PhysicsCBSE Class 11 · Thermal Properties of Matter

Thermal Expansion and Heat Conduction Class 11

Updated 17 October 2026 · Fact-checked

On heating, a solid's length increases by ΔL = αLΔT (area by 2α, volume by 3α); a liquid's apparent expansion is γ_apparent = γ_liquid − γ_vessel. In steady conduction through a slab, the rate of heat flow is H = KAΔT/L, where K is the thermal conductivity; slabs in series carry the same H and their thermal resistances L/KA add.

Understand Thermal Expansion and Heat Conduction

Linear, area and volume expansion. For a small temperature change $Δ T$ :
$ $Δ L = α L Δ T, Δ A = β A Δ T, Δ V = γ V Δ T,$ $
with $β \approx 2 α$ and $γ \approx 3 α$ for an isotropic solid. A hole in a plate expands as if the hole were filled with the same material. Two rods of lengths $L_{1}$ , $L_{2}$ and coefficients $α_{1}$ , $α_{2}$ keep a constant difference in length at every temperature if $L_{1} α_{1} = L_{2} α_{2}$ . A pendulum clock's period is proportional to $L$ , so it runs slow when heated.

Liquids. A liquid in a vessel shows an *apparent* expansion $γ_{a} = γ_{l i q u i d} - γ_{v esse l}$ ; the overflow from a full vessel is $V (γ_{l} - γ_{v}) Δ T$ . Water contracts between 0 °C and 4 °C (maximum density at 4 °C); this anomalous behaviour lets lakes freeze from the top.

Thermal stress. A rod prevented from expanding develops $σ = Y α Δ T$ .

Conduction. In steady state, the rate of heat flow through a slab of area $A$ and thickness $L$ with faces at temperatures difference $Δ T$ is
$ $H = \frac{Q}{t} = \frac{K A Δ T}{L} = \frac{Δ T}{R _{t h}}, R_{t h} = \frac{L}{K A} .$ $
$K$ is the thermal conductivity (W m⁻¹ K⁻¹): large for metals, small for air and wood. For slabs in series, $H$ is the same through each, thermal resistances add, $R = \sum L_{i} / K_{i} A$ , and the temperature drop across each is proportional to its resistance. For slabs in parallel, the conductances add.

Linking to latent heat. The heat conducted along a rod from a hot end to a block of ice melts a mass $m = H t / L_{f}$ .

Thermal Expansion and Heat Conduction: formulas and key facts

Linear expansion: $Δ L = α L Δ T$; $α$ in K⁻¹.
Area and volume: $β \approx 2 α, γ \approx 3 α$; Isotropic solids.
Apparent expansion of a liquid: $γ_{a} = γ_{l} - γ_{v}$; Overflow $= V γ_{a} Δ T$ .
Thermal stress: $σ = Y α Δ T$; Rod held rigidly.
Conduction: $H = \frac{K A Δ T}{L}$; Steady state.
Series slabs: $H = \frac{Δ T}{\sum L _{i} / K _{i} A}$; Same $H$ through each.

How to solve Thermal Expansion and Heat Conduction questions

For expansion, identify which dimension is asked and use the matching coefficient; for conduction, treat the slabs as thermal resistors.

1Convert temperature change to ΔT (a difference, same in °C and K).
2For solids, choose α (length), 2α (area) or 3α (volume).
3For liquids in vessels, subtract the vessel's γ from the liquid's γ.
4For conduction, compute each slab's resistance L/KA and add for series.
5Find H from the total ΔT, then the interface temperature from the drop across the first slab.
6For ice-melting problems, equate Ht to mL_f.

Thermal Expansion and Heat Conduction in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Gaps in a railway track

Suppose steel rails 25 m long (α = 1.2 × 10⁻⁵ K⁻¹) are laid when the temperature is 15 °C and the track reaches 45 °C in summer.

Find the expansion of one rail.

Try it first, then show the solutionSolution

ΔL = αLΔT = 1.2 × 10⁻⁵ × 25 × 30 = 9.0 × 10⁻³ m.

Answer: 9 mm, so gaps of at least 9 mm are needed.

Memory hook: The longer the rail, the greater the expansion.

A cold window pane

Suppose a window pane of area 1.5 m² and thickness 5 mm (K = 1.0 W m⁻¹ K⁻¹) has faces at 18 °C and 8 °C.

Find the rate of heat flow through it.

Try it first, then show the solutionSolution

H = KAΔT/L = 1.0 × 1.5 × 10/0.005 = 3000 W.

Answer: 3000 W.

Memory hook: Thin conducting layers lose heat quickly.

Tricks to remember Thermal Expansion and Heat Conduction

Short memory aids. Say each one aloud once, then use it on a question the same day.

α, 2α, 3α
Length, area, volume coefficients.
Use it when: Expansion type.
Constant difference
L₁α₁ = L₂α₂ keeps the length difference constant.
Use it when: Two-rod problems.
Resistors in series
R_th = L/KA; add for series slabs.
Use it when: Compound slabs.
Liquid minus vessel
Overflow uses γ_liquid − γ_vessel.
Use it when: Full-vessel problems.

Thermal Expansion and Heat Conduction: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Linear expansion of a rod · 2 marks · Very short answerBoard level

A brass rod 2.0 m long at 20 °C is heated to 120 °C (α = 1.9 × 10⁻⁵ K⁻¹). Find the increase in length.

Try it first, then show the solutionSolution

ΔL = αLΔT = 1.9 × 10⁻⁵ × 2.0 × 100. (1)
= 3.8 × 10⁻³ m = 3.8 mm. (1)

Answer: 3.8 mm.

Example 2: Mercury overflow · 3 marks · Short answerHarder board level

A glass flask holds 500 cm³ at 20 °C and is completely filled with mercury. It is heated to 70 °C (γ_mercury = 1.8 × 10⁻⁴ K⁻¹, γ_glass = 2.7 × 10⁻⁵ K⁻¹). Find the volume of mercury that overflows.

Try it first, then show the solutionSolution

Apparent expansion coefficient = 1.8 × 10⁻⁴ − 2.7 × 10⁻⁵ = 1.53 × 10⁻⁴ K⁻¹. (1)
Overflow = V γ_a ΔT = 500 × 1.53 × 10⁻⁴ × 50. (1)
= 3.8 cm³. (1)

Answer: About 3.8 cm³.

Example 3: Rods with a constant difference in length · 3 marks · Short answerHarder board level

A copper rod (α = 1.7 × 10⁻⁵ K⁻¹) and a steel rod (α = 1.1 × 10⁻⁵ K⁻¹) must differ in length by 6 cm at every temperature. Find the lengths of the two rods.

Try it first, then show the solutionSolution

L_cu α_cu = L_st α_st, so L_st/L_cu = 17/11. (1)
L_st − L_cu = 0.06 gives L_cu(17/11 − 1) = 0.06, L_cu = 0.11 m. (1)
L_st = 0.11 × 17/11 = 0.17 m. (1)

Answer: Copper 11 cm, steel 17 cm.

Example 4: Compound slab · 4 marks · Long answerHarder board level

Two slabs of equal area 1.0 m² and thickness 10 cm each are placed face to face. Their conductivities are 0.8 and 0.4 W m⁻¹ K⁻¹. The outer faces are held at 100 °C (first slab) and 0 °C (second slab). Find the rate of heat flow and the temperature of the interface.

Try it first, then show the solutionSolution

R₁ = L/K₁A = 0.1/0.8 = 0.125 K/W; R₂ = 0.1/0.4 = 0.25 K/W; total 0.375 K/W. (1)
H = ΔT/R = 100/0.375 = 267 W. (1)
Drop across slab 1 = H R₁ = 267 × 0.125 = 33.3 K. (1)
Interface temperature = 100 − 33.3 = 66.7 °C. (1)

Answer: 267 W; 66.7 °C.

Example 5: Ice melted by a copper rod · 4 marks · Long answerHOTS / competency

One end of a copper rod of length 0.5 m and cross-section 2.0 cm² (K = 400 W m⁻¹ K⁻¹) is in steam at 100 °C and the other end is in melting ice at 0 °C. The rod is well insulated along its sides (L_f = 3.35 × 10⁵ J/kg). Find (a) the rate of heat flow, (b) the temperature at the midpoint, (c) the mass of ice melted in 1 minute.

Try it first, then show the solutionSolution

(a) H = KAΔT/L = 400 × 2.0 × 10⁻⁴ × 100/0.5 = 16 W. (1)
(b) The temperature falls uniformly along the rod, so the midpoint is at 50 °C. (1)
(c) Q = 16 × 60 = 960 J. (1)
m = Q/L_f = 960/3.35 × 10⁵ = 2.9 × 10⁻³ kg ≈ 2.9 g. (1)

Answer: (a) 16 W; (b) 50 °C; (c) about 2.9 g.

Where students lose marks in Thermal Expansion and Heat Conduction

The slips examiners see most often, and what to write instead.

Mistake: Using ΔL = αΔT without L.
Forgetting the original length.
Do this: ΔL = αLΔT.
Usually costs: The answer mark.
Mistake: Using γ of the liquid alone for the overflow.
Ignoring the vessel's expansion.
Do this: Overflow = V(γ_l − γ_v)ΔT.
Usually costs: 1 to 2 marks.
Mistake: Adding conductivities for slabs in series.
Confusing series with parallel.
Do this: Add resistances L/KA for series.
Usually costs: Method marks.
Mistake: Using °C difference as °F or mixing K and °C absolutes.
Unit confusion.
Do this: ΔT is the same in °C and K.
Usually costs: Accuracy mark.

Exam tips

Name the coefficient used (α, β or γ) before substituting.
For compound slabs draw the temperature profile and label the interface.
Write the steady-state condition (same H through each slab).
Convert area to m² and thickness to m.
Link melting-ice questions through Q = Ht = mL.

Practice questions on Thermal Expansion and Heat Conduction

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The coefficient of area expansion of a solid is approximately
(a) α (b) 2α (c) 3α (d) α/2
Hint
Length is squared.
Answer
(b) 2α.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When a metal plate with a circular hole is heated, the diameter of the hole increases.
Reason (R): The hole expands as if it were made of the same material as the plate.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Imagine the hole filled.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
An aluminium rod 1.5 m long at 0 °C is heated to 100 °C (α = 2.3 × 10⁻⁵ K⁻¹). Find its new length.
Hint
L = L₀(1 + αΔT).
Answer
L = 1.5(1 + 2.3 × 10⁻³) = 1.50345 m.
Q4 · 3 marks · Short answerHarder board level
A steel ring has an inner diameter of 5.994 cm at 20 °C and must slip over a 6.000 cm rod with a clearance of 0.002 cm. To what temperature must the ring be heated (α = 1.2 × 10⁻⁵ K⁻¹)?
Hint
The diameter grows like a length.
Answer
Required diameter = 6.002 cm, so ΔD = 0.008 cm = 5.994 × 1.2 × 10⁻⁵ × ΔT, ΔT ≈ 111 K; T ≈ 131 °C.
Q5 · 4 marks · Long answerHOTS / competency
A pond is covered by a 5 cm layer of ice (K = 2.2 W m⁻¹ K⁻¹, density 920 kg/m³, L_f = 3.35 × 10⁵ J/kg). The upper surface is at −10 °C and the water below is at 0 °C. (a) Find the heat flux through the ice. (b) Find the rate at which the thickness increases, assuming all the flux freezes water at the lower surface.
Hint
Heat flux = KΔT/x; freezing x per second needs ρL x.
Answer
(a) H/A = KΔT/x = 2.2 × 10/0.05 = 440 W/m². (b) ρL dx/dt = 440, so dx/dt = 440/(920 × 3.35 × 10⁵) = 1.43 × 10⁻⁶ m/s ≈ 5.1 mm/hour.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.ΔL formula?Show answer
αLΔT.
2.γ in terms of α?Show answer
3α.
3.Conduction rate?Show answer
KAΔT/L.
4.Thermal resistance?Show answer
L/KA.
5.Thermal stress?Show answer
YαΔT.
6.Maximum density of water at?Show answer
4 °C.

Thermal Expansion and Heat Conduction: frequently asked questions

Why do railway tracks have gaps?

So that expansion in summer does not buckle the rails.

Does a hole in a metal plate shrink on heating?

No, it expands along with the plate.

Why are metals good conductors?

Free electrons carry heat rapidly in addition to lattice vibrations.

What is steady state in conduction?

The temperature at each point stays constant, so the heat flow through every section is the same.

Why does the lake bottom stay at 4 °C?

Water is densest at 4 °C, so colder water rises and freezes at the top while the bottom stays at 4 °C.