MHT-CET Physics · Thermal Properties of Matter

Temperature Scales and Thermal Expansion

A solid grows in every dimension when heated: its length by αLΔT, its area by 2αAΔT and its volume by 3αVΔT, where α is the coefficient of linear expansion; the Celsius, Fahrenheit and Kelvin scales are linear relabellings of one another.

All Thermal Properties of Matter notes MHT-CET Physics strategy

Why this matters

19 PYQs, 2 of them HARD. Two convert between temperature scales. Eleven are linear expansion — the gap left between rails, a rod's α from its growth, and the favourite: two rods whose difference in length stays the same at every temperature. Six are area and volume expansion, including a thermometer's mercury column. Three cards.

Concept 1 of 3: Converting Between Temperature Scales

Every temperature scale is a straight line through two fixed points. Celsius puts ice at 0 and steam at 100; Fahrenheit puts them at 32 and 212; Kelvin is Celsius shifted by 273. So C/5 = (F − 32)/9 and K = C + 273. A question that asks where two scales read the same number sets them equal and solves: C = F at −40, and a body reads the same number in Kelvin and Fahrenheit at about 574.

Definition

$\dfrac{C}{5} = \dfrac{F - 32}{9}$ , $K = C + 273$ .
140 °F = 60 °C; C = F at −40; K = F at ≈ 574.

Scale conversion

\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273}{5}

Worked example

A room is at 25 °C. What is that in Fahrenheit and in kelvin?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q14Moderate

Example 1 · Thermal Properties of Matter · Thermal Expansion — Linear, Surface, Volumetric

The temperature of a body on Kelvin scale is '

x

' K. When it is measured by a Fahrenheit thermometer, it is found to be ' x '

\ ^{\circ}F

. The value of '

x

' is (nearly)

Forgetting the 32 in Fahrenheit

Fahrenheit is not a scaled Celsius: it starts at 32. Subtract 32 before multiplying by 5/9, not after.

Drill 1 more on converting between temperature scales

Concept 2 of 3: Linear Expansion and Rods That Keep Their Difference

A rod of length L heated through ΔT grows by ΔL = αLΔT. A rail laid with a gap needs that gap to equal the growth between the laying temperature and the hottest day. Two rods of different materials keep a constant difference in length only if they grow by the same amount, αₐLₐ = α_bL_b, so the longer rod must be made of the material that expands less. Once that ratio is fixed, the given difference fixes each length.

Definition

$\Delta L = \alpha L \Delta T$ ; $\alpha = \dfrac{\Delta L}{L\,\Delta T}$ (2 m, 1.6 mm over 60 °C ⇒ $1.33\times10^{-5}$ /°C).
Rail gap: 10 m, $\alpha = 1.3\times10^{-5}$ , 17 → 45 °C ⇒ 3.64 mm.
Constant difference: $\alpha_A L_A = \alpha_B L_B$ , $\dfrac{L_A}{L_B} = \dfrac{\alpha_B}{\alpha_A}$ .
Two rods joined, equal growth: $\dfrac{L_1}{L_1 + L_2} = \dfrac{\alpha_2}{\alpha_1 + \alpha_2}$ .

Linear expansion

\Delta L = \alpha L \Delta T

Worked example

A 0.5 m brass rod (α = 2 × 10⁻⁵ /°C) is heated from 20 °C to 120 °C. Increase in length?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q28Hard

Example 2 · Thermal Properties of Matter · Thermal Expansion — Linear, Surface, Volumetric

The difference in length between two rods A and B is 60 cm at all temperatures. If

\alpha_{A}= 18 \times10^{- 6}/\ ^{\circ}C

and

\alpha_{B}= 27 \times10^{- 6}/\ ^{\circ}C

, then the length of

rodA

and

rodB

0^{\circ}C

is respectively

Pairing the longer rod with the larger α

Equal growth needs αL equal, so the longer rod has the SMALLER coefficient: L_A/L_B = α_B/α_A, not α_A/α_B.

Using the final length in ΔL = αLΔT

L is the original length. For the small changes asked here the difference is negligible, but a question that gives length 'at 0 °C' means use that one.

Drill 10 more on linear expansion and rods that keep their difference

Concept 3 of 3: Area and Volume Expansion

Each dimension grows by the same fraction αΔT, so an area grows by twice that fraction and a volume by three times: β = 2α and γ = 3α. A percentage increase in volume over a temperature rise gives γ directly; divide by 3 for α. Two rods of the same material heated equally grow in volume in proportion to their volumes. A liquid-in-glass thermometer works because the liquid's volume grows by γVΔT and has nowhere to go but up a thin stem, so the column rises by that volume divided by the stem's cross-section.

Definition

$\Delta A = 2\alpha A\Delta T$ , $\Delta V = 3\alpha V\Delta T$ (β = 2α, γ = 3α).
From a percentage: $\gamma = \dfrac{\Delta V / V}{\Delta T}$ (0.225% over 30 °C ⇒ $\alpha = 2.5\times10^{-5}$ ).
Cube of side 1 m, α = 18 × 10⁻⁶, 100 °C: $\Delta V = 54\times10^{-4}$ m³.
Thermometer column: $h = \dfrac{\gamma V \Delta T}{A_{\text{stem}}}$ .

Area and volume

\Delta A = 2\alpha A\,\Delta T, \qquad \Delta V = 3\alpha V\,\Delta T

Worked example

A steel plate 20 cm × 10 cm (α = 1.2 × 10⁻⁵ /°C) is heated by 50 °C. Increase in area?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q28Easy

Example 3 · Thermal Properties of Matter · Thermal Expansion — Linear, Surface, Volumetric

The volume of a metal sphere increases by

0.33\%

when its temperature is raised by

50^{\circ}C

. The coefficient of linear expansion of the metal is

Using α for a volume

A volume grows by 3αVΔT. A percentage change in VOLUME gives γ; dividing by 3 gives α, and the options include γ itself.

Drill 5 more on area and volume expansion

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 Temperature Scales
Scale conversion
$\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273}{5}$
Linear Expansion and Rods That Keep Their Difference
Linear expansion
$\Delta L = \alpha L \Delta T$
Area and Volume Expansion
Area and volume
$\Delta A = 2\alpha A\,\Delta T, \qquad \Delta V = 3\alpha V\,\Delta T$

Watch out for (4)

Forgetting the 32 in Fahrenheit
→ Converting Between Temperature Scales
Pairing the longer rod with the larger α
→ Linear Expansion and Rods That Keep Their Difference
Using the final length in ΔL = αLΔT
→ Linear Expansion and Rods That Keep Their Difference
Using α for a volume
→ Area and Volume Expansion

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