Chapter 01 · Thermal Properties of Matter · Full book

Thermal Properties of Matter: The Physics of Hotness

Core intuition: Chapter Roadmap

This chapter is about hotness—what it is, how we measure it, how it moves, and how matter responds to it. We build the story in a deliberate order, each idea earning the next.

Temperature and its measurement. The Zeroth Law makes temperature meaningful; scales (Celsius, Fahrenheit, Kelvin) and the triple point make it precise. Kelvin is the scale physics actually uses.
Thermal expansion. Why solids and liquids swell when heated—linear, area, and volume coefficients, the “hole expands” rule, bimetallic strips, and real-versus-apparent expansion of liquids.
The anomaly of water. The density maximum at \(4^\circ\mathrm{C}\) and why it lets ponds freeze top-down.
Calorimetry and specific heat. \(Q = mc\,\Delta T\), the method of mixtures, water equivalent, and the Dulong–Petit rule.
Latent heat and phase change. The heating curve, its flat plateaus, and why a phase change can stall a temperature rise.
Heat transfer. Conduction (Fourier's law, thermal resistance, composite and radial geometries), convection, and radiation (Stefan–Boltzmann, Wien, emissivity, net exchange).
Newton's law of cooling. The small-excess limit of radiative and convective loss—and its limits.
Thermal stress. What happens when expansion is prevented.

Throughout, watch the single recurring trap: know which temperature unit each formula demands. Differences are the same in kelvin and Celsius, but absolute temperatures—everywhere a \(T\) stands alone, as in radiation and Clausius–Clapeyron—must be in kelvin.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Loose Lid: Railway tracks are laid with small gaps between segments. Bridges have expansion joints. The designers are not being careless—they are being careful. But careful about what, exactly? And why does a tight lid on a glass jar come loose if you run hot water over it, even though both the jar and the lid expand?
The Ice Paradox: Water is one of the most studied substances on Earth, yet it behaves unlike almost every other liquid. When you cool most liquids, they contract and sink. Water does this too—until \(4^\circ\mathrm{C}\), after which further cooling makes it expand and float. If water behaved normally, lakes would freeze solid from the bottom up. Life in cold climates would be impossible. What makes water so strange?
The Styrofoam Illusion: On a cold morning you pick up a metal rod and a styrofoam block that have been sitting in the same room all night. The metal feels colder. But a thermometer tells you both are at exactly the same temperature. Which one is lying—the thermometer or your hand?
The Cooling Coffee: You pour yourself a cup of coffee. You want to drink it in 5 minutes but it is too hot. Should you add cold milk now, or just before you drink it, to have the coolest cup after 5 minutes? The answer is not obvious, and it depends on a law you will derive in this chapter.

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why Thermal Physics Deserves Its Own Chapter

Every chapter so far has described the world in mechanical terms: forces, velocities, accelerations, energies. A block slides down an incline. A spring stores energy and releases it. A planet orbits a star. The underlying variables are positions and momenta—and they obey Newton's Laws.

Now heat a block of iron in a furnace. Its temperature rises. It expands. Eventually it glows red. None of these phenomena involve a force in the Newtonian sense. No acceleration can be assigned to “heat flowing into the iron.” No free-body diagram captures what is happening.

Something new is needed.

What Is Different About Thermal Phenomena?

Consider two observations that Newton's Laws alone cannot explain.

Observation 1. Place a hot iron block in contact with a cold one. Heat flows from the hot block to the cold one—never the other way. Both processes would conserve energy; yet nature chooses only one direction. Why?

Observation 2. Rub your hands together. Work is done against friction; the hands warm up. The reverse—cool hands spontaneously doing work and pushing each other apart—never happens, even though it would not violate energy conservation.

These observations reveal two things: first, there is a form of energy called thermal energy (or internal energy) that is distinct from mechanical kinetic and potential energy; second, thermal processes have a preferred direction, unlike purely mechanical ones.

The first point is the domain of this chapter: thermal properties, temperature, heat transfer, and the behaviour of matter when energy is exchanged thermally. The second point—directionality—leads to the Second Law of Thermodynamics, which belongs to the Laws of Thermodynamics chapter.

Heat and Temperature Are Not the Same Thing

The single most important conceptual distinction in this entire chapter is this:

Core intuition: Heat Is Not Temperature

Temperature is a property of a body. It measures, loosely, the average kinetic energy of the microscopic constituents of that body. It is a number you can assign to a system in equilibrium. Heat is energy in transit. It is not stored in a body; it flows between bodies when they are at different temperatures. The moment heat stops flowing and temperatures equalise, there is no more “heat”—only internal energy, now distributed differently. Saying a body “contains heat” is as imprecise as saying a lake “contains rain.” Once the rain falls into the lake, it is lake water. Once heat flows into a body, it is internal energy.

This distinction resolves the Styrofoam puzzle immediately. The metal rod and the styrofoam block are at the same temperature. But when you touch them, heat flows from your hand into each object. The metal conducts heat away from your skin far more rapidly than styrofoam does; your skin loses heat faster, so it cools down faster, so the metal feels colder. Your hand is not measuring temperature—it is measuring the rate of heat transfer. The thermometer measures temperature. Neither is lying; they are measuring different things.

The Roadmap

This chapter develops four ideas in sequence.

Temperature and its measurement. We define temperature operationally, introduce the Celsius and Kelvin scales, and understand why absolute zero is not merely “very cold” but physically special.

Thermal expansion. Solids, liquids, and gases all change size with temperature. We quantify this with coefficients of linear, area, and volume expansion, and examine the extraordinary exception: water between \(0^\circ\mathrm{C}\) and \(4^\circ\mathrm{C}\).

Calorimetry and phase transitions. When heat flows into a body, either its temperature rises (specific heat) or its state changes without any temperature change at all (latent heat). We build the tools to track both.

Heat transfer mechanisms. Heat moves by conduction, convection, and radiation. Each mechanism has its own law, its own characteristic situations, and its own set of exam traps.

✓ Checkpoint · Why Thermal Physics?

A student says: “I added \(500\,\mathrm{J}\) of heat to this iron block, so the block now contains \(500\,\mathrm{J}\) more heat than before.” Identify the error in this statement and rewrite it correctly.
“Heat” is energy in transit, not a stored quantity. Correct: the internal energy of the block has increased by \(500\,\mathrm{J}\) (assuming no work is done).
A metal spoon and a wooden spoon are both left in a pot of boiling water for one minute. Both reach \(100^\circ\mathrm{C}\). You touch each one briefly. The metal spoon burns you; the wooden one barely feels warm. Are the two spoons at different temperatures? What quantity actually differs between them?
Both spoons are at \(100^\circ\mathrm{C}\)—same temperature. What differs is thermal conductivity: metal transfers heat to your skin at a much higher rate than wood, causing a burn.
(Olympiad) A perfectly insulating room contains two identical iron blocks at temperatures \(T_1 = 400\,\mathrm{K}\) and \(T_2 = 300\,\mathrm{K}\). They are placed in contact and reach thermal equilibrium. The final temperature is not \(350\,\mathrm{K}\) but slightly below it. Explain why, without invoking any formula.
Total internal energy is conserved, so by symmetry the average ought to be \(350\,\mathrm{K}\). However, specific heat of metals increases slightly with temperature; the hotter block contains more internal energy per degree than the colder block does. To conserve total internal energy the equilibrium temperature shifts below \(350\,\mathrm{K}\). (At JEE level, specific heats are treated as constant and the answer is exactly \(350\,\mathrm{K}\); the correction is an Olympiad-level subtlety.)

Temperature and Its Measurement

◈ Temperature scales

live bench

Slide one thermometer and the others track it: Celsius, Fahrenheit and Kelvin are one line read three ways, meeting at absolute zero.

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Fig 13.1 The Celsius, Kelvin and Fahrenheit scales aligned at water’s freezing and boiling points. The same interval is divided into 100, 100 and 180 degrees respectively — which is why conversions are just a linear stretch and shift.

Temperature is one of those quantities that feels intuitive until you try to define it precisely. Everyone knows that boiling water is hotter than ice. But what does “hotter” mean in a way that can be measured, compared across materials, and used in a formula?

The answer requires two steps: a definition of thermal equilibrium, and an operational procedure for assigning numbers to temperatures.

Thermal Equilibrium and the Zeroth Law

Place a cold iron block in contact with a warm copper block. Over time, heat flows from copper to iron until the two blocks no longer exchange energy. At this point the blocks are in thermal equilibrium: any measurable thermal property of either block has stopped changing.

Now add a third body—a thermometer. If the thermometer reaches equilibrium with the copper block (reads the same value and stops changing), and then is placed against the iron block and reads the same value again, what can you conclude?

The copper and iron blocks must be in equilibrium with each other.

This is the content of the Zeroth Law of Thermodynamics:

Core intuition: The Zeroth Law

If body \(A\) is in thermal equilibrium with body \(C\), and body \(B\) is also in thermal equilibrium with body \(C\), then \(A\) and \(B\) are in thermal equilibrium with each other. The Zeroth Law is what makes thermometry possible. The thermometer is the “body \(C\)” that allows you to compare the temperatures of any two objects without bringing them into direct contact.

The Zeroth Law is called “zeroth” because it was recognised as logically prior to the First and Second Laws—which were already named—but was only made explicit later. Without it, the concept of temperature would have no operational meaning.

Temperature Scales

A thermometer works by identifying a physical property that changes measurably and reproducibly with temperature—the length of a mercury column, the electrical resistance of a wire, the pressure of a fixed volume of gas—and assigning numbers to values of that property.

The assignment of numbers requires two fixed points: two reproducible temperatures that any thermometer anywhere in the world can be calibrated against.

The Celsius Scale

Anders Celsius (1742) chose:

\(0^\circ\mathrm{C}\): the melting point of ice at standard atmospheric pressure.
\(100^\circ\mathrm{C}\): the boiling point of water at standard atmospheric pressure.

The interval between them is divided into 100 equal parts. This is the scale used in everyday life and in most laboratory work.

The Fahrenheit Scale

The Fahrenheit scale sets the freezing point of water at \(32^\circ\mathrm{F}\) and the boiling point at \(212^\circ\mathrm{F}\), giving 180 equal divisions between them. The conversion is: \[ T_F = \frac{9}{5}\,T_C + 32 \] The Fahrenheit scale is used primarily in the United States and in some medical contexts. For JEE, NEET, and all international physics, you will never need it in a calculation—but you should know it exists and know the conversion.

The Kelvin Scale and Absolute Zero

The Celsius scale has a problem. Its zero point is arbitrary: the freezing point of water is convenient, but physically special only in a kitchen. Below \(0^\circ\mathrm{C}\) temperatures become negative, which creates awkward algebra in formulas involving temperature ratios.

More fundamentally: is there a natural lowest possible temperature?

The answer is yes. As a gas is cooled, its pressure (at constant volume) decreases linearly with temperature. If you extrapolate this line to zero pressure—the point at which the gas would exert no force at all, meaning its molecules have ceased all thermal motion—you reach the same temperature regardless of which gas you use: \[ T_{\text{abs zero}} = -273.15^\circ\mathrm{C} \]

This temperature is absolute zero. It is the coldest possible temperature: below it, nothing can exist, because there is no such thing as negative thermal motion.

The Kelvin scale places its zero at absolute zero and uses the same size degree as Celsius: \[ \boxed{T(\mathrm{K}) = T({}^\circ\mathrm{C}) + 273.15} \]

For all JEE and NEET calculations use \(273\) (not \(273.15\)) unless the problem specifies otherwise.

Birth of the formula: Why Kelvin Is the Natural Scale for Physics

Consider the ideal gas law, which you will meet properly later: \[ pV = nRT \] Here \(p\) is pressure, \(V\) is volume, \(n\) is the number of moles, and \(R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}\) is the universal gas constant. If \(T\) is in Celsius, then at \(T_C = 0\) the gas has \(pV = 0\), which would mean either zero pressure or zero volume. But we know a gas at \(0^\circ\mathrm{C}\) is not at zero pressure. The formula is simply wrong if you substitute Celsius temperatures. If \(T\) is in Kelvin, then at \(T = 0\,\mathrm{K}\) (absolute zero) we get \(pV = 0\), which is physically correct: at absolute zero the gas would have no thermal motion and hence no pressure. Rule: in every physics formula involving temperature (\(pV = nRT\), Stefan's Law, Wien's Law, the Boltzmann distribution), \(T\) must be in Kelvin. Substituting Celsius into these formulas produces wrong answers. This is one of the most common errors in thermodynamics problems.

The Triple Point: A Better Fixed Point

Modern metrology does not use the ice point and steam point to define the Kelvin scale, because both depend on pressure and are affected by dissolved impurities in water.

Instead, the Kelvin is defined using the triple point of water: the unique combination of temperature and pressure at which solid ice, liquid water, and water vapour coexist in equilibrium. This is a far more reproducible and pressure-independent reference. The triple point is defined as exactly \(273.16\,\mathrm{K}\) (\(0.01^\circ\mathrm{C}\)).

For competitive exams: the triple point of water is \(273.16\,\mathrm{K}\), not \(273.15\,\mathrm{K}\). The difference (\(0.01\,\mathrm{K}\)) is small but sometimes tested as a precision question.

Thermometers and Their Ranges

Any physical property that varies monotonically and reproducibly with temperature can serve as a thermometer. In practice:

p{2.5cm} p{3.5cm}} Type	Range	Principle
Mercury-in-glass	\(-39\) to \(357^\circ\mathrm{C}\)	Liquid expansion
Alcohol-in-glass	\(-115\) to \(78^\circ\mathrm{C}\)	Liquid expansion
Platinum resistance	\(-200\) to \(850^\circ\mathrm{C}\)	Resistance \(\propto T\)
Thermocouple	\(-270\) to \(2300^\circ\mathrm{C}\)	Seebeck EMF
Constant-volume gas	\(-270\) to \(1500^\circ\mathrm{C}\)	Pressure \(\propto T\)
Pyrometer (optical)	\(700^\circ\mathrm{C}\) and above	Radiation intensity

The constant-volume gas thermometer is the closest to an ideal thermometer because its readings depend only on the fundamental properties of an ideal gas, not on the specific material of the thermometer itself. It defines the Kelvin scale operationally.

Misconception alert: Celsius and Kelvin Are Interchangeable

They are not interchangeable in formulas involving ratios or products of temperature. When a formula involves a difference of temperatures (\(\Delta T\)), Celsius and Kelvin give identical results because \(\Delta T_C = \Delta T_K\) (the scales have the same degree size). When a formula involves \(T\) itself—as in \(pV = nRT\), Stefan's radiation law \(P = \sigma A T^4\), Wien's displacement law \(\lambda_{\max} T = b\)—you must use Kelvin. Substituting \(T_C = 27\) when the formula needs \(T_K = 300\) introduces a factor of \(\sim 11\) error. Safe habit: convert to Kelvin at the start of every problem, before substituting into any formula.

✓ Checkpoint · Temperature and Scales

Convert the following temperatures to Kelvin: (a) \(27^\circ\mathrm{C}\), (b) \(-173^\circ\mathrm{C}\), (c) \(-273^\circ\mathrm{C}\). What is physically special about the answer to (c)?
(a) \(300\,\mathrm{K}\). (b) \(100\,\mathrm{K}\). (c) \(0\,\mathrm{K}\) — absolute zero; the lowest possible temperature, at which classical thermal motion ceases entirely.
A student measures the boiling point of a liquid using a mercury thermometer (range \(-39\) to \(357^\circ\mathrm{C}\)) and reports \(355^\circ\mathrm{C}\). Why should you be sceptical of this reading?
\(355^\circ\mathrm{C}\) is within \(2^\circ\) of the mercury thermometer's upper limit (\(357^\circ\mathrm{C}\), the boiling point of mercury itself). Near the limit, the mercury column expands nonlinearly and the thermometer becomes unreliable. A thermocouple or resistance thermometer should be used instead.
(Trap) Two thermometers, \(A\) and \(B\), are both calibrated using the ice point and steam point of water. Thermometer \(A\) uses mercury (which expands non-uniformly with temperature); thermometer \(B\) uses an ideal gas. At \(50^\circ\mathrm{C}\) on the ideal-gas scale, thermometer \(A\) reads \(51^\circ\mathrm{C}\). Is thermometer \(A\) broken?
No. Non-ideal thermometric substances (like mercury) expand non-uniformly, so they agree with the ideal-gas thermometer only at the calibration points (\(0^\circ\mathrm{C}\) and \(100^\circ\mathrm{C}\)) and may diverge in between. Thermometer \(A\) is working correctly — it is simply not a perfect replica of the ideal-gas (Kelvin) scale. This is why the constant-volume gas thermometer defines the scale and all others are calibrated against it.

Thermal Expansion

◈ Linear, area & volume expansion

live bench

Heat the bar and it grows by \(\Delta L=L\alpha\Delta T\); areas and volumes follow with \(2\alpha\) and \(3\alpha\). Even a hole expands as if it were filled.

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Fig 13.2 Linear expansion \(\Delta L=\alpha L_0\,\Delta T\) lengthens a heated rod (left). Bond two metals of different \(\alpha\) and the strip must curl on heating (right) — the more-expansive metal on the outside of the bend, the basis of the thermostat.

Fig 13.3 A rod clamped between rigid walls cannot lengthen when heated, so the expansion it is denied reappears as a compressive thermal stress \(\sigma=Y\alpha\,\Delta T\) — remarkably, independent of the rod’s length. This is what buckles rails and bridges.

Perplexing questions

A steel railway track is laid in winter at \(5\,{}^{\circ}\mathrm{C}\). Each rail is \(25\,\mathrm{m}\) long. In summer, the track temperature reaches \(55\,{}^{\circ}\mathrm{C}\). How much longer does each rail become—and what happens if the engineers forget to leave gaps?

When a solid is heated, its atoms vibrate more vigorously about their equilibrium positions. The asymmetry of the inter-atomic potential energy curve means the average separation increases with amplitude: the solid expands. This is not an accident of geometry; it is a direct consequence of the anharmonic shape of the bond potential.

Linear Expansion

Consider a rod of natural length \(L_0\) at temperature \(T_0\). When its temperature changes by \(\Delta T\), its length becomes \(L\):

\[ \Delta L \;=\; \alpha\, L_0\, \Delta T, \qquad L \;=\; L_0(1 + \alpha\,\Delta T). \]

The quantity \(\alpha\) is the coefficient of linear thermal expansion, with SI unit \(\mathrm{K^{-1}}\). It is a material property, independent of the shape of the object.

Core intuition: Why \(\alpha\) is the same in every direction for isotropic solids

An isotropic material has no preferred direction. If it expanded more along \(x\) than along \(y\), rotating the sample would change its expansion behaviour—contradicting isotropy. Therefore \(\alpha\) is the same along every axis, and a cube remains a cube (just a larger one) when heated uniformly.

Area and Volume Expansion

The same temperature change \(\Delta T\) acts on every dimension simultaneously. For a rectangular block of dimensions \(L_x \times L_y \times L_z\):

\[\begin{aligned} L_x' &= L_x(1+\alpha\Delta T), \quad L_y' = L_y(1+\alpha\Delta T), \quad L_z' = L_z(1+\alpha\Delta T). \notag \end{aligned}\]

Area \(A = L_x L_y\) expands to \(A' = L_x' L_y' = A(1+\alpha\Delta T)^2 \approx A(1 + 2\alpha\Delta T)\), since \((\alpha\Delta T)^2 \ll 1\) for all practical cases. Volume \(V = L_x L_y L_z\) expands to \(V' \approx V(1 + 3\alpha\Delta T)\).

Defining the coefficient of area expansion \(\beta\) and the coefficient of volume expansion \(\gamma\):

\[ \beta \;=\; 2\alpha, \qquad \gamma \;=\; 3\alpha. \]

These relations hold for isotropic solids. For liquids and gases, only \(\gamma\) is meaningful (they have no fixed shape), and the approximation \(\gamma = 3\alpha\) does not apply; \(\gamma\) must be measured directly.

Misconception alert: A hole in a plate expands, not contracts

Students frequently argue: “the material expands inward, so the hole shrinks.” This is wrong. Every linear dimension of the solid—including the diameter of any hole—scales by the factor \((1 + \alpha\Delta T)\). Think of it as a photographic enlargement: a photo scaled up by \(5\%\) enlarges holes and solid regions by the same factor. The hole expands as if it were filled with the same material.

Apparent vs. Real Expansion of a Liquid in a Vessel

When a liquid is heated inside a vessel, both the liquid and the vessel expand. The apparent rise in liquid level reflects only the difference between the two expansions.

Let the vessel have volume expansion coefficient \(\gamma_v = 3\alpha_v\), and the liquid have coefficient \(\gamma_L\). The volume of liquid that apparently overflows when \(\Delta T\) is applied is:

\[ \Delta V_{\text{app}} = V_0\,(\gamma_L - \gamma_v)\,\Delta T = V_0\,(\gamma_L - 3\alpha_v)\,\Delta T. \]

Hence \(\gamma_{\text{app}} = \gamma_L - 3\alpha_{\text{vessel}}\). Only when \(\gamma_L \gt 3\alpha_v\) does the liquid appear to expand.

Birth of the formula: Thermal stress in a constrained rod

A rod of length \(L_0\), Young's modulus \(Y\), and linear expansion coefficient \(\alpha\) is clamped at both ends so it cannot change length. The temperature is raised by \(\Delta T\). Step 1 — Free expansion. If unconstrained, the rod would expand by \(\Delta L = \alpha L_0\Delta T\). Step 2 — Compression by clamps. The clamps impose a compressive strain \(\varepsilon = \Delta L / L_0 = \alpha\Delta T\) to restore the original length. Step 3 — Stress. By Hooke's law, \(\sigma = Y\varepsilon\): \[ \sigma \;=\; Y\alpha\Delta T. \] For steel (\(Y = 2\times10^{11}\,\mathrm{Pa}\), \(\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}}\)) and \(\Delta T = 40\,\mathrm{K}\): \(\sigma = 2\times10^{11}\times1.2\times10^{-5}\times40 = 9.6\times10^7\,\mathrm{Pa} \approx 100\,\mathrm{MPa}\). This approaches the yield strength of mild steel (\(\sim250\,\mathrm{MPa}\)), explaining why expansion joints are non-negotiable in bridges and railway tracks.

Bimetallic Strip

Two metals with different \(\alpha\) values are bonded together. When heated, the metal with larger \(\alpha\) would expand more but is constrained by the other; the strip bends so that the high-\(\alpha\) side is on the outside of the curve.

The degree of bending is proportional to \((\alpha_1 - \alpha_2)\Delta T\). This is the operating principle of thermostats, bimetallic thermometers, and circuit-breaker thermal trips.

Worked Examples

Example 1

A steel railway rail of length \(25\,\mathrm{m}\) is laid at \(5\,{}^{\circ}\mathrm{C}\). The coefficient of linear expansion of steel is \(\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}}\). (a) Find the increase in length when the rail reaches \(55\,{}^{\circ}\mathrm{C}\). (b) If no expansion gap is left and the rail is rigidly constrained, find the compressive stress. (\(Y_{\text{steel}} = 2\times10^{11}\,\mathrm{Pa}\).)

Solution. \(\Delta T = 55 - 5 = 50\,\mathrm{K}\). (a) \(\Delta L = \alpha L_0\Delta T = 1.2\times10^{-5}\times25\times50 = 1.5\times10^{-2}\,\mathrm{m} = 1.5\,\mathrm{cm}\). (b) \(\sigma = Y\alpha\Delta T = 2\times10^{11}\times1.2\times10^{-5}\times50 = 1.2\times10^{8}\,\mathrm{Pa} = 120\,\mathrm{MPa}\). This compressive stress is nearly half the tensile strength of mild steel; without gaps, the track would buckle laterally within a few hot summers.

Example 2

A glass flask of volume \(500\,\mathrm{cm^3}\) is filled to the brim with mercury at \(0\,{}^{\circ}\mathrm{C}\). How much mercury overflows when the system is heated to \(100\,{}^{\circ}\mathrm{C}\)?
(\(\gamma_{\text{mercury}} = 1.82\times10^{-4}\,\mathrm{K^{-1}}\), \(\alpha_{\text{glass}} = 9\times10^{-6}\,\mathrm{K^{-1}}\).)

Solution. \(\gamma_{\text{glass}} = 3\alpha_{\text{glass}} = 3\times9\times10^{-6} = 2.7\times10^{-5}\,\mathrm{K^{-1}}\). Apparent expansion coefficient of mercury: \(\gamma_{\text{app}} = \gamma_{\text{Hg}} - \gamma_{\text{glass}} = 1.82\times10^{-4} - 2.7\times10^{-5} = 1.55\times10^{-4}\,\mathrm{K^{-1}}\). Volume overflow: \(\Delta V = V_0\,\gamma_{\text{app}}\,\Delta T = 500\times1.55\times10^{-4}\times100 = 7.75\,\mathrm{cm^3}\).

Checkpoint

✓ Checkpoint · Thermal Expansion

A copper rod (\(\alpha = 1.7\times10^{-5}\,\mathrm{K^{-1}}\), \(L_0 = 2\,\mathrm{m}\)) and a steel rod (\(\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}}\), same \(L_0\)) are joined end-to-end between rigid walls. When heated by \(\Delta T = 80\,\mathrm{K}\), what compressive force do the walls exert? (Cross-sectional area \(A = 2\,\mathrm{cm^2}\) for both; \(Y_{\text{Cu}} = 1.2\times10^{11}\,\mathrm{Pa}\), \(Y_{\text{steel}} = 2\times10^{11}\,\mathrm{Pa}\).)
The composite rod cannot change total length. Free expansion: \(\delta_{\text{Cu}} + \delta_{\text{steel}} = (1.7+1.2)\times10^{-5}\times2\times80 = 4.64\times10^{-3}\,\mathrm{m}\). Compressive force \(F\) produces compressions \(FL_0/(Y_{\text{Cu}}A)\) and \(FL_0/(Y_{\text{steel}}A)\). Setting total compression equal to free expansion: \(F\cdot2\!\left(\tfrac{1}{Y_{\text{Cu}}}+\tfrac{1}{Y_{\text{steel}}}\right)/A = 4.64\times10^{-3}\). \(F = 4.64\times10^{-3}\times A\,/\!\left[2\left(\tfrac{1}{1.2}+\tfrac{1}{2}\right)\times10^{-11}\right] \approx 20.1\,\mathrm{kN}\).
A circular disc of brass (\(\alpha = 1.9\times10^{-5}\,\mathrm{K^{-1}}\), diameter \(10\,\mathrm{cm}\)) has a circular hole of diameter \(3\,\mathrm{cm}\) at its centre. By how much does the hole diameter change when the disc is heated by \(120\,\mathrm{K}\)? Does the hole expand or contract?
Hole diameter increases: \(\Delta d = \alpha\,d_0\,\Delta T = 1.9\times10^{-5}\times0.03\times120 = 6.84\times10^{-5}\,\mathrm{m} = 0.068\,\mathrm{mm}\). The hole expands—every linear dimension scales by \((1+\alpha\Delta T)\).
A mercury thermometer has a bulb of volume \(V_b = 0.20\,\mathrm{cm^3}\) and a capillary of cross-sectional area \(A_c = 6\times10^{-4}\,\mathrm{cm^2}\). Estimate the length of mercury column that rises per kelvin of temperature change. (\(\gamma_{\text{Hg}} = 1.82\times10^{-4}\,\mathrm{K^{-1}}\); neglect glass expansion for this estimate.)
\(\Delta V \approx V_b\,\gamma_{\text{Hg}}\,\Delta T = 0.20\times1.82\times10^{-4}\times1 = 3.64\times10^{-5}\,\mathrm{cm^3}\) per K. Rise in column: \(\Delta h = \Delta V/A_c = 3.64\times10^{-5}/(6\times10^{-4}) \approx 0.061\,\mathrm{cm} = 0.61\,\mathrm{mm\,K^{-1}}\).

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Where the Scales Cross

At what temperature do the Celsius and Fahrenheit scales read the same number?

Solution. Given: The conversion \(T_F = \tfrac{9}{5}T_C + 32\).
Find: The value \(x\) at which \(T_C = T_F = x\).
Setup: Set the two readings equal and solve the single linear equation.
Solve: \[ x = \tfrac{9}{5}x + 32 \quad\Longrightarrow\quad x - \tfrac{9}{5}x = 32 \quad\Longrightarrow\quad -\tfrac{4}{5}x = 32 \quad\Longrightarrow\quad x = -40. \] Answer: \(\boxed{-40^\circ\mathrm{C} = -40^\circ\mathrm{F}}\)
Check: Substitute back into the conversion: \(T_F = \tfrac{9}{5}(-40) + 32 = -72 + 32 = -40^\circ\mathrm{F}\). ✓ The two scales meet exactly once because they are straight lines of different slope.

Example 2 · A Railway Rail in Summer

A steel rail is \(12.0\,\mathrm{m}\) long when laid at \(20^\circ\mathrm{C}\). On a hot day its temperature rises to \(50^\circ\mathrm{C}\). By how much does it lengthen? Take \(\alpha_{\text{steel}} = 1.2\times10^{-5}\,\mathrm{K^{-1}}\).

Solution. Given: \(L_0 = 12.0\,\mathrm{m}\), \(\Delta T = 50 - 20 = 30\,\mathrm{K}\) (a difference, so K and \(^\circ\mathrm{C}\) are interchangeable), \(\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}}\).
Find: The increase in length \(\Delta L\).
Setup: Linear expansion, \(\Delta L = \alpha L_0 \Delta T\).
Solve: \[ \Delta L = (1.2\times10^{-5})(12.0)(30) = 4.32\times10^{-3}\,\mathrm{m}. \] Answer: \(\boxed{\Delta L = 4.32\,\mathrm{mm}}\)
Check: Dimensions: \([\mathrm{K^{-1}}][\mathrm{m}][\mathrm{K}] = [\mathrm{m}]\). ✓ The fractional change is \(\Delta L/L_0 = \alpha\Delta T = 3.6\times10^{-4}\)—tiny, as expected, yet enough across many rails to demand expansion gaps.

Example 3 · Does the Hole Grow or Shrink?

A steel plate has a circular hole of diameter \(2.000\,\mathrm{cm}\) at \(20^\circ\mathrm{C}\). The plate is heated to \(220^\circ\mathrm{C}\). Find the new diameter of the hole. Take \(\alpha_{\text{steel}} = 1.2\times10^{-5}\,\mathrm{K^{-1}}\).

Solution. Given: \(d_0 = 2.000\,\mathrm{cm}\), \(\Delta T = 220 - 20 = 200\,\mathrm{K}\), \(\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}}\).
Find: New hole diameter \(d\).
Setup: A hole expands exactly as if it were filled with the surrounding metal: every linear dimension scales by \((1+\alpha\Delta T)\). The hole grows.
Solve: \[ d = d_0(1 + \alpha\Delta T) = 2.000\bigl(1 + (1.2\times10^{-5})(200)\bigr) = 2.000(1 + 2.4\times10^{-3}). \] \[ d = 2.000 + 4.8\times10^{-3} = 2.0048\,\mathrm{cm}. \] Answer: \(\boxed{d = 2.0048\,\mathrm{cm}\ \text{(it grows)}}\)
Check: Limiting case \(\alpha\to 0\) gives \(d\to d_0\). ✓ The common “hole shrinks” intuition is wrong: imagine the absent disc restored—it would expand outward, so the boundary it leaves behind must move outward too.

Example 4 · Heating a Copper Block

How much heat is required to raise the temperature of a \(0.500\,\mathrm{kg}\) copper block from \(25^\circ\mathrm{C}\) to \(75^\circ\mathrm{C}\)? Take \(c_{\text{Cu}} = 385\,\mathrm{J\,kg^{-1}K^{-1}}\).

Solution. Given: \(m = 0.500\,\mathrm{kg}\), \(\Delta T = 75 - 25 = 50\,\mathrm{K}\), \(c = 385\,\mathrm{J\,kg^{-1}K^{-1}}\).
Find: Heat \(Q\).
Setup: Sensible heating, \(Q = mc\,\Delta T\) (no phase change).
Solve: \[ Q = (0.500)(385)(50) = 9625\,\mathrm{J}. \] Answer: \(\boxed{Q = 9.63\,\mathrm{kJ}}\)
Check: Dimensions: \([\mathrm{kg}][\mathrm{J\,kg^{-1}K^{-1}}][\mathrm{K}] = [\mathrm{J}]\). ✓ The same energy would raise only \(Q/(m\,c_{\text{water}}) = 9625/(0.500\times4186) \approx 4.6\,\mathrm{K}\) in water—copper's small specific heat means metals heat fast.

Example 5 · Melting a Block of Ice

How much heat is needed to melt \(0.250\,\mathrm{kg}\) of ice already at \(0^\circ\mathrm{C}\)? Take \(L_f = 3.34\times10^{5}\,\mathrm{J\,kg^{-1}}\).

Solution. Given: \(m = 0.250\,\mathrm{kg}\), ice at the melting point, \(L_f = 3.34\times10^{5}\,\mathrm{J\,kg^{-1}}\).
Find: Heat \(Q\) for the phase change alone.
Setup: Melting at constant temperature: \(Q = mL_f\) (no \(\Delta T\) term—all the energy goes into breaking the lattice).
Solve: \[ Q = mL_f = (0.250)(3.34\times10^{5}) = 8.35\times10^{4}\,\mathrm{J}. \] Answer: \(\boxed{Q = 83.5\,\mathrm{kJ}}\)
Check: The same \(83.5\,\mathrm{kJ}\) would raise the resulting water by \(Q/(mc_{\text{water}}) = 83500/(0.250\times4186) \approx 80\,\mathrm{K}\). ✓ Melting ice “costs” as much as heating the melt-water by \(80\,\mathrm{K}\)—the large latent heat of fusion in one familiar comparison.

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Scale Conversion
Convert the following to Kelvin: (a) \(37^\circ\mathrm{C}\), (b) \(-40^\circ\mathrm{C}\), (c) \(1064^\circ\mathrm{C}\) (melting point of gold). Which of these can be directly substituted into \(P = \sigma A T^4\)?
(a) \(310\,\mathrm{K}\). (b) \(233\,\mathrm{K}\). (c) \(1337\,\mathrm{K}\). All three Kelvin values can be substituted into \(P = \sigma AT^4\); the Celsius values cannot.
Linear Expansion
An iron rod is \(2\,\mathrm{m}\) long at \(20^\circ\mathrm{C}\). By how much does it expand when heated to \(120^\circ\mathrm{C}\)? How does this compare to the thickness of a human hair (\(\approx 70\,\mu\mathrm{m}\))? Take \(\alpha_{\text{Fe}} = 12\times10^{-6}\,\mathrm{K^{-1}}\).
\(\Delta L = 12\times10^{-6}\times2\times100 = 2.4\,\mathrm{mm} = 2400\,\mu\mathrm{m} \approx 34\) hair-widths.
Specific Heat
How much heat is required to raise the temperature of \(3\,\mathrm{kg}\) of aluminium from \(20^\circ\mathrm{C}\) to \(320^\circ\mathrm{C}\)? Take \(c_{\text{Al}} = 900\,\mathrm{J\,kg^{-1}K^{-1}}\).
\(Q = 3\times900\times300 = 810000\,\mathrm{J} = 810\,\mathrm{kJ}\).
Expansion Mismatch
A brass rod (\(\alpha = 19\times10^{-6}\,\mathrm{K^{-1}}\), length \(1.2\,\mathrm{m}\)) and a steel rod (\(\alpha = 12\times10^{-6}\,\mathrm{K^{-1}}\), length \(1.2\,\mathrm{m}\)) are the same length at \(20^\circ\mathrm{C}\). At what temperature will the brass rod be exactly \(1.5\,\mathrm{mm}\) longer than the steel rod?
\(\Delta L_{\text{brass}} - \Delta L_{\text{steel}} = (19-12)\times10^{-6}\times1.2\times\Delta T = 1.5\times10^{-3}\). \(\Delta T = 1.5\times10^{-3}/(7\times10^{-6}\times1.2) = 178.6\,\mathrm{K}\). \(T = 20 + 178.6 \approx 199^\circ\mathrm{C}\).
Parallel Conduction Paths
A wall section has two parallel paths between the same hot and cold faces: brick (\(k_1 = 0.8\), area \(A_1\), thickness \(L\)) and an air cavity (\(k_2 = 0.026\), area \(A_2\), thickness \(L\)), with \(A_1+A_2 = 1\,\mathrm{m^2}\) and \(A_1 = 3A_2\). Find the effective conductivity.
\(A_1 = 0.75\), \(A_2 = 0.25\,\mathrm{m^2}\). Parallel paths share \(\Delta T\): \(H = (k_1A_1+k_2A_2)\Delta T/L\), and \(k_{\rm eff} = (k_1A_1+k_2A_2)/A = (0.8\times0.75+0.026\times0.25)/1 = 0.607\,\mathrm{W\,m^{-1}K^{-1}}\). The cavity contributes only \(\sim1\%\) of the conductance; brick dominates.