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Free Thermal Expansion Calculator

This thermal expansion calculator finds how much a solid object grows or shrinks in length when its temperature changes. Enter the original length L₀, the material's linear expansion coefficient α (for example, steel ≈ 1.2×10⁻⁵ per °C), and the temperature change ΔT, and it returns the change in length ΔL in both metres and millimetres along with the final length. Linear thermal expansion is the everyday reason that bridges have expansion joints, railway tracks need gaps, and metal lids loosen under hot water — almost every solid expands when heated and contracts when cooled, and this tool quantifies that effect exactly.

Enter the original length, the linear expansion coefficient α (steel ≈ 0.000012 /°C), and the temperature change ΔT to find the change in length and the new length.

Result
Change in length ΔL0.006 m
Change in length ΔL6 mm
New length L₀ + ΔL10.006 m

Quick answer

Linear thermal expansion is calculated with ΔL = α·L₀·ΔT, where ΔL is the change in length, α is the material's linear expansion coefficient (per °C), L₀ is the original length, and ΔT is the temperature change. For example, a 10 m steel bar with α = 1.2×10⁻⁵ /°C heated by 50 °C expands by ΔL = 0.000012 × 10 × 50 = 0.006 m, or 6 mm, giving a new length of 10.006 m.

Formula & method

ΔL = α · L₀ · ΔT     •     L_new = L₀ + ΔL     •     ΔT = T_final − T_initial
  • ΔL — Change in length (positive = expansion, negative = contraction)
  • α — Coefficient of linear thermal expansion, per °C (steel ≈ 1.2×10⁻⁵)
  • L₀ — Original length before the temperature change
  • ΔT — Temperature change, final minus initial (°C or K)

ΔL = change in length, α = linear expansion coefficient (per °C or per K — numerically identical for a temperature difference), L₀ = original length, ΔT = temperature change. ΔL comes out in the same length unit as L₀.

Examples

Example 1: Default — steel bar heated by 50 °C
Input
L₀ = 10 m, α = 1.2×10⁻⁵ /°C, ΔT = 50 °C
Result
ΔL = 0.006 m = 6 mm, new length = 10.006 m
Why
Apply ΔL = α·L₀·ΔT = 0.000012 × 10 × 50 = 0.006 m. Converting to millimetres, 0.006 m × 1000 = 6 mm. The new length is L₀ + ΔL = 10 + 0.006 = 10.006 m.
Example 2: Railway rail expanding on a hot day
Input
L₀ = 25 m, α = 1.2×10⁻⁵ /°C (steel), ΔT = 30 °C
Result
ΔL = 0.009 m = 9 mm, new length = 25.009 m
Why
ΔL = 0.000012 × 25 × 30 = 0.009 m = 9 mm. A single 25 m rail therefore stretches 9 mm when it warms 30 °C, which is exactly why continuous track is laid with expansion gaps or pre-stressed.
Example 3: Aluminium bridge span in summer heat
Input
L₀ = 50 m, α = 2.3×10⁻⁵ /°C (aluminium), ΔT = 40 °C
Result
ΔL = 0.046 m = 46 mm, new length = 50.046 m
Why
ΔL = 0.000023 × 50 × 40 = 0.046 m = 46 mm. Aluminium expands roughly twice as much as steel for the same conditions, so a 50 m span grows almost 5 cm — a real expansion joint must absorb this movement.
Example 4: Copper pipe cooling (contraction)
Input
L₀ = 20 m, α = 1.7×10⁻⁵ /°C (copper), ΔT = −60 °C
Result
ΔL = −0.0204 m = −20.4 mm, new length = 19.9796 m
Why
With a negative temperature change, ΔL = 0.000017 × 20 × (−60) = −0.0204 m = −20.4 mm. The pipe shrinks by 20.4 mm, so the new length is 20 − 0.0204 = 19.9796 m. A negative ΔT always gives contraction.

When to use this tool

  • Sizing expansion joints or allowable gaps for pipes, rails, bridges, and long metal runs that see seasonal temperature swings.
  • Checking the clearance or fit between mating parts, such as a shaft in a bearing or a shrink-fit ring, after a temperature change.
  • Estimating sag, buckling risk, or stress in beams, cables, and ducts when ambient or process temperatures rise.
  • Solving physics and engineering homework on linear thermal expansion where you know α, L₀, and ΔT.
  • Comparing how different materials (steel vs. aluminium vs. copper) move under the same heating or cooling conditions.

Common mistakes

  • Forgetting the negative power of ten in the coefficient. Steel's α is 1.2×10⁻⁵, i.e. 0.000012 per °C, not 1.2 or 12. Entering 1.2 overstates the expansion by 100,000 times.
  • Using a temperature instead of a temperature change. The formula needs ΔT = T_final − T_initial (e.g. 80 − 20 = 60 °C), not the absolute final temperature on its own.
  • Mixing up length and area/volume expansion. This tool computes linear (1-D) expansion; area expansion uses about 2α and volume expansion about 3α, so do not apply the linear result directly to a surface or a volume.
  • Mismatching length units. ΔL is returned in the same unit you enter L₀ in (here, metres). If you enter L₀ in millimetres, ΔL will also be in millimetres — keep units consistent throughout.
  • Assuming α is constant over huge temperature ranges. Published coefficients are room-temperature averages; near phase changes or at cryogenic temperatures, α itself varies and the linear formula becomes an approximation.

Frequently asked questions

What is the formula for linear thermal expansion?

The linear thermal expansion formula is ΔL = α·L₀·ΔT, where ΔL is the change in length, α is the coefficient of linear expansion (per °C), L₀ is the original length, and ΔT is the temperature change. The new length is simply L₀ + ΔL. For example, 10 m of steel (α = 1.2×10⁻⁵) heated 50 °C grows 0.006 m, or 6 mm.

What is the coefficient of thermal expansion for common materials?

Typical room-temperature linear coefficients (per °C) are: steel ≈ 1.2×10⁻⁵, stainless steel ≈ 1.6×10⁻⁵, copper ≈ 1.7×10⁻⁵, brass ≈ 1.9×10⁻⁵, aluminium ≈ 2.3×10⁻⁵, glass ≈ 0.9×10⁻⁵, concrete ≈ 1.2×10⁻⁵, and Invar ≈ 0.1×10⁻⁵. A larger α means the material moves more for the same temperature change.

Do I use Celsius or Kelvin for the temperature change?

Either works, because a temperature change of 1 °C equals a change of 1 K — only the difference matters, not the zero point. So ΔT = 50 °C and ΔT = 50 K give the same expansion. Just make sure your α is expressed in the matching unit (per °C or per K); for linear expansion the numerical value is identical.

What happens when the object cools instead of heats?

When the temperature falls, ΔT is negative, so ΔL is negative and the object contracts. For instance, a 20 m copper pipe cooled by 60 °C gives ΔL = 0.000017 × 20 × (−60) = −0.0204 m, meaning it shrinks by about 20.4 mm. The same formula handles both expansion and contraction automatically.

How is linear expansion different from area and volume expansion?

Linear expansion describes the change in one dimension (length). For the same material and temperature change, the area expansion coefficient is approximately 2α and the volume (or cubic) expansion coefficient is approximately 3α. This calculator uses the linear form ΔL = α·L₀·ΔT, so use the doubled or tripled coefficient if you need area or volume change.

Why do bridges and railway tracks need expansion gaps?

Because they are long and made of metals with appreciable α, even modest temperature swings produce real movement. A 50 m aluminium span warming 40 °C expands about 46 mm. Without expansion joints or gaps, that growth would create enormous compressive stress, leading to buckling or cracking — so engineers design clearances using exactly this formula.

Sources & references

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