Thermal Diffusivity Calculator
Enter thermal conductivity, density, and specific heat capacity to calculate thermal diffusivity. Switch the solve mode to find any one of those three properties from the others. The calculator also outputs penetration depth, thermal response time, and the Fourier number for a given sample length and time, with a step-by-step breakdown of every formula used.
What is thermal diffusivity?
Thermal diffusivity (symbol alpha, units m²/s) measures how quickly a temperature change at one point in a material propagates to another point. It is defined as the ratio of thermal conductivity to the volumetric heat capacity: alpha = k / (rho × cp). A high value means the material equalises temperatures quickly; a low value means it resists transient heat flow and acts more like a thermal buffer. Copper and aluminium, with their high conductivity and relatively modest heat capacity, spread heat through themselves in milliseconds. Rubber and wood, despite some conductivity, store so much energy per unit volume that waves of heat take much longer to travel the same distance.
How to use this calculator
Select a material preset or choose "Custom" and enter values for thermal conductivity (k in W/(m·K)), density (rho in kg/m³), and specific heat capacity (cp in J/(kg·K)). The "Solve for" dropdown lets you run the calculation in any direction: solve for alpha from the three properties, or solve for any one property if you know alpha and the other two. Enter a characteristic length (the half-thickness or radius of your sample) and an elapsed time to also get penetration depth, thermal response time, and the Fourier number. The diffusivity unit selector lets you switch the output between m²/s, mm²/s, cm²/s, ft²/s, and ft²/hr.
Penetration depth, response time and the Fourier number
Beyond the primary diffusivity value, this calculator provides three applied outputs. Penetration depth (delta = sqrt(pi × alpha × t)) tells you how far a thermal wave has diffused into the material after a given time, useful for estimating the time needed to heat or cool a part through its thickness. Thermal response time (tau = L² / (pi × alpha)) is the characteristic time for heat to travel across a length L; it answers the question of how long it takes for a slab of given thickness to equilibrate. The Fourier number (Fo = alpha × t / L²) is the dimensionless ratio of the heat diffusion rate to the heat storage rate for the geometry. When Fo > 0.2, the temperature profile inside the body is close to its steady shape and a lumped capacitance approach gives good accuracy. When Fo < 0.2, significant internal gradients remain and a distributed-parameter model is more appropriate.
Thermal diffusivity vs thermal conductivity
Engineers sometimes confuse thermal diffusivity with thermal conductivity. Conductivity (k) describes steady-state heat transfer: how many watts flow through a slab when there is a fixed temperature difference across it. Diffusivity (alpha) describes the transient response: how fast the temperature inside a body changes when boundary conditions shift. A material can have high conductivity but low diffusivity if it also has a large volumetric heat capacity, which is approximately the case for water. Conversely, air has very low conductivity yet a moderate diffusivity because it also has a tiny volumetric heat capacity (low density). For any problem involving time-varying temperatures, diffusivity is the governing parameter; for steady-state insulation problems, conductivity is what matters.
Typical thermal diffusivity values for common materials
| Material | k (W/(m·K)) | rho (kg/m³) | cp (J/(kg·K)) | alpha (mm²/s) |
|---|---|---|---|---|
| Silver | 429 | 10500 | 235 | 173.7 |
| Copper | 401 | 8960 | 385 | 116.2 |
| Gold | 318 | 19300 | 128 | 128.7 |
| Aluminium | 205 | 2700 | 900 | 84.4 |
| Iron (pure) | 80 | 7874 | 449 | 22.6 |
| Carbon steel | 50 | 7850 | 490 | 13.0 |
| Titanium | 22 | 4507 | 520 | 9.4 |
| Concrete | 1.7 | 2300 | 880 | 0.84 |
| Glass | 1 | 2500 | 840 | 0.48 |
| Wood (pine) | 0.12 | 550 | 1700 | 0.13 |
| Engine oil | 0.145 | 884 | 1900 | 0.086 |
| Water (25 °C) | 0.598 | 997 | 4182 | 0.143 |
| Air (0 °C) | 0.024 | 1.29 | 1005 | 18.5 |
| Rubber | 0.16 | 1200 | 2000 | 0.067 |
All values at approximately 20-25 °C. Actual values vary with temperature and purity.
Frequently asked questions
What is the formula for thermal diffusivity?
Thermal diffusivity alpha = k / (rho × cp), where k is thermal conductivity in W/(m·K), rho is density in kg/m³, and cp is specific heat capacity in J/(kg·K). The product rho × cp is the volumetric heat capacity, which represents the energy stored per unit volume per kelvin. A higher denominator means the material buffers heat more effectively and diffuses it more slowly.
What units does thermal diffusivity use?
The SI unit is m²/s, but values for most solids are so small (on the order of 10⁻⁵ to 10⁻⁷ m²/s) that mm²/s and cm²/s are often more convenient in practice. This calculator supports m²/s, mm²/s, cm²/s, ft²/s, and ft²/hr.
Which material has the highest thermal diffusivity?
Among common engineering materials, silver has the highest thermal diffusivity at about 174 mm²/s, followed closely by copper (116 mm²/s) and gold (129 mm²/s). Diamond has an even higher value (around 1000 mm²/s) but is rarely used in bulk structural applications. At the other extreme, polymers, rubber, and wood have diffusivities below 0.15 mm²/s.
What does the Fourier number tell you?
The Fourier number (Fo = alpha × t / L²) is a dimensionless measure of how far the transient heat conduction process has progressed. When Fo is large (typically greater than 0.2), the temperature inside the body has had time to redistribute and a simplified lumped capacitance model is valid. When Fo is small, steep internal temperature gradients still exist and you need a full partial differential equation solution (or charts for standard geometries such as slabs, cylinders, and spheres).
How is penetration depth calculated?
Penetration depth delta = sqrt(pi × alpha × t). It represents the distance a thermal disturbance has diffused into the material in time t. For example, with aluminium (alpha approximately 84.4 mm²/s), after 60 seconds of heating the surface, the thermal wave reaches about sqrt(pi × 84.4e-6 m²/s × 60) = 126 mm into the material.
Can I solve for thermal conductivity or specific heat if I know alpha?
Yes. Use the "Solve for" dropdown to select which property you want to find, then enter the known value of alpha along with the other two known properties. The calculator rearranges the formula: k = alpha × rho × cp when solving for conductivity; rho = k / (alpha × cp) for density; and cp = k / (alpha × rho) for specific heat.
Why does water have low thermal diffusivity despite good conductivity?
Water has a thermal conductivity of about 0.60 W/(m·K), which is moderate for a liquid. However, it also has an exceptionally high specific heat capacity (4182 J/(kg·K)) and moderate density (997 kg/m³). The product rho × cp (the denominator in the formula) is very large at about 4.17 × 10⁶ J/(m³·K), so alpha = 0.60 / 4.17 × 10⁶ = 0.143 mm²/s, which is low. This is why water is excellent at absorbing and storing heat without its temperature rising much.