The thermal diffusivity calculator will assist you in estimating the rate of heat transfer from one point to another for different substances. The thermal diffusivity parameter is a function of thermal conductivity, specific heat, and material density.
Thermal diffusivity is the parameter that explains the penetration of heat in a substance. It is used to characterize the unsteady flow of heat and to estimate the temperature field, so you can understand the cooling process inside a material. You can also use it to determine the Prandtl number that appears in several heat-transfer processes.
This parameter is also used in modern non-destructive techniques (NDT) such as thermography, or thermographic damage detection in composite structures. This method is crucial for monitoring the health of different aircraft structures. The sections below explain what thermal diffusivity is and how to calculate the thermal diffusivity of substances.
What is thermal diffusivity?
Thermal diffusivity definition — it is the property of a substance that tells us how heat would flow through it from one point to another. Say you heat one end of a steel plate; the heat travels towards the cooler end. The substance's thermal diffusivity tells us how fast it travels, i.e., the rate of heat transfer across two points. It is measured in units of area per unit time, i.e., mm²/s or ft²/s. The SI unit of thermal diffusivity is m²/s.
Most heat-conducting solids have a high thermal diffusivity. For instance, the thermal diffusivity of copper at 25 °C is about 111 mm²/s, whereas the value for a carbon/carbon composite can be around 216.5 mm²/s. This means heat would flow more rapidly through the carbon/carbon composite than through copper.
The thermal diffusivity (α) depends on three properties of the material: the thermal conductivity k, the specific heat capacity Cp, and the density ρ.
How to calculate thermal diffusivity
Thermal conductivity and thermal diffusivity are related using the thermal diffusivity formula:
α = k / (ρ · Cp)
where:
- α — the thermal diffusivity, in m²/s (or mm²/s, ft²/s);
- k — the thermal conductivity of the material, in W/(m·K);
- ρ — the density of the material, in kg/m³; and
- Cp — the specific heat capacity, in J/(kg·K).
The denominator ρ · Cp is the volumetric heat capacity — the amount of heat needed to raise the temperature of a unit volume by one degree. A material conducts heat quickly (high α) when it transfers heat well (high k) but stores little of it (low ρ·Cp). With this calculator you can rearrange the formula to solve for k, ρ, or Cp as well — just pick the matching mode.
Example: Using the thermal diffusivity calculator
- Select the measurement system — metric (SI) or American (Imperial).
- Choose what you want to find. Keep Thermal diffusivity (α) selected for this example.
- Pick a material preset, or enter the values yourself. For copper: k = 401 W/(m·K), ρ = 8960 kg/m³ and Cp = 385 J/(kg·K).
- The calculator applies the formula: α = k / (ρ · Cp) = 401 / (8960 × 385) ≈ 1.162 × 10⁻⁴ m²/s = 116.2 mm²/s, which is close to the textbook value of ~111 mm²/s for copper.
Thermal diffusivity of substances
Typical thermal diffusivity values at room temperature:
| Substance | α (mm²/s) | α (ft²/s) |
|---|---|---|
| Carbon/carbon composite | 216.5 | 2.33 × 10⁻³ |
| Copper | 111 | 1.19 × 10⁻³ |
| Gold | 127 | 1.37 × 10⁻³ |
| Aluminum | 97 | 1.04 × 10⁻³ |
| Iron | 23 | 2.48 × 10⁻⁴ |
| Carbon steel | 13 | 1.40 × 10⁻⁴ |
| Stainless steel | 3.75 | 4.04 × 10⁻⁵ |
| Air (300 K) | 21.1 | 2.27 × 10⁻⁴ |
| Brick | 0.5 | 5.4 × 10⁻⁶ |
| Glass | 0.34 | 3.7 × 10⁻⁶ |
| Water (25 °C) | 0.14 | 1.5 × 10⁻⁶ |
| Wood (pine) | 0.12 | 1.3 × 10⁻⁶ |
This calculator supports both the metric (SI) system — k in W/(m·K), ρ in kg/m³, Cp in J/(kg·K), α in m²/s and mm²/s — and the American/Imperial system — k in BTU/(h·ft·°F), ρ in lb/ft³, Cp in BTU/(lb·°F), α in ft²/h and ft²/s. Pick the system that fits your data and the units update automatically.
Thermographic damage detection
Because thermal diffusivity governs how fast a temperature change propagates through a material, it is the basis of thermographic damage detection — a non-destructive testing (NDT) method. A surface is heated with a flash or a lamp, and an infrared camera records how the heat diffuses inward over time. A hidden defect, delamination, or void changes the local thermal diffusivity, so the surface above it cools at a different rate and shows up as a hot or cold spot in the thermal image.
This technique is widely used to inspect composite aircraft structures, wind-turbine blades, pipes, and coatings, where it can reveal internal damage without cutting the part open. The same physics — captured by the α = k/(ρ·Cp) relationship — links the measured cooling rate back to the health of the structure.
FAQs
What is the difference between thermal conductivity and thermal diffusivity?
Thermal conductivity (k) measures how well a material transfers heat in a steady state. Thermal diffusivity (α) measures how quickly a material responds to a change in temperature — it combines conductivity with how much heat the material stores (ρ·Cp). A material can conduct heat well yet still respond slowly if it stores a lot of heat.
What is the SI unit of thermal diffusivity?
The SI unit is the square meter per second (m²/s). Because real values are small, they are often quoted in mm²/s (1 m²/s = 1,000,000 mm²/s) or, in imperial work, in ft²/s and ft²/h.
Why does copper have a high thermal diffusivity?
Copper has a very high thermal conductivity and a moderate volumetric heat capacity, so heat spreads through it quickly. That is why heat reaches the cool end of a copper rod much faster than through, say, glass or water.
How is thermal diffusivity used to find the Prandtl number?
The Prandtl number is the ratio of momentum diffusivity (kinematic viscosity ν) to thermal diffusivity (α): Pr = ν / α. Once you know α from this calculator, you can combine it with the fluid's kinematic viscosity to get the Prandtl number used in convective heat-transfer correlations.