JEE Mains Physics · Thermal Properties of Matter

Temperature Scales, Expansion and Thermal Stress

Any two linear temperature scales agree on the fraction of the way from the ice point to the steam point; a heated solid grows by ΔL = LαΔT in each direction, and a solid that is not allowed to grow carries a stress YαΔT instead.

All Thermal Properties of Matter notes JEE Mains Physics strategy

Why this matters

Twenty-three PYQs, six of them asking for a number, and four from 2026. Five convert a reading between temperature scales, twelve find how much a length, an area or a volume grows, and six find the stress, force or stored energy in a rod that is held so it cannot grow. Most need one line of arithmetic once the right coefficient, α, 2α or 3α, is chosen.

Concept 1 of 3: Converting between linear temperature scales

Every linear thermometer scale is fixed by two points: the reading in melting ice and the reading in steam. A temperature sits a certain fraction of the way between those two points, and that fraction is the same on every scale. So to convert, find the fraction on one scale and put it into the other. A faulty thermometer is just another linear scale with its own two fixed points.

Definition

Any two linear scales: $\dfrac{X - X_{ice}}{X_{steam} - X_{ice}}$ has the same value on both.
Celsius, Fahrenheit, kelvin: $\dfrac{C}{100} = \dfrac{F - 32}{180} = \dfrac{K - 273}{100}$ , so $\dfrac{C}{5} = \dfrac{F - 32}{9}$ .
A CHANGE carries no offset: $\Delta F = 1.8\,\Delta C$ and $\Delta K = \Delta C$ .
Celsius against Fahrenheit is a straight line: $C = \tfrac{5}{9}F - \tfrac{160}{9}$ , slope $\tfrac{5}{9}$ , crossing the F-axis at 32 and the C-axis below the origin.
The two scales read the same number at −40.
A scale with 150 divisions between its fixed points has divisions $\tfrac{100}{150}$ of a Celsius degree.

Two linear scales

\frac{X - X_{ice}}{X_{steam} - X_{ice}} = \frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}

Worked example

A temperature scale Y reads

20^{\circ}Y

at the ice point and

220^{\circ}Y

at the steam point. What is

70^{\circ}Y

in degrees Celsius and in degrees Fahrenheit?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q16Moderate

Example 1 · Thermal Properties of Matter · Temperature Scales, Expansion and Thermal Stress

On a temperature scale '

X

'. The boiling point of water is

65^{\circ}X

and the freezing point is

-15^{\circ}X

. Assume that the

X

scale is linear. The equivalent temperature corresponding to

-95^{\circ}X

on the Fahrenheit scale would be:

Adding 32 to a temperature change

The 32 shifts a reading, not a difference. A rise of 25 Celsius degrees is a rise of 45 Fahrenheit degrees, not 77.

Using 0 and 100 for a faulty thermometer

A faulty thermometer has its own ice and steam readings. Measure the fraction between those readings, then put it on the true scale.

Drill 4 more on converting between linear temperature scales

Concept 2 of 3: Linear, area and volume expansion of solids and gases

Heat a solid and every length in it grows by the same fraction, αΔT. An area has two lengths, so it grows by about 2αΔT; a volume has three, so about 3αΔT. A hole grows too, exactly as the piece of metal that would fill it. For a gas the volume coefficient comes from the gas law: at constant pressure V is proportional to T, so the fraction it grows per kelvin is 1/T.

Definition

Length: $\Delta L = L\alpha\Delta T$ . Area: $\Delta A = A(2\alpha)\Delta T$ . Volume: $\Delta V = V(3\alpha)\Delta T$ .
A hole, or a gap in a ring, expands as if it were filled with the same material.
Rods joined end to end: add each piece's own $L\alpha\Delta T$ .
A second rise in temperature starts from the longer length $L_0 + \Delta L_1$ , not from $L_0$ .
Two rods whose difference in length never changes: $L_1\alpha_1 = L_2\alpha_2$ .
If the heat supplied is given, find $\Delta T = Q/(ms)$ first.
A bimetallic strip bends with the larger-α metal on the outside when heated, and on the inside when cooled.
Ideal gas: $\gamma = \dfrac{1}{V}\dfrac{dV}{dT}$ . At constant pressure $\gamma = 1/T$ . If $PT^{n}$ is constant, $V \propto T^{n+1}$ and $\gamma = (n + 1)/T$ .

Thermal expansion

\Delta L = L\alpha\Delta T \qquad \Delta A = A(2\alpha)\Delta T \qquad \Delta V = V(3\alpha)\Delta T \qquad \gamma_{gas} = \frac{1}{V}\frac{dV}{dT}

Worked example

A brass plate has a circular hole of diameter 4.00 cm at

20^{\circ}C

. The plate is heated to

220^{\circ}C

. Find the new diameter of the hole and the percentage increase in its area. (

\alpha_{brass} = 1.9 \times 10^{-5}\ ^{\circ}C^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q15Moderate

Example 2 · Thermal Properties of Matter · Temperature Scales, Expansion and Thermal Stress

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at

30^{\circ}C

. The coefficient of linear expansion of aluminium and steel are

24 \times 10^{- 6}/\ ^{\circ}C

and

1.2 \times 10^{- 5}/\ ^{\circ}C

, respectively. The length of this composite rod when its temperature is raised to

100^{\circ}C

, is

\_\_\_\_

cm.

Using α for an area or a volume

An area grows by 2αΔT and a volume by 3αΔT. Using α alone gives an answer two or three times too small, and that value is usually an option.

Thinking a hole shrinks when the plate is heated

A hole grows exactly as a disc of the same metal would. The metal around it expands outward, carrying the edge of the hole with it.

Reporting the rise instead of the final temperature

ΔL = LαΔT gives the rise ΔT. If the question asks for the temperature to heat to, add the starting temperature. Options often include both.

Drill 11 more on linear, area and volume expansion of solids and gases

Concept 3 of 3: Thermal stress in a rod that cannot expand

A rod lying free simply grows when heated and feels no stress. Clamp its ends and it cannot grow, so the supports squeeze it back by exactly the strain it wanted, αΔT. Hooke's law then gives the stress, Y times that strain. A wire held taut between supports and then cooled is the same story the other way: it wants to shrink, cannot, and is pulled into tension.

Definition

Forced strain $= \alpha\Delta T$ ; thermal stress $\sigma = Y\alpha\Delta T$ ; force $F = YA\alpha\Delta T$ .
The length of the rod cancels: the force does not depend on it.
Heating a clamped rod gives compression; cooling a clamped wire gives tension.
The force is proportional to the temperature change measured from the stress-free state.
A hanging rod cooled by ΔT is stretched back to its old length by a load $Mg = YA\alpha\Delta T$ .
Stored elastic energy: per unit volume $\tfrac{1}{2}Y(\alpha\Delta T)^{2}$ ; per unit length multiply by A.
A rod free to expand has no thermal stress at all.

Thermal stress

\sigma = Y\alpha\Delta T \qquad F = YA\alpha\Delta T \qquad u = \tfrac{1}{2}Y(\alpha\Delta T)^{2}

Worked example

A steel rod of cross-section

2\ \text{cm}^{2}

is clamped between rigid walls and heated by

50^{\circ}C

. Find the stress, the force on the walls and the elastic energy stored per unit volume. (

Y = 2 \times 10^{11}\ \text{N m}^{-2}

\alpha = 1.2 \times 10^{-5}\ ^{\circ}C^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q14Moderate

Example 3 · Thermal Properties of Matter · Temperature Scales, Expansion and Thermal Stress

A brass wire of length 2 m and radius 1 mm at

27^{\circ}C

is held taut between two rigid supports. Initially it was cooled to a temperature of

-43^{\circ}C

creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4 T , is

\_\_\_\_

\ ^{\circ}C

Measuring the temperature change from the wrong state

Thermal tension is proportional to the change from the stress-free temperature, not from the last temperature mentioned. To raise the tension by 40%, the total cooling from the stress-free state must rise by 40%.

Putting the length into the force

Force = YAαΔT. The length appears in both the forced extension and the strain, and cancels. A rod twice as long pushes on its clamps with the same force.

Drill 5 more on thermal stress in a rod that cannot expand

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Converting between linear temperature scales
Two linear scales
$\frac{X - X_{ice}}{X_{steam} - X_{ice}} = \frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}$
Linear, area and volume expansion of solids and gases
Thermal expansion
$\Delta L = L\alpha\Delta T \qquad \Delta A = A(2\alpha)\Delta T \qquad \Delta V = V(3\alpha)\Delta T \qquad \gamma_{gas} = \frac{1}{V}\frac{dV}{dT}$
Thermal stress in a rod that cannot expand
Thermal stress
$\sigma = Y\alpha\Delta T \qquad F = YA\alpha\Delta T \qquad u = \tfrac{1}{2}Y(\alpha\Delta T)^{2}$

Watch out for (7)

Adding 32 to a temperature change
→ Converting between linear temperature scales
Using 0 and 100 for a faulty thermometer
→ Converting between linear temperature scales
Using α for an area or a volume
→ Linear, area and volume expansion of solids and gases
Thinking a hole shrinks when the plate is heated
→ Linear, area and volume expansion of solids and gases
Reporting the rise instead of the final temperature
→ Linear, area and volume expansion of solids and gases
Measuring the temperature change from the wrong state
→ Thermal stress in a rod that cannot expand
Putting the length into the force
→ Thermal stress in a rod that cannot expand

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