Chapter 10 of 14

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West Bengal (WBBSE)Class 11 Physics★ Unit VII · part of the 20-mark block covering Units VII to IX in CBSE Class 11 Physics (70 marks total); calorimetry and thermal expansion are steady sources of JEE Main numericals

Thermal Properties of Matter

Common sense calls both "hotness" and "heat" the same thing. This chapter separates them properly — temperature is a state a body is in, heat is energy on its way somewhere. From there it explains why lakes freeze from the top down, why a steel tape lies to you on a hot day, and why the coldest place in a room and the warmest place trade positions when you switch from heating to cooling. CBSE Class 11 Physics Chapter 10.

⏱ 17 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

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Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Distinguish temperature from heat, and explain why the triple point of water replaced the old ice and steam points
2Apply linear, area and volume thermal expansion, including the expansion of a hole
3Explain water's anomalous expansion between 0C and 4C and its effect on freezing lakes
4Say when a thermal stress develops and when it does not
5Use Q = ms(delta T) and solve calorimetry problems with a calorimeter's water equivalent
6Apply Q = mL across a change of state, including partial melting or partial condensation
7Apply conduction, and describe convection and radiation qualitatively
8Use Stefan's law and Wien's displacement law, and state Kirchhoff's law of radiation
9Apply Newton's law of cooling in its exponential form

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Why this chapter matters

Common sense treats hotness and heat as the same idea. This chapter separates them properly, and that one distinction explains why a steel measuring tape lies to you on a hot day, why lakes freeze from the top down rather than the bottom up, and why a steam burn does far more damage than a splash of boiling water at the same temperature.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Mechanical Properties of Solids

Young's modulus — needed to turn a blocked thermal expansion into a real tension

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Class 11 Physics: Mechanical Properties of Fluids

Density, and the same rho g h reasoning used for pressure-based gas thermometers

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Class 11 Physics: Work, Energy and Power

Energy conservation — the basis of every heat-lost-equals-heat-gained calorimetry problem

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Class 9 Science: Force and Laws of Motion

Basic mechanics used in the pendulum-clock and pressure arguments in this chapter

Thermal Properties of Matter

1. What this chapter covers

Textbook section	Topic
10.1	Introduction
10.2	Temperature and heat
10.3	Measurement of temperature
10.4	Ideal-gas equation and absolute temperature
10.5	Thermal expansion
10.6	Specific heat capacity
10.7	Calorimetry
10.8	Change of state
10.9	Heat transfer — conduction, convection, radiation
10.10	Newton's law of cooling

Two things worth knowing before you revise

Kirchhoff's law of radiation is absent. The chapter never states that a good absorber is also a good emitter, though this is exactly what Exercise 10.19(a) needs to explain why a highly reflective body is a poor emitter. Section 9 below supplies the missing piece.

One exercise needs Chapter 12. Exercise 10.15 asks you to explain why diatomic gases have larger molar specific heats than monatomic ones, and why chlorine's is larger still. That explanation rests on degrees of freedom, a concept this chapter never introduces — it belongs to Kinetic Theory, two chapters ahead. Section 6 below borrows just enough of it to answer the question.

Anomalous expansion of water — listed by CBSE by that exact name — is in the chapter, just under different wording: "water exhibits an anomalous behaviour." No gap there.

2. Temperature and heat are not the same thing

Temperature is what a thermometer reads — a measure of how hot or cold a body is, and a rough indicator of which way heat will flow if the body is put in contact with something else.

Heat is energy transferred between two systems, or between a system and its surroundings, because of a temperature difference.

The distinction settles a question everyone has actually asked. Take a glass tumbler of ice-cold water and a cup of hot tea, both left on a table. Heat flows into the tumbler from the room, and out of the tea into the room. Both processes stop at the same destination — thermal equilibrium with the surroundings — but the direction of energy flow was opposite. Temperature told you which body was hotter; heat is the energy that actually moved.

$[Heat] = joule [Temperature] = kelvin$

3. Measuring temperature

Any physical property that changes measurably and reproducibly with temperature can be used to build a thermometer — mercury's length in a capillary, a wire's electrical resistance, a gas's pressure at fixed volume.

The constant-volume gas thermometer

Hold a gas's volume fixed and its pressure grows with temperature:

$T \propto p ⟹ T = 273.16 K \times \frac{p}{p _{t r i pl e}}$

calibrated against the triple point of water, fixed by definition at 273.16 K.

Why the triple point, and not the melting and boiling points of water that the original Celsius scale used? Both of those depend on pressure — boiling point especially, as anyone who has cooked at altitude knows. The triple point occurs at exactly one pressure, so it needs no further specification. That is the whole reason modern thermometry abandoned the old fixed points.

Why 273.15, not 273.16, in the Celsius conversion

$t_{C} = T - 273.15$

These are two different physical references, and mixing them up is an easy mistake. 273.16 K is the triple point of water — the calibration point that defines the Kelvin scale. 273.15 is the ordinary melting point of ice at standard atmospheric pressure, about 0.01 K lower, and it is that point the everyday Celsius zero is pinned to.

Real gases are not quite ideal

Different gases, used as the working substance in a constant-volume thermometer, give slightly different readings for the same physical temperature at ordinary pressures. Neither instrument is faulty — $T \propto p$ is only exactly true for an ideal gas, and real gases deviate from ideal behaviour differently from each other.

Take the readings at successively lower pressure and extrapolate to zero pressure, and the disagreement vanishes — both gases converge to the same ideal-gas limit. Exercise 10.5 is built entirely around this.

4. Thermal expansion

Most substances expand on heating. There are three kinds, and the names describe exactly what changes:

Kind	What grows	Coefficient
Linear	length	$α$
Area (superficial)	area	$β \approx 2 α$
Volume (cubical)	volume	$γ \approx 3 α$

$Δ L = L_{0} α Δ T$

A hole in a sheet of metal is not an exception. Heat the sheet and the hole expands, exactly as though it were a solid disc of the surrounding material — a fact that feels wrong until you have solved Exercise 10.8 once.

And a measuring tape lies to you when it is at the wrong temperature. A steel tape correctly calibrated at 27°C has, by 45°C, expanded along with everything else — its own centimetre marks are now slightly longer than a true centimetre. A reading taken at 45°C therefore needs correcting upward to recover the true length.

Exercise 10.6 turns on exactly this, and reveals something neat in passing: if the tape and the object it measures are made of the same material, the two expansions very nearly cancel once the object is brought back to the calibration temperature.

The anomalous behaviour of water

Nearly everything contracts on cooling. Water does the opposite between 0°C and 4°C — it contracts on heating over that narrow range, reaching maximum density at 4°C.

This has a real environmental consequence. As a lake cools toward 4°C, the colder surface water is denser, sinks, and is replaced by warmer water rising from below — ordinary convective mixing. But once the surface drops below 4°C it becomes less dense and stays on top, where it freezes. A lake therefore freezes from the top down, not the bottom up, leaving liquid water — and the fish in it — insulated beneath the ice all winter.

The stress a constrained expansion produces

A rod that is free to change length develops no stress at all, whatever its coefficient of expansion — it simply grows or shrinks (Exercise 10.10).

A rod rigidly clamped at both ends cannot change length at all. Cool it, and the contraction it "wants" to undergo instead shows up entirely as a tensile strain, which Young's modulus converts into a real, calculable tension (Exercise 10.9):

$strain = α Δ T ⟹ tension = Y A α Δ T$

The lesson: a thermal stress requires a constraint. Different expansion coefficients meeting at a junction, on their own, produce nothing if both ends are free.

5. Specific heat capacity

$Q = m s Δ T$

Specific heat capacity is heat per unit mass per unit temperature rise — a property of the substance. For gases, it matters how you heat them:

$C_{p} > C_{v}$

Heating at constant pressure means the gas also does work expanding, so it takes more heat to raise the temperature by the same amount than heating at constant volume does.

Why diatomic gases have larger molar specific heats — borrowed from Chapter 12

Exercise 10.15 asks you to explain the gap between monatomic and diatomic molar specific heats. This chapter never derives it, but here is the minimum needed.

A monatomic gas molecule can only move — three independent directions in space, three degrees of freedom. Each contributes $\frac{1}{2} R$ to $C_{v}$ , by the equipartition of energy:

$C_{v} = \frac{3}{2} R \approx 2.98 cal/mol K$

very close to the chapter's quoted 2.92. A diatomic molecule can also rotate, about two axes perpendicular to the bond — two more degrees of freedom:

$C_{v} = \frac{5}{2} R \approx 4.97 cal/mol K$

which matches nitrogen, oxygen, nitric oxide and carbon monoxide closely. Chlorine's value, 6.17, is higher still — a sign that its vibrational mode, the two atoms oscillating along the bond, is already thermally active at room temperature, storing extra energy the other listed gases do not yet access.

6. Calorimetry

An isolated system exchanges no heat with its surroundings. Inside one, heat lost by the hotter part exactly equals heat gained by the colder part:

$heat lost = heat gained$

That single equation solves every mixture and calorimeter problem in this chapter. The calorimeter itself absorbs heat too — accounted for by its water equivalent, an equivalent mass of water with the same heat capacity, added alongside the actual water in the "heat gained" side.

And real calorimeters leak heat to the room. If some of the heat a hot object releases escapes before it can be measured, the specific heat you calculate from the water and calorimeter alone comes out smaller than the true value — because you have divided a real temperature drop by less heat than the object actually gave up. Exercise 10.14 asks exactly this.

7. Change of state

Heat a solid steadily and its temperature climbs — until it starts to melt. Then the temperature holds constant at the melting point for as long as any solid remains, however much heat continues to flow in. The same happens again at the boiling point.

All the heat during a plateau goes into breaking bonds, not raising temperature. That hidden energy is the latent heat:

$Q = m L$

$L_{f}$ for fusion, $L_{v}$ for vaporisation. Water's $L_{v}$ (2256 kJ/kg) dwarfs the specific heat needed to warm the same water by a few degrees — which is the whole reason a steam burn is worse than a hot-water splash at the same temperature, and why steam heating systems outperform hot-water ones (Exercise 10.19(e)): condensing steam dumps its huge latent heat into the room on top of whatever sensible heat the water would give up alone.

8. Heat transfer: conduction, convection, radiation

Conduction

Heat moves between adjacent parts of a body in direct contact, without any bulk motion of matter — put one end of a rod in a flame and the other end eventually burns your hand.

$H = \frac{K A Δ T}{d}$

$K$ is the thermal conductivity — a property of the material, spanning an enormous range: metals like silver and copper are hundreds of times better conductors than wood, glass or water.

This range is why a brass tumbler feels colder than a wooden tray on a chilly day, though both sit at the same room temperature. Brass conducts heat away from your hand far faster than wood does — what you feel is the rate of heat leaving your skin, not any real difference in the objects' temperatures.

Convection

Heat moves with the bulk motion of a fluid — warmer, less dense fluid rises, cooler, denser fluid sinks, setting up a circulating current. This is how a room heater warms an entire room, and how ocean and atmospheric currents redistribute heat around the planet.

Radiation

Every body above absolute zero radiates electromagnetic energy, needing no medium at all — it is how the Sun's heat crosses empty space to reach us.

$H = A e σ T^{4} σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$

$e$ is emissivity: $e = 1$ for a perfect radiator, and less for everything else.

Wien's displacement law ties an object's temperature to the wavelength at which it radiates most strongly:

$λ_{m} T = constant \approx 2.9 \times 1 0^{- 3} m K$

which is why iron heated in a flame visibly shifts from dull red through yellow to white as it gets hotter — its peak wavelength is sliding down as $T$ climbs.

9. The absorption-emission link the chapter never states

Kirchhoff's law of radiation: at a given temperature, a good absorber of a particular wavelength is equally a good emitter of it. This is not stated anywhere in the 2026-27 chapter, but Exercise 10.19(a) depends on it.

A body with large reflectivity absorbs very little — reflectivity and absorptivity are complementary for an opaque surface. By Kirchhoff's law, a poor absorber is equally a poor emitter, so a highly polished, reflective surface radiates heat away only slowly. This is why a dull black surface cools faster than a shiny one at the same starting temperature.

It also explains the optical pyrometer puzzle in Exercise 10.19(c). A pyrometer calibrated against an ideal black body ( $e = 1$ ) reads a red-hot iron piece too low in the open, because real iron has $e < 1$ and so emits less than a perfect radiator at the same true temperature. Inside a furnace, though, repeated reflection between the hot walls makes the whole cavity behave as a near-perfect black body, and the pyrometer reads correctly.

And it is the physics behind Exercise 10.19(d). The atmosphere absorbs and re-radiates much of the infrared the Earth's surface emits, warming the surface. Strip the atmosphere away and that radiation escapes unimpeded, leaving the surface far colder.

10. Newton's law of cooling

For a modest temperature difference, the rate of cooling is proportional to the excess over the surroundings — which integrates to an exponential decay:

$T - T_{s} = (T_{0} - T_{s}) e^{- k t}$

Two consequences worth having ready:

Because the decay is exponential, equal ratios of excess temperature always take equal time — halving the excess temperature twice (a factor of 4) takes exactly twice as long as halving it once.
The law only holds for a small temperature difference from the surroundings; radiation losses, which go as $T^{4}$ , make it break down for large differences.

Summary

Temperature is a state; heat is energy in transit because of a temperature difference — heat can flow either way between two bodies depending on which is hotter.
The triple point of water, 273.16 K, is the modern fixed point because it needs no pressure specification, unlike the melting or boiling points of water.
$t_{C} = T - 273.15$ : the 273.15 is the ordinary ice point, a different reference from the 273.16 triple point used to define the Kelvin scale.
Real gas thermometers agree only in the limit of zero pressure — extrapolate readings there to remove instrument-to-instrument disagreement.
Linear, area and volume expansion: $α$ , roughly $2 α$ , roughly $3 α$ .
A hole in a sheet expands like a solid disc of the same material; it never shrinks on heating.
A tape and the object it measures, made of the same material, very nearly cancel their expansion errors at the calibration temperature.
Water contracts between 0°C and 4°C, reaching maximum density at 4°C — why lakes freeze from the top down.
A thermal stress needs a constraint against expansion; free ends produce no stress regardless of differing coefficients.
$Q = m s Δ T$ ; $C_{p} > C_{v}$ for a gas, since constant-pressure heating also does work.
Diatomic gases have $C_{v} = \frac{5}{2} R$ against a monatomic gas's $\frac{3}{2} R$ , from two extra rotational degrees of freedom — material properly covered in Chapter 12.
In an isolated system, heat lost equals heat gained; a calorimeter's own heat capacity is folded in as its water equivalent.
Unaccounted heat loss to the surroundings makes a measured specific heat come out smaller than the true value.
During a change of state, temperature holds constant while $Q = m L$ is absorbed or released.
$H = K A Δ T / d$ for conduction; convection needs bulk fluid motion; radiation needs no medium at all.
$H = A e σ T^{4}$ , and Wien's law $λ_{m} T =$ constant links peak radiated wavelength to temperature.
Kirchhoff's law — good absorbers are good emitters — is absent from the chapter but needed for Exercise 10.19(a), (c) and (d).
Newton's law of cooling gives exponential decay of the excess temperature; equal ratios of excess temperature take equal time.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Kelvin to Celsius

t_C = T - 273.15

273.15 is the ordinary ice point; 273.16 is the triple point of water, a different reference used to define the Kelvin scale itself.

Constant-volume gas thermometer

T = 273.16 K x (p / p_triple)

Real gases agree with each other only in the limit of zero pressure. Extrapolate readings there to remove instrument-to-instrument disagreement.

Linear, area, volume expansion

dL = L0 alpha dT;  beta = 2 alpha;  gamma = 3 alpha

A hole in a sheet expands like a solid disc of the same material — it enlarges, never shrinks, on heating.

Thermal stress in a constrained rod

tension = Y A alpha (dT)

Only develops when expansion or contraction is blocked. Free ends produce zero stress, however different the coefficients.

Specific heat capacity

Q = m s (dT)

Cp is greater than Cv for a gas, because constant-pressure heating also does work as the gas expands.

Molar specific heat, monatomic and diatomic gases

Cv = (3/2)R monatomic;  Cv = (5/2)R diatomic

From degrees of freedom (Chapter 12): 3 translational for any gas, plus 2 rotational for a diatomic molecule. A larger measured value, as for chlorine, signals an active vibrational mode.

Calorimetry

heat lost = heat gained (isolated system)

Include the calorimeter's own heat capacity via its water equivalent. Unaccounted heat loss makes a measured specific heat come out smaller than the true value.

Latent heat

Q = m L

Temperature holds constant during a change of state while this heat is absorbed or released. Lf for melting, Lv for boiling.

Conduction

H = K A (dT) / d

K is thermal conductivity, a property of the material. Metals conduct hundreds of times faster than wood, glass or water — why metal feels colder to the touch.

Stefan's law of radiation

H = A e sigma T^4,  sigma = 5.67e-8 W/m^2K^4

e = 1 for a perfect radiator (a black body), less for everything else. By Kirchhoff's law, a poor absorber is equally a poor emitter — not stated in the chapter but needed for several 'explain why' exercises.

Wien's displacement law

lambda_m T = constant = 2.9e-3 m K

The peak radiated wavelength falls as temperature rises — why heated iron shifts from dull red to yellow to white.

Newton's law of cooling

T - T_s = (T0 - T_s) e^(-kt)

Exponential decay of the excess temperature. Equal ratios of excess temperature always take equal time, and the law only holds for small differences from the surroundings.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using 273.16 in the everyday Celsius-Kelvin conversion

✓ The conversion t_C = T - 273.15 uses 273.15, the ordinary ice point. 273.16 is a separate reference, the triple point of water, used only to define the Kelvin scale's calibration.

WATCH OUT

✗ Assuming a hole in a heated metal sheet shrinks

✓ A hole expands exactly like a solid disc of the surrounding material would — it always enlarges on heating, never closes in.

WATCH OUT

✗ Using a temperature reading from a tape or scale without correcting for its own thermal expansion

✓ A tape calibrated at one temperature has itself expanded or contracted by the time it is used at another, so its markings no longer represent true length until that is corrected for.

WATCH OUT

✗ Believing different expansion coefficients at a junction always create a thermal stress

✓ A stress needs a constraint against expansion. With both ends free, differing coefficients only produce different unimpeded elongations, and no internal force develops anywhere.

WATCH OUT

✗ Forgetting the calorimeter's own heat capacity in a mixture problem

✓ Add the calorimeter's water equivalent to the heat-gained side of the balance; omitting it understates the heat absorbed and distorts the calculated specific heat.

WATCH OUT

✗ Assuming water always contracts on cooling

✓ Water contracts on HEATING between 0C and 4C, reaching maximum density at 4C, before expanding normally above that. This anomalous range is why lakes freeze from the top down.

WATCH OUT

✗ Reporting a negative bulk-type answer for density change without checking direction

✓ As with any expansion problem, heating increases volume and decreases density; the fractional change in density should come out negative, equal in size to the fractional volume change.

WATCH OUT

✗ Assuming metal feels colder than wood because it truly is a lower temperature

✓ Both are at the same room temperature. Metal conducts heat away from the skin far faster than wood, which is what produces the sensation of cold — a conductivity effect, not a temperature difference.

WATCH OUT

✗ Treating a thermos or furnace pyrometer reading as automatically correct

✓ An optical pyrometer calibrated for an ideal black body reads a real object's temperature too low if its emissivity is below 1, unless the object sits inside an enclosed cavity that itself behaves like a black body.

WATCH OUT

✗ Applying Newton's law of cooling to a large temperature difference

✓ The law's simple exponential form assumes the body's excess temperature over its surroundings stays modest. For large differences, radiation losses growing as the fourth power of temperature make the simple law inaccurate.

WATCH OUT

✗ Using specific heat instead of latent heat when a substance changes state during the calculation

✓ While a substance is melting or boiling its temperature does not change at all, so Q = ms(dT) gives zero for that step. Use Q = mL for the phase-change step, and only apply Q = ms(dT) to the periods before and after it.

WATCH OUT

✗ Explaining Exercise 10.15 using only material from this chapter

✓ The gap between monatomic and diatomic molar specific heats needs the concept of degrees of freedom, which belongs to Chapter 12 (Kinetic Theory) and is not derived here.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Temperature scales and gas thermometers, thermal expansion of solids and the anomalous behaviour of water, calorimetry, latent heat, conduction, convection and radiation, and Newton's law of cooling (Q10.1-10.20)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Thermal Properties of Matter?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~20 marks in West Bengal (WBBSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Temperature is a state; heat is energy transferred because of a temperature difference.
•The triple point of water, 273.16 K, replaced the ice and steam points because it needs no pressure specification.
•t_C = T - 273.15; the 273.15 is the ordinary ice point, distinct from the 273.16 triple point.
•Real gas thermometers agree only in the limit of zero pressure; extrapolate readings there.
•Linear expansion dL = L0(alpha)(dT); area and volume coefficients are roughly 2 alpha and 3 alpha.
•A hole in a heated sheet always enlarges, behaving like a solid disc of the same material.
•A tape at the wrong temperature has itself expanded or contracted, and readings need correcting for it.
•Water contracts on heating between 0C and 4C, reaching maximum density at 4C, which is why lakes freeze top-down.
•A thermal stress needs a constraint against expansion; free ends develop no stress at all.
•Q = ms(dT); Cp exceeds Cv for a gas because constant-pressure heating also does work.
•Diatomic gases have Cv = (5/2)R against a monatomic gas's (3/2)R, from two extra rotational degrees of freedom.
•In an isolated system, heat lost equals heat gained; include the calorimeter's water equivalent.
•Unaccounted heat loss to the surroundings makes a measured specific heat smaller than the true value.
•During a change of state, temperature holds constant while Q = mL is absorbed or released.
•Always check the total heat available in a phase-change mixture before assuming the final state.
•Conduction: H = KA(dT)/d; metals conduct hundreds of times faster than wood, glass or water.
•A metal object feels colder than wood at the same temperature because it conducts heat away from skin faster.
•Convection needs bulk fluid motion; radiation needs no medium at all.
•Stefan's law: H = Ae(sigma)T^4; emissivity e = 1 for a perfect (black body) radiator.
•Kirchhoff's law of radiation — a good absorber is equally a good emitter — is not stated in the chapter but is needed for several explain-why exercises.
•Wien's displacement law: peak radiated wavelength times temperature is constant, which is why hot iron shifts from red to yellow to white.
•An optical pyrometer under-reads a real object's temperature in the open (emissivity below 1) but reads correctly inside an enclosed cavity.
•Newton's law of cooling gives exponential decay of excess temperature, valid only for a modest temperature difference from the surroundings.

West Bengal (WBBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit VII sits inside the 20-mark block covering Units VII to IX (CBSE Class 11 Physics, 70 marks)

Question type	Marks each	Typical count	What it tests
Thermal expansion	3-5	1	Linear, area and volume expansion; constrained versus free expansion
Calorimetry	3-5	1	Heat lost equals heat gained, water equivalent, phase-change problems
Heat transfer	3-5	1	Conduction numericals, and qualitative radiation and convection reasoning
Temperature and heat	2-3	1	Scale conversions, gas thermometers, Newton's law of cooling

Prep strategy

Learn Kirchhoff's law even though the chapter omits it — several explain-why exercises depend on it
Always check whether a calorimetry problem needs a phase-change step before assuming a single mixing temperature
Keep 273.15 (ice point) and 273.16 (triple point) as two distinct numbers, not interchangeable roundings
Practise the free-versus-constrained expansion distinction; it decides whether a thermal stress exists at all
Borrow the degrees-of-freedom idea from Chapter 12 for molar specific heat questions in this chapter

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Bimetallic strip thermostats

Two metals with different expansion coefficients bonded together bend by an amount proportional to temperature, which is used to switch a circuit on or off at a set point.

Expansion joints in bridges and railway tracks

Gaps are deliberately left between sections so that thermal expansion has somewhere to go, preventing the enormous constrained-expansion stresses this chapter calculates.

Why lakes support life through winter

Water's anomalous expansion between 0C and 4C means ice forms at the surface first, insulating the liquid water and any life in it from the cold air above.

Vacuum flasks

A silvered, reflective inner surface minimises radiative heat loss by minimising emissivity, while the vacuum between the walls eliminates conduction and convection entirely.

Steam heating systems

Circulating steam rather than hot water delivers a large amount of extra energy as latent heat when it condenses in the radiator, warming a building more efficiently per kilogram of fluid moved.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

For any 'explain why' radiation question, check whether it needs Kirchhoff's law (good absorber equals good emitter) — the chapter never states it, but several standard questions assume it.

In a mixture or calorimetry problem, calculate the two extreme heat totals first (fully melting/condensing versus none at all) before deciding what the final state actually is.

State explicitly whether a rod's ends are free or constrained before deciding whether a thermal stress exists — this is usually worth a mark on its own.

Keep 273.15 and 273.16 straight: one is the everyday ice point, the other is the triple point used to define the Kelvin scale.

For conduction through composite rods, set the heat current equal on both sides rather than averaging the two temperatures.

When a molar specific heat question appears, remember the degrees-of-freedom reasoning from Chapter 12, since this chapter alone does not supply it.

Check the sign of any fractional density change: heating always decreases density, so the answer should come out negative.

For Newton's law of cooling, use the full exponential relation rather than a crude average rate whenever two separate time intervals are compared.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Derive the bimetallic strip radius-of-curvature formula from first principles using the differing arc lengths of the two bonded layers.

STRETCH

Analyse a gas thermometer's departure from ideal behaviour using the van der Waals equation, and estimate the correction needed at a given pressure.

STRETCH

Derive Newton's law of cooling from Stefan's law in the small-temperature-difference limit, and find the temperature range over which the approximation holds to within 5%.

STRETCH

Work out the equilibrium temperature of a planet from radiative balance alone, using Stefan's law for both absorbed sunlight and emitted infrared radiation.

STRETCH

Investigate the triple point and critical point on a full phase diagram, and explain why some substances (like carbon dioxide) never show a stable liquid phase at atmospheric pressure.

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainThermal expansion and pendulum clocksNumerical

A pendulum clock has a steel pendulum with coefficient of linear expansion $α = 1.2 \times 1 0^{- 5} K^{- 1}$ . If the temperature rises by $10°$ C, find how many seconds the clock gains or loses in one day.

Stuck? Show the approach

The period of a simple pendulum is $T = 2 π L / g$ . Heating the pendulum lengthens it, which changes the period. Find the fractional change in period first, then convert that fraction into seconds over a full day.

Show the full solution

Step 1 — Find the fractional change in length.

$\frac{Δ L}{L} = α Δ T = (1.2 \times 1 0^{- 5}) (10) = 1.2 \times 1 0^{- 4}$

Step 2 — Find the fractional change in period. Since $T \propto L$ , for a small fractional change,

$\frac{Δ T}{T} = \frac{1}{2} \frac{Δ L}{L} = \frac{1}{2} (1.2 \times 1 0^{- 4}) = 6.0 \times 1 0^{- 5}$

Step 3 — Interpret the sign. The rod gets longer, so the period gets longer too — each swing takes slightly more time than it should. A clock whose pendulum swings slower loses time; it falls behind the correct time.

Step 4 — Convert to seconds lost per day. One day has 86400 seconds, and the clock loses this fraction of every second of running time:

$Δ t = \frac{Δ T}{T} \times 86400 = (6.0 \times 1 0^{- 5}) (86400) = 5.184 s$

Answer: The clock loses about 5.18 seconds per day

The trap

Forgetting the factor of one half from the square root in $T \propto L$ , which doubles the answer to about 10.4 s. Also, concluding the clock gains time — a longer pendulum swings slower, so it falls behind, it does not run fast.

JEE MainConduction through composite rodsNumerical

A copper rod of length $1.0$ m and a steel rod of length $0.5$ m, both of the same cross-sectional area, are joined end to end. The free end of the copper rod is kept at $100°$ C and the free end of the steel rod at $0°$ C. Find the temperature of the copper-steel junction in the steady state. Take $K_{co pp er} = 385$ W m⁻¹K⁻¹ and $K_{s t ee l} = 50$ W m⁻¹K⁻¹.

Stuck? Show the approach

In the steady state, the same heat current must flow through both rods — none is stored anywhere in between. Set the conduction rate through the copper equal to the conduction rate through the steel, with the unknown junction temperature appearing in both.

Show the full solution

Step 1 — Write the steady-state condition. Since no heat accumulates at the junction, the rate of heat flow into it from the copper side equals the rate flowing out into the steel side:

$\frac{K _{C u} A ( 100 - T _{j} )}{L _{C u}} = \frac{K _{s t ee l} A ( T _{j} - 0 )}{L _{s t ee l}}$

The area $A$ is common to both and cancels.

Step 2 — Substitute the numbers.

$\frac{385 ( 100 - T _{j} )}{1.0} = \frac{50 T _{j}}{0.5} = 100 T_{j}$

Step 3 — Solve for $T_{j}$ .

$38500 - 385 T_{j} = 100 T_{j}$ $38500 = 485 T_{j}$ $T_{j} = \frac{38500}{485} \approx 79.4° C$

Step 4 — Sense-check. Copper conducts far better than steel (385 against 50), so it should take only a small temperature drop across the copper to drive the same heat current that needs a much larger drop across the poorly-conducting steel. The junction sitting close to the copper's own end temperature (100°C) rather than in the middle is exactly what that asymmetry predicts.

Answer: T_j ≈ 79.4°C

The trap

Assuming the junction sits at the simple average of the two end temperatures, 50°C. That would only be true if the two rods had equal $K / L$ ; here copper's much higher conductivity means the junction sits far closer to copper's own temperature.

JEE AdvancedBimetallic stripNumerical

A bimetallic strip is made of two metal strips of equal thickness $0.5$ mm each, bonded face to face, with coefficients of linear expansion $α_{1} = 2.0 \times 1 0^{- 5} K^{- 1}$ (brass) and $α_{2} = 2.0 \times 1 0^{- 6} K^{- 1}$ (invar). If the strip is heated through $50°$ C, find the radius of curvature it bends into, using the approximate formula $R = d / [(α_{1} - α_{2}) Δ T]$ , where $d$ is the total thickness.

Stuck? Show the approach

This uses a standard result for bimetallic strips that is not derived in the NCERT chapter: unequal expansion of the two bonded layers forces the strip to bend into an arc, with the faster-expanding metal on the outside (convex side) of the curve. Apply the given formula directly, being careful with units.

Show the full solution

Step 1 — Identify the given quantities.

$d = 0.5 mm + 0.5 mm = 1.0 mm = 1.0 \times 1 0^{- 3} m$ $α_{1} - α_{2} = 2.0 \times 1 0^{- 5} - 2.0 \times 1 0^{- 6} = 1.8 \times 1 0^{- 5} K^{- 1}$ $Δ T = 50 K$

Step 2 — Substitute into the formula.

$R = \frac{d}{( α _{1} - α _{2} ) Δ T} = \frac{1.0 \times 1 0 ^{- 3}}{( 1.8 \times 1 0 ^{- 5} ) ( 50 )} = \frac{1.0 \times 1 0 ^{- 3}}{9.0 \times 1 0 ^{- 4}}$

$R \approx 1.11 m$

Step 3 — State which way it bends. Brass expands more than invar for the same temperature rise, so the brass layer must stretch further along the outside of the curve — the strip bends with brass on the convex (outer) side and invar on the concave (inner) side, curling toward the invar.

Step 4 — Why this matters. This is the mechanism behind a bimetallic thermostat: the curvature is directly proportional to $Δ T$ , so the amount of bending can be used to switch a circuit at a chosen temperature.

Answer: R ≈ 1.11 m, curving with brass on the outside

The trap

Getting the direction of curvature backwards. The metal with the LARGER expansion coefficient ends up on the outside of the curve, since it must cover a longer arc length than the metal it is bonded to.

JEE AdvancedCalorimetry with partial phase changeNumerical

20 g of ice at $- 10°$ C is mixed with 10 g of steam at $100°$ C in an insulated container. Find the final temperature and composition of the mixture. (Specific heat of ice = 2100 J kg⁻¹K⁻¹, specific heat of water = 4186 J kg⁻¹K⁻¹, latent heat of fusion of ice = 3.35×10⁵ J kg⁻¹, latent heat of vaporisation of steam = 2.256×10⁶ J kg⁻¹.)

Stuck? Show the approach

Don't assume the final state before checking. First calculate the heat needed to bring all the ice up to 100°C as liquid water, then compare that against the maximum heat the steam could possibly release by condensing completely. Whichever is smaller tells you what actually happens.

Show the full solution

Step 1 — Find the heat needed to bring all the ice to 100°C water.

Warming ice from −10°C to 0°C: $Q_{1} = (0.020) (2100) (10) = 420 J$

Melting the ice at 0°C: $Q_{2} = (0.020) (3.35 \times 1 0^{5}) = 6700 J$

Warming the melted water from 0°C to 100°C: $Q_{3} = (0.020) (4186) (100) = 8372 J$

$Q_{n ee d e d} = 420 + 6700 + 8372 = 15492 J$

Step 2 — Find the maximum heat the steam could release. If all 10 g of steam condensed completely at 100°C:

$Q_{ma x} = (0.010) (2.256 \times 1 0^{6}) = 22560 J$

Step 3 — Compare, and decide the final state. Since $Q_{ma x} (22560 J) > Q_{n ee d e d} (15492 J)$ , the steam has more than enough heat to bring all the ice to 100°C water — so the final state must be at exactly 100°C, with only part of the steam having condensed. (If $Q_{ma x}$ had been smaller than $Q_{n ee d e d}$ , the final state would instead be water below 100°C with no steam left.)

Step 4 — Find how much steam actually condenses. Only enough steam condenses to supply exactly $Q_{n ee d e d}$ :

$m_{co n d e n se d} = \frac{Q _{n ee d e d}}{L _{v}} = \frac{15492}{2.256 \times 1 0 ^{6}} = 6.87 \times 1 0^{- 3} kg = 6.87 g$

Step 5 — State the final composition. Liquid water present: $20 + 6.87 = 26.87$ g. Steam remaining: $10 - 6.87 = 3.13$ g.

Answer: Final temperature 100°C; about 26.9 g of liquid water and 3.1 g of steam remain, in equilibrium

The trap

Assuming all the steam condenses and solving for a single final temperature above 100°C — which is physically impossible, since water cannot exist as a liquid above 100°C at atmospheric pressure. Always check the two extreme heat totals first before assuming which phase-change outcome actually occurs.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Thermal Properties)High

NEET PhysicsMedium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Temperature tells you how hot or cold a body is, and predicts which way heat will flow if it touches another body. Heat is the energy that actually flows between two systems because of that temperature difference. A large iceberg and a cup of hot coffee can contain vastly different total thermal energy despite the iceberg being far colder — temperature and total heat content are simply not the same quantity. When two bodies are placed in contact, heat always flows from the higher temperature to the lower one until they reach the same temperature, at which point the flow stops even though both bodies still contain plenty of internal energy.

This exercise relies on Kirchhoff's law of radiation, which states that at a given temperature, a good absorber of radiation is equally a good emitter of it, and a poor absorber is equally a poor emitter. The 2026-27 NCERT chapter never states this law explicitly, even though the exercise assumes it. A body with high reflectivity absorbs very little incoming radiation, since most of it bounces off, so by Kirchhoff's law it also emits very little — which is why polished, shiny surfaces stay hot longer than dull, dark ones at the same starting temperature, and why thermos flasks are silvered on the inside.

Ice floats because water's anomalous expansion means it is less dense as a solid than as a liquid near its freezing point — the opposite of almost every other substance. As a lake cools toward 4°C, the colder surface water is denser and sinks, mixing normally with the water below. But below 4°C the water becomes progressively less dense again, so it stays at the surface and freezes there first. The floating ice layer then insulates the liquid water beneath from further cooling, which is why lakes freeze from the top down and aquatic life can survive underneath through winter.

Both are at 100°C, but steam carries an enormous amount of extra energy that plain boiling water does not — its latent heat of vaporisation, about 2256 kJ per kilogram. When steam condenses on skin, it releases all of that latent heat first, before even beginning to cool down as liquid water, delivering far more total energy to the skin than an equal mass of water at the same 100°C, which can only give up the comparatively small amount of energy involved in cooling down from 100°C.

Check the temperatures and substances involved before assuming a single mixing temperature. If the problem includes ice, calculate the heat needed to warm and then melt all of it; if it includes steam, calculate the maximum heat it could release by condensing completely. Compare that against the heat available from the other substance. If one substance has more than enough heat to fully convert the other's phase, the final state sits exactly at the phase-change temperature (0°C or 100°C) with only part of one substance changing phase — not a simple weighted average of the starting temperatures.

Exercise 10.15 asks why diatomic gases like nitrogen and oxygen have larger molar specific heats than monatomic gases, and this chapter never explains it, because the explanation rests on degrees of freedom — the independent ways a molecule can store energy — which is introduced properly in the kinetic theory of gases, two chapters later. A monatomic gas can only store energy in translational motion, giving Cv = (3/2)R. A diatomic molecule can also rotate about two axes, adding two more degrees of freedom and giving Cv = (5/2)R, which matches most of the gases listed in that exercise.

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