is why a tight metal lid loosens under hot water.

✏️ Example 5 — Rail gaps and a shrink fit
(a) Steel rails, each $25.0\ \text{m}$ long, are laid when the temperature is $10\ ^\circ\text{C}$. The hottest the rails get is $45\ ^\circ\text{C}$. What gap must be left between neighbouring rails? (b) A steel ring has an inner diameter of $4.990\ \text{cm}$ at $20\ ^\circ\text{C}$. To what temperature must it be heated to slip over a steel shaft of diameter $5.000\ \text{cm}$ (kept at $20\ ^\circ\text{C}$)? Take $\alpha_\text{steel}=1.2\times10^{-5}\ \text{K}^{-1}$.
(a) $\Delta L=\alpha L\Delta T=1.2\times10^{-5}\times25.0\times35=1.05\times10^{-2}\ \text{m}$: leave a gap of at least $10.5\ \text{mm}$ — in practice about $1\ \text{cm}$ or a little more.
(b) The hole's diameter grows like any length: $\Delta d=\alpha d\,\Delta T$, and we need $\Delta d=0.010\ \text{cm}$. $\ \Delta T=\dfrac{0.010}{1.2\times10^{-5}\times4.990}=167\ \text{K}$, so heat the ring to $20+167\approx1.9\times10^2\ ^\circ\text{C}$.
Check Strain $\alpha\Delta T=1.2\times10^{-5}\times167=2.0\times10^{-3}$, and $2.0\times10^{-3}\times4.990\ \text{cm}=0.0100\ \text{cm}$ ✓.
Why it works When the ring cools it tries to shrink back to $4.990\ \text{cm}$ and grips the shaft with a huge force — the way railway wheels' steel tyres and gear rings are fitted without bolts.

Thermal stress. If a rod is prevented from expanding — welded rails, a pipe fixed between two walls, a concrete slab without joints — the walls must squeeze it back by exactly the length it would have gained. The compressive strain is $\dfrac{\Delta L}{L}=\alpha\Delta T$, and by Hooke's law (Section 8.2) the stress is Young's modulus times that strain.

📐 Result 10.5 — Thermal stress in a clamped rod

A rod of Young's modulus $Y$, cross-section $A$ and expansion coefficient $\alpha$, held between rigid supports and heated by $\Delta T$ (within the proportional limit), carries a compressive thermal stress and pushes on the supports with force

$$\sigma=Y\alpha\,\Delta T,\qquad F=YA\alpha\,\Delta T .$$

Both are independent of the rod's length. Cooling a rod fixed at both ends gives a tensile stress of the same size.

✏️ Example 6 — Continuous welded rail
Modern high-speed lines use rails welded into lengths of several kilometres with no gaps; the rails are clamped so that they cannot expand. A rail of cross-section $7.5\times10^{-3}\ \text{m}^2$ is clamped at $20\ ^\circ\text{C}$. Take $Y=2.0\times10^{11}\ \text{Pa}$ and $\alpha=1.2\times10^{-5}\ \text{K}^{-1}$. Find (a) the stress and the force on the clamps on a day when the rail reaches $50\ ^\circ\text{C}$, and (b) the stress on a winter night at $5\ ^\circ\text{C}$. Steel of this kind yields at about $2.5\times10^8\ \text{Pa}$; is the rail safe?
(a) $\sigma=Y\alpha\Delta T=2.0\times10^{11}\times1.2\times10^{-5}\times30=7.2\times10^7\ \text{Pa}$, compressive. $\ F=\sigma A=7.2\times10^7\times7.5\times10^{-3}=5.4\times10^5\ \text{N}$ — the weight of about $55$ tonnes.
(b) $\Delta T=-15\ \text{K}$: $\sigma=2.0\times10^{11}\times1.2\times10^{-5}\times15=3.6\times10^7\ \text{Pa}$, now tensile.
Safe? Both are well below $2.5\times10^8\ \text{Pa}$, so the steel is not overstressed. The real danger is different: a long rail under large compression can buckle sideways, so engineers clamp it firmly to heavy sleepers and lay it at a "neutral" temperature near the middle of the yearly range.
Check Dimensions: $[\text{Pa}][\text{K}^{-1}][\text{K}]=[\text{Pa}]$ ✓, and the length of the rail never entered — exactly as Result 10.5 says.

The bimetallic strip. Rivet a strip of brass to a strip of steel of the same length. On heating, brass ($\alpha=1.9\times10^{-5}\ \text{K}^{-1}$) tries to grow more than steel ($1.2\times10^{-5}\ \text{K}^{-1}$), but the rivets keep them together, so the pair curls into an arc with brass on the outside (the longer, convex side). Cooled below the riveting temperature it curls the other way. A bimetallic strip that bends to open or close an electrical contact is the thermostat in irons, kettles, room heaters and the flashing indicator of older cars; a coiled one turns the pointer of a dial thermometer.

Interactive 2 — Thermal Expansion Lab

Choose an experiment. Rod: a $1.000\ \text{m}$ rod at $20\ ^\circ\text{C}$ is heated by $\Delta T$; the growth (drawn magnified, in red) is $\alpha L\Delta T$, and the last readout is the stress if the rod were clamped. Bimetallic strip: a $10\ \text{cm}$ brass–steel strip, each layer $0.50\ \text{mm}$ thick, curls as the temperature changes (bending exaggerated $\times3$, simple geometric model $R=d/(\Delta\alpha\,\Delta T)$). Water: the density of water from $0$ to $10\ ^\circ\text{C}$ (a smooth fit to measured values) — watch where it peaks. Press Play to heat and cool.

+100 K
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——
——

Liquids, gases and the anomalous expansion of water

When a body of fixed mass $m$ expands, its density $\rho=m/V$ falls. Since $V$ becomes $V(1+\gamma\Delta T)$,

$$\rho=\frac{\rho_0}{1+\gamma\,\Delta T}\approx\rho_0\left(1-\gamma\,\Delta T\right),$$

the approximation holding when $\gamma\Delta T\ll1$. Warm water rising above cold, hot air rising in a chimney and the convection currents of Section 10.3 all start here. For a gas the expansion is much larger, and the ideal-gas equation gives it at once.

📐 Result 10.6 — Volume expansion of an ideal gas

At constant pressure $V=\dfrac{nR}{p}T$, so $\Delta V=\dfrac{nR}{p}\Delta T=\dfrac{V}{T}\Delta T$ and the coefficient of volume expansion is

$$\gamma=\frac1T .$$

It is not a constant: $3.66\times10^{-3}\ \text{K}^{-1}$ at $0\ ^\circ\text{C}$, $3.33\times10^{-3}\ \text{K}^{-1}$ at $27\ ^\circ\text{C}$ — some ten times that of a liquid and a hundred times that of a solid.

The anomalous expansion of water. Water breaks the rule. Heated from $0\ ^\circ\text{C}$, it contracts until about $4\ ^\circ\text{C}$ and only then begins to expand; its density is greatest, about $1000\ \text{kg/m}^3$, at $4\ ^\circ\text{C}$ (more precisely $3.98\ ^\circ\text{C}$). Ice is less dense still, about $917\ \text{kg/m}^3$, so water expands by about $9\%$ when it freezes. The reason is the open, hydrogen-bonded structure of ice: some of that structure survives in cold liquid water, and melting it lets the molecules pack more closely — an effect that wins over normal thermal expansion just above $0\ ^\circ\text{C}$.

The consequences are large. As a lake cools in winter, the surface water cools, becomes denser and sinks, until the whole lake is at $4\ ^\circ\text{C}$. Further cooling makes the surface water lighter, so it stays on top, reaches $0\ ^\circ\text{C}$ and freezes. The ice floats and insulates the water below, which stays near $4\ ^\circ\text{C}$ at the bottom: lakes freeze from the top down, and fish survive the winter. The same expansion on freezing bursts water pipes and splits rocks in cold mountains.

⚠️ Common mistakes

1. Celsius inside a ratio. $\dfrac{p_2}{p_1}=\dfrac{T_2}{T_1}$ and $pV=nRT$ need kelvin. Heating a gas from $10\ ^\circ\text{C}$ to $20\ ^\circ\text{C}$ does not double its pressure — it raises it by $\dfrac{10}{283}\approx3.5\%$.

2. Offsets on differences. A rise of $10\ ^\circ\text{C}$ is a rise of $10\ \text{K}$ and $18\ ^\circ\text{F}$ — never add $273$ or $32$ to a difference.

3. Holes shrinking. On heating, a hole in a plate grows, exactly as if it were filled with the same material.

4. $\beta=\alpha^2$ or $\gamma=\alpha^3$. The coefficients add, they do not multiply: $\beta=2\alpha$, $\gamma=3\alpha$ for an isotropic solid.

5. Length in thermal stress. $\sigma=Y\alpha\Delta T$ does not depend on the rod's length or thickness; the force $YA\alpha\Delta T$ depends on the area only.

6. "Water always expands when heated." Not between $0$ and $4\ ^\circ\text{C}$.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — apparent expansion of a liquid in a container

In JEE Main, NEET and JEE Advanced (as part of thermal expansion of liquids); not required by the CBSE syllabus.

A liquid is always held in a vessel, and the vessel expands too. What you see — the rise in a thermometer column or the overflow from a full flask — is the apparent expansion, with coefficient

$$\gamma_\text{app}=\gamma_\text{liquid}-\gamma_\text{vessel},$$

where $\gamma_\text{vessel}=3\alpha_\text{vessel}$ for a solid container. The overflow from a vessel filled to the brim with volume $V$ is $V\gamma_\text{app}\,\Delta T$.

Example. A soda-glass flask holds exactly $500\ \text{cm}^3$ of mercury at $20\ ^\circ\text{C}$. Heated to $100\ ^\circ\text{C}$: $\gamma_\text{app}=1.8\times10^{-4}$ $-\,3\times0.90\times10^{-5}$ $=1.53\times10^{-4}\ \text{K}^{-1}$, and the overflow is $500\times1.53\times10^{-4}\times80$ $=6.1\ \text{cm}^3$. Ignoring the flask's expansion would give $7.2\ \text{cm}^3$. A mercury thermometer works only because $\gamma_\text{Hg}$ is about seven times $\gamma_\text{glass}$.

Exercises

Use $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$, $T=t_C+273$ unless more figures are needed, $1\ \text{atm}=1.013\times10^5\ \text{Pa}$, and the expansion coefficients in the table of this section.

I
Temperature scales.
  • 1 Convert $25\ ^\circ\text{C}$ to degrees Fahrenheit and to kelvin.
  • 2 Paper catches fire at about $451\ ^\circ\text{F}$. Express this in degrees Celsius.
  • 3 Liquid nitrogen boils at $77\ \text{K}$ at atmospheric pressure. Express this in $^\circ\text{C}$ and $^\circ\text{F}$.
  • 4 The temperature of a room rises by $18\ ^\circ\text{F}$. By how many degrees Celsius, and by how many kelvin, has it risen?
  • 5 At what temperature is the Fahrenheit reading five times the Celsius reading?
  • 6 A thermometer with its own scale $X$ reads $20$ at the ice point and $180$ at the steam point. (a) What is the Celsius temperature when it reads $60$? (b) What does it read at $60\ ^\circ\text{C}$?
  • 7 The mercury column of an unmarked thermometer is $4.0\ \text{cm}$ long in melting ice and $24.0\ \text{cm}$ long in steam. Find the temperature when the column is $10.0\ \text{cm}$ long.
  • 8 Explain why a temperature can be negative on the Celsius scale but not on the kelvin scale, and why a temperature difference has the same value in both.
1 $t_F=1.8(25)+32=77\ ^\circ\text{F}$; $\ T=25+273.15=298\ \text{K}$ ($298.15$).
2 $t_C=\tfrac59(451-32)=\tfrac59(419)=233\ ^\circ\text{C}$.
3 $t_C=77-273.15=-196\ ^\circ\text{C}$; $\ t_F=1.8(-196.15)+32=-321\ ^\circ\text{F}$.
4 A difference: $\Delta t_C=\dfrac{18}{1.8}=10\ ^\circ\text{C}$, which is also $10\ \text{K}$.
5 $1.8t+32=5t$ gives $3.2t=32$, $t=10\ ^\circ\text{C}$ ($=50\ ^\circ\text{F}$).
6 $\dfrac{X-20}{160}=\dfrac{t_C}{100}$. (a) $t_C=\dfrac{100(60-20)}{160}=25\ ^\circ\text{C}$. (b) $X=20+0.60\times160=116$.
7 Fraction of the way from ice to steam: $\dfrac{10.0-4.0}{24.0-4.0}=0.30$, so $t=30\ ^\circ\text{C}$ (assuming the column length is linear in temperature).
8 The Celsius zero is just a convenient reference (melting ice), and many things are colder. The kelvin zero is absolute zero, the lowest temperature possible, so nothing has $T\lt0$. Both scales use the same size of degree and differ only by a constant, $273.15$, which cancels in a difference.
II
The ideal-gas equation and the gas thermometer.
  • 1 Find the volume occupied by $2.0\ \text{mol}$ of an ideal gas at $27\ ^\circ\text{C}$ and $1.013\times10^5\ \text{Pa}$.
  • 2 A rigid oxygen cylinder is filled to $1.50\times10^7\ \text{Pa}$ at $20\ ^\circ\text{C}$. What is its pressure when it is left in a hot van at $50\ ^\circ\text{C}$?
  • 3 A balloon holds $3.0\ \text{L}$ of air at $27\ ^\circ\text{C}$. It is dipped in liquid nitrogen at $77\ \text{K}$, the pressure staying constant. Find its new volume (assume the air stays a gas).
  • 4 Estimate the mass of air in a classroom $5.0\ \text{m}\times4.0\ \text{m}\times3.0\ \text{m}$ at $27\ ^\circ\text{C}$ and $1.013\times10^5\ \text{Pa}$. The molar mass of air is $0.029\ \text{kg/mol}$.
  • 5 A constant-volume gas thermometer reads $5.00\times10^4\ \text{Pa}$ at the triple point of water and $7.30\times10^4\ \text{Pa}$ at the boiling point of a liquid. Find the boiling point in K and $^\circ\text{C}$.
  • 6 (Modelling) An air bubble of volume $1.0\ \text{cm}^3$ is released at the bottom of a lake $20\ \text{m}$ deep, where the water is at $7\ ^\circ\text{C}$. Find its volume just below the surface, where the water is at $27\ ^\circ\text{C}$. The pressure at depth $h$ is $p_0+\rho gh$, with $p_0=1.013\times10^5\ \text{Pa}$, $\rho=1.0\times10^3\ \text{kg/m}^3$ and $g=9.8\ \text{m/s}^2$.
  • 7 Different gases in a constant-volume gas thermometer give slightly different readings for the same temperature at ordinary pressures. How is a single, gas-independent temperature obtained?
1 $V=\dfrac{nRT}{p}=\dfrac{2.0\times8.31\times300}{1.013\times10^5}=4.9\times10^{-2}\ \text{m}^3$ (about $49\ \text{L}$).
2 $p_2=p_1\dfrac{T_2}{T_1}=1.50\times10^7\times\dfrac{323}{293}=1.65\times10^7\ \text{Pa}$.
3 $V_2=V_1\dfrac{T_2}{T_1}=3.0\times\dfrac{77}{300}=0.77\ \text{L}$.
4 $V=60\ \text{m}^3$; $n=\dfrac{1.013\times10^5\times60}{8.31\times300}=2.44\times10^3\ \text{mol}$; $m=nM=2438\times0.029=71\ \text{kg}$ — heavier than most students in the room.
5 $T=273.16\times\dfrac{7.30}{5.00}=398.8\ \text{K}$, i.e. $125.7\ ^\circ\text{C}$.
6 $p_1=1.013\times10^5+1.0\times10^3\times9.8\times20=2.97\times10^5\ \text{Pa}$, $T_1=280\ \text{K}$; $p_2=1.013\times10^5\ \text{Pa}$, $T_2=300\ \text{K}$. $\ V_2=V_1\dfrac{p_1T_2}{p_2T_1}=1.0\times\dfrac{2.973\times300}{1.013\times280}=3.1\ \text{cm}^3$.
7 Measure with less and less gas in the bulb (lower $p_\text{tp}$) and extrapolate the temperature reading to zero gas pressure. In that limit every gas behaves ideally and all give the same value — the ideal-gas temperature, which agrees with the kelvin scale.
III
Expansion of solids and thermal stress.
  • 1 An aluminium rod is $2.00\ \text{m}$ long at $20\ ^\circ\text{C}$. By how much does it lengthen at $120\ ^\circ\text{C}$?
  • 2 A steel bridge is $300\ \text{m}$ long. Its temperature ranges from $-10\ ^\circ\text{C}$ to $45\ ^\circ\text{C}$ over the year. How much room must its expansion joints allow?
  • 3 A brass plate has a circular hole of diameter $10.00\ \text{cm}$ at $20\ ^\circ\text{C}$. Find the diameter of the hole at $220\ ^\circ\text{C}$.
  • 4 A copper sheet of area $0.50\ \text{m}^2$ is heated through $80\ \text{K}$. Find the increase in its area.
  • 5 A steel ball has radius $2.00\ \text{cm}$. Find its volume and the increase in its volume when it is heated through $100\ \text{K}$.
  • 6 A steel measuring tape is correct at $25\ ^\circ\text{C}$. On a cold day at $5\ ^\circ\text{C}$ it gives the length of a rod as $85.00\ \text{cm}$. What is the actual length of the rod? (Treat the rod's own expansion as negligible.)
  • 7 A steel ring has an inner diameter of $7.980\ \text{cm}$ at $20\ ^\circ\text{C}$. To what temperature must it be heated to slide onto a shaft of diameter $8.000\ \text{cm}$ at $20\ ^\circ\text{C}$?
  • 8 A brass rod of cross-section $2.0\ \text{cm}^2$ is fixed between two rigid walls and heated through $40\ \text{K}$. Taking $Y_\text{brass}=1.0\times10^{11}\ \text{Pa}$, find the thermal stress and the force on each wall.
1 $\Delta L=2.3\times10^{-5}\times2.00\times100=4.6\times10^{-3}\ \text{m}=4.6\ \text{mm}$.
2 $\Delta L=1.2\times10^{-5}\times300\times55=0.198\ \text{m}\approx0.20\ \text{m}$.
3 The hole expands like brass: $\Delta d=1.9\times10^{-5}\times10.00\times200=0.038\ \text{cm}$, so $d=10.04\ \text{cm}$.
4 $\Delta A=2\alpha A\Delta T=2\times1.7\times10^{-5}\times0.50\times80=1.4\times10^{-3}\ \text{m}^2$ ($14\ \text{cm}^2$).
5 $V=\tfrac43\pi(2.00)^3=33.5\ \text{cm}^3$; $\ \Delta V=3\alpha V\Delta T=3.6\times10^{-5}\times33.5\times100=0.121\ \text{cm}^3$.
6 At $5\ ^\circ\text{C}$ each "centimetre" of the tape has shrunk to $1-\alpha(20)$ of a true centimetre, so the tape reads too high. True length $=85.00\times(1-1.2\times10^{-5}\times20)=84.98\ \text{cm}$.
7 $\Delta T=\dfrac{0.020}{1.2\times10^{-5}\times7.980}=209\ \text{K}$: heat to about $229\ ^\circ\text{C}\approx2.3\times10^2\ ^\circ\text{C}$.
8 $\sigma=Y\alpha\Delta T=1.0\times10^{11}\times1.9\times10^{-5}\times40=7.6\times10^7\ \text{Pa}$; $F=\sigma A=7.6\times10^7\times2.0\times10^{-4}=1.5\times10^4\ \text{N}$.
IV
Liquids, gases, water and applications.
  • 1 A car's fuel tank is filled with $50\ \text{L}$ of petrol at $15\ ^\circ\text{C}$ ($\gamma=9.5\times10^{-4}\ \text{K}^{-1}$). The car is parked in the sun and the petrol warms to $35\ ^\circ\text{C}$. Ignoring the expansion of the tank, how much petrol overflows if the tank was full?
  • 2 Mercury has density $1.360\times10^4\ \text{kg/m}^3$ at $0\ ^\circ\text{C}$. Find its density at $100\ ^\circ\text{C}$.
  • 3 Starting from $pV=nRT$, show that an ideal gas at constant pressure has $\gamma=1/T$, and evaluate it at $27\ ^\circ\text{C}$.
  • 4 In a very cold winter, the surface of a deep lake is frozen. What is the temperature (a) just under the ice, (b) at the bottom of the lake? Explain.
  • 5 A sealed bottle full of water is warmed from $0\ ^\circ\text{C}$ to $8\ ^\circ\text{C}$. Describe how the volume of the water changes along the way, and compare its volume at $8\ ^\circ\text{C}$ with that at $0\ ^\circ\text{C}$ (use Interactive 2).
  • 6 A straight brass–steel bimetallic strip is heated. Which metal ends up on the convex side? What happens if the strip is cooled instead? Why is invar useful in clock parts and measuring tapes?
  • 7 (Modelling) Boiling water poured into a cold, thick soda-glass tumbler can crack it, but a borosilicate (Pyrex) dish with $\alpha=3.2\times10^{-6}\ \text{K}^{-1}$ survives. Suppose the inner surface heats by $80\ \text{K}$ while the outer surface stays cold. Estimate the thermal strain and stress in the inner layer for each glass ($Y\approx6.5\times10^{10}\ \text{Pa}$ for both), and compare with the tensile strength of ordinary glass, roughly $5\times10^7\ \text{Pa}$.
1 $\Delta V=\gamma V\Delta T=9.5\times10^{-4}\times50\times20=0.95\ \text{L}$. (A real steel tank expands a little too, so slightly less spills.)
2 $\rho=\dfrac{\rho_0}{1+\gamma\Delta T}=\dfrac{1.360\times10^4}{1+1.8\times10^{-4}\times100}=\dfrac{1.360\times10^4}{1.018}=1.336\times10^4\ \text{kg/m}^3$.
3 $V=\dfrac{nR}{p}T$, so $\dfrac{\Delta V}{\Delta T}=\dfrac{nR}{p}=\dfrac VT$ and $\gamma=\dfrac{1}{V}\dfrac{\Delta V}{\Delta T}=\dfrac1T$. At $300\ \text{K}$: $\gamma=3.33\times10^{-3}\ \text{K}^{-1}$.
4 (a) $0\ ^\circ\text{C}$ — water in contact with melting/freezing ice. (b) About $4\ ^\circ\text{C}$: water at $4\ ^\circ\text{C}$ is densest and sinks to the bottom; colder water is lighter and stays on top, where it freezes.
5 The volume decreases from $0$ to about $4\ ^\circ\text{C}$ (density rising to its maximum) and then increases. In the fit of Interactive 2 the density at $8\ ^\circ\text{C}$ equals that at $0\ ^\circ\text{C}$ ($999.84\ \text{kg/m}^3$), so the volume ends up almost exactly where it started; measured values agree to within about $0.01\ \text{kg/m}^3$.
6 Brass ($\alpha=1.9\times10^{-5}$) grows more than steel ($1.2\times10^{-5}$), so brass is on the outside (convex side) of the curve. Cooled, brass shrinks more and the strip curls with steel on the convex side. Invar's $\alpha$ is about ten times smaller than steel's, so lengths made of it hardly change with temperature.
7 Soda glass: strain $\alpha\Delta T=9.0\times10^{-6}\times80=7.2\times10^{-4}$, stress up to $Y\alpha\Delta T\approx4.7\times10^7\ \text{Pa}$ — about the breaking stress, so it may crack. Pyrex: $3.2\times10^{-6}\times80\times6.5\times10^{10}\approx1.7\times10^7\ \text{Pa}$, about a third of the strength: it survives. (The estimate treats the hot layer as fully restrained, an upper bound.)
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks What is meant by the anomalous expansion of water? Use it to explain why aquatic animals survive in a lake whose surface has frozen.

Meaning Unlike most liquids, water contracts when heated from $0\ ^\circ\text{C}$ to $4\ ^\circ\text{C}$ and expands only above $4\ ^\circ\text{C}$; its density is a maximum at $4\ ^\circ\text{C}$. [1]
Lake As the lake cools, surface water sinks until all of it is at $4\ ^\circ\text{C}$. Colder water is less dense and stays at the top, where it freezes; ice is lighter still and floats. The ice layer insulates the water below, which stays at about $4\ ^\circ\text{C}$ near the bottom, so animals survive. [1]
Exam tip Mention both facts — the density maximum at $4\ ^\circ\text{C}$ and ice floating and insulating — for full marks.

E2. NEET At what temperature does a Fahrenheit thermometer show a reading twice that of a Celsius thermometer? (a) $80\ ^\circ\text{C}$ (b) $160\ ^\circ\text{C}$ (c) $-24.6\ ^\circ\text{C}$ (d) $320\ ^\circ\text{C}$

Idea Put $t_F=2t_C$ in $t_F=1.8t_C+32$: $2t_C=1.8t_C+32$, $0.2t_C=32$, $t_C=160\ ^\circ\text{C}$ (and $t_F=320\ ^\circ\text{F}$).
Answer (b).
Exam tip (d) is the Fahrenheit reading, not the temperature asked for; (c) comes from the reversed condition $t_C=2t_F$. Always write which reading is twice which before solving.

E3. JEE Main (Numerical value) A steel rod and a brass rod are to be cut so that the steel rod is always $10.0\ \text{cm}$ longer than the brass rod, at every temperature. With $\alpha_\text{steel}=1.2\times10^{-5}\ \text{K}^{-1}$ and $\alpha_\text{brass}=1.9\times10^{-5}\ \text{K}^{-1}$, find the length of the steel rod in cm, to one decimal place.

Condition The difference stays fixed only if both rods grow by the same amount for every $\Delta T$: $L_s\alpha_s=L_b\alpha_b$.
Solve With $L_s-L_b=10.0$: $L_s\alpha_s=(L_s-10.0)\alpha_b$, so $L_s=\dfrac{10.0\,\alpha_b}{\alpha_b-\alpha_s}=\dfrac{10.0\times1.9}{0.7}=27.1\ \text{cm}$ (and $L_b=17.1\ \text{cm}$).
Answer $27.1$.
Exam tip The longer rod must be the one with the smaller $\alpha$. Check: $27.14\times1.2=32.6$ and $17.14\times1.9=32.6$ (in units of $10^{-5}\ \text{cm K}^{-1}$) ✓.

E4. JEE Advanced (One or more correct) A uniform rod of length $L$, cross-sectional area $A$, Young's modulus $Y$ and linear expansion coefficient $\alpha$ is fitted without stress between two rigid walls and is then heated through $\Delta T$. Which statements are correct? (A) The stress in the rod is $Y\alpha\Delta T$. (B) The force on each wall is $YA\alpha\Delta T$. (C) A rod of the same material twice as long would carry twice the stress. (D) If one wall is removed so that one end is free, no thermal stress develops.

(A) The walls prevent the free expansion $\alpha L\Delta T$, so the compressive strain is $\alpha\Delta T$ and the stress $Y\alpha\Delta T$ ✓.
(B) $F=\sigma A=YA\alpha\Delta T$ ✓.
(C) The strain $\alpha\Delta T$ does not involve $L$, so the stress is the same ✗.
(D) With one end free the rod simply expands; nothing compresses it ✓.
Answer (A), (B), (D).
Exam tip Thermal stress is a "prevented strain" problem: find the strain the supports must undo, then multiply by $Y$.

E5. JEE Main NEET The pressure of an ideal gas in a closed rigid vessel rises by $0.50\%$ when it is heated through $1\ \text{K}$. The initial temperature of the gas was (a) $100\ \text{K}$ (b) $200\ \text{K}$ (c) $273\ \text{K}$ (d) $500\ \text{K}$

Idea At constant volume $p\propto T$, so $\dfrac{\Delta p}{p}=\dfrac{\Delta T}{T}$: $0.0050=\dfrac{1}{T}$.
Answer $T=200\ \text{K}$: (b).
Exam tip Convert the percentage first ($0.50\%=0.0050$). (c) is the "$0\ ^\circ\text{C}$ reflex": at $273\ \text{K}$ a $1\ \text{K}$ rise gives only $0.37\%$. The colder the gas, the larger its fractional change per kelvin.

E6. Board · Case study In a laboratory, a constant-volume gas thermometer containing helium at low pressure reads $4.00\times10^4\ \text{Pa}$ with its bulb at the triple point of water. (i) The bulb is placed in a bath and the pressure becomes $5.50\times10^4\ \text{Pa}$. Find the temperature of the bath in kelvin and in degrees Celsius. [2] (ii) What pressure would the gas exert at absolute zero if it remained an ideal gas? Why is this point never actually reached? [1] (iii) Why is the gas kept at low pressure? [1]

(i) $T=273.16\times\dfrac{5.50}{4.00}=375.6\ \text{K}$; $\ t_C=375.6-273.15=102.4\ ^\circ\text{C}$. [2]
(ii) $p\propto T$, so $p=0$ at $T=0$. Real gases liquefy (helium at about $4\ \text{K}$) long before, and absolute zero itself cannot be reached; the zero is found by extrapolating the straight line. [1]
(iii) At low pressure (low density) every real gas behaves very nearly as an ideal gas, so the reading does not depend on which gas is used. [1]
Exam tip Use the triple point, $273.16\ \text{K}$, as the reference when it is given; convert to Celsius with $273.15$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How much of the observed rise in sea level comes simply from seawater expanding as the oceans warm?

Why it matters. Rising seas threaten coastal cities from Mumbai and Dhaka to Miami, Lagos and Jakarta, and small island nations in the Pacific and Indian Oceans. Two effects raise the sea: water added by melting glaciers and ice sheets, and thermal expansion of the water already there. Climate scientists separate the two using ocean temperature measurements from thousands of free-drifting profiling floats.

What to do. Model a column of ocean of depth $h$ whose temperature rises by $\Delta T$. Because the sides of the ocean cannot move outward, all of the volume increase shows up as a rise in height.

  1. Show that a column of height $h$ and fixed base area rises by $\Delta h=\gamma h\,\Delta T$.
  2. Take $\gamma\approx2.0\times10^{-4}\ \text{K}^{-1}$ for seawater near the surface, and suppose the top $700\ \text{m}$ of the ocean warms by $0.3\ \text{K}$. Estimate $\Delta h$.
  3. The coefficient for seawater depends strongly on temperature — roughly $0.5\times10^{-4}\ \text{K}^{-1}$ near $0\ ^\circ\text{C}$ and $3\times10^{-4}\ \text{K}^{-1}$ near $25\ ^\circ\text{C}$. Would the same warming raise the sea more in the tropics or near the poles? By roughly what factor?
  4. The deep ocean is near $2\ ^\circ\text{C}$. Repeat step 2 for a warming of $0.05\ \text{K}$ spread over $3000\ \text{m}$ of deep water, using the cold-water value of $\gamma$.
  5. Compare your total with published estimates of the "thermosteric" contribution to sea-level rise, and with the total observed rise of the last few decades.

Tools: Calculator or spreadsheet; open ocean-temperature data and sea-level reports from international climate assessments and ocean-observing programmes.

Hand in: A one-page report: your model and its assumptions, the calculations, a comparison table (your estimate versus published values) and two sentences on which assumption matters most.

Going further. Water is not the only thing that expands: find out how engineers allow for thermal expansion in long pipelines, bridges and railways in places with large temperature swings, such as deserts and Himalayan passes.

Instructor note: step 1 from $\Delta V=\gamma V\Delta T$ with $V=Ah$. Step 2: $2.0\times10^{-4}\times700\times0.3=0.042\ \text{m}\approx4\ \text{cm}$. Step 3: the tropics, by a factor of about $6$. Step 4: $0.5\times10^{-4}\times3000\times0.05\approx0.008\ \text{m}$, under $1\ \text{cm}$. Published thermosteric rates are of order $1\ \text{mm}$ per year, a large fraction of the total observed rise of a few mm per year; students should find their estimate of the right order of magnitude.

🧪 Self-check quiz

0/5 correct

Q1. The Celsius and Fahrenheit scales give the same reading at

Q2. Heating a metal cube increases each of its edges by $0.10\%$. Its volume increases by about

Q3. A metal plate with a circular hole in it is heated uniformly. The hole

Q4. The density of water is greatest at about

Q5. An ideal gas is heated at constant pressure from $27\ ^\circ\text{C}$ to $327\ ^\circ\text{C}$. Its volume