Thermal Diffusivity Calculator

Created by Rahul Dhari
Reviewed by Steven Wooding
Last updated: Dec 13, 2021

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 fluid density.

Thermal diffusivity is the parameter that explains the penetration of heat in a substance and is used to characterize the unsteady flow of heat and estimate the temperature field to understand the cooling process in a substance. You can use it to determine the Prandtl number used for several heat transfer processes.

This parameter is also used in modern non-destructive techniques (NDT) like thermography or thermographic damage detection in composite structures. This method is crucial to monitor the health of different aircraft structures. The article further explains what thermal diffusivity is and the calculation of thermal diffusivity of substances.

What is thermal diffusivity?

Thermal diffusivity definition — It is the property of a substance that tells us how the heat would flow through it from one point to another. Say you heat one end of a steel plate; the heat would travel towards the cooler end. The entity's thermal diffusivity would tell us how fast it would go, i.e., the rate of heat transfer across two points. It is measured in the units of area per unit time, i.e., mm2/s\text{mm}^2\text{/s} or ft2/s\text{ft}^2\text{/s}. The SI unit for thermal diffusivity is m2/s\text{m}^2\text{/s}.

Most heat-conducting solids have high thermal diffusivity; for instance, the thermal diffusivity of copper at 25 °C25 \ °\text{C} is 111 mm2/s111 \text{ mm}^2\text{/s} whereas the value of thermal diffusivity for a carbon/carbon composite is 216.5 mm2/s216.5 \text{ mm}^2\text{/s}. This means the heat would flow rapidly in the carbon/carbon composite compared to copper.

The thermal diffusivity (α\alpha) depends on three properties of material, i.e., thermal conductivity k, specific heat capacity, Cp\text{C}_p, and density, ρ\rho. The thermal conductivity and thermal diffusivity are related using the thermal diffusivity formula:

α=kρ Cp\quad \alpha = \frac{k}{\rho \ C_p}

Here, the term ρ Cp\rho \ C_p is also known as volumetric heat capacity with units of J/m3K\text{J/m}^3\cdot\text{K}. The volumetric heat capacity is the heat needed to raise the temperature of a material having unit volume by one unit.

You can use the thermal diffusivity and kinematic viscosity (ν\nu) to find the substance's Prandtl number (Pr\text{Pr}). Mathematically, that's:

 Pr=Momentum transportThermal (or heat) transport=να\ \footnotesize \text{Pr} = \frac{\text{Momentum transport}}{\text{Thermal (or heat) transport}} = \frac{\nu}{\alpha}

How to calculate thermal diffusivity

Follow the steps below to use the thermal diffusivity formula:

  1. Fill in the thermal conductivity kk for the material.
  2. Enter the density ρ\rho for the material.
  3. Insert the material's specific heat capacity, CpC_p.
  4. The calculator will return the thermal diffusivity of the material.

💡 You can utilize the preloaded data of common materials from the list to directly obtain the thermal diffusivity.

Example: Using the thermal diffusivity calculator

Find the thermal diffusivity of water considering the density as 997 kg/m3997 \text{ kg/m}^3, specific heat capacity as 4184 J/kg⋅K4184 \text{ J/kg⋅K}, and thermal conductivity, 0.607 W/m⋅K0.607 \text{ W/m⋅K}.

To find the thermal diffusivity:

  1. Fill in the thermal conductivity, k=0.607 W/m⋅Kk = 0.607 \text{ W/m⋅K}.
  2. Enter the density, ρ=997 kg/m3\rho = 997 \text{ kg/m}^3.
  3. Insert the specific heat capacity, Cp=4184 J/kg⋅KC_p = 4184 \text{ J/kg⋅K}.
  4. Using the thermal diffusivity equation:
 α=0.6079974184=0.14558 mm2/s\quad \ \footnotesize \alpha = \frac{0.607}{997 * 4184} = 0.14558 \text{ mm}^2\text{/s}

Thermal diffusivity of substances

The table below has the list of thermal diffusivity of common substances:

Thermal diffusivity of materials.


Thermal diffusivity (mm2/s)







Air (at 300 K)






Wood (pine)


Engine oil (unused, 300 K)


Thermographic damage detection

The process of thermographic damage detection is used to monitor the health of structures. This method is part of several non-destructive testing techniques used to detect defects or hidden subsurface damage in the structure. By non-destructive means, the testing does not interfere with the structure's structural integrity. Examples of other such techniques include ultrasound and X-ray testing.

During thermographic damage detection, flash lamps heat one side of the structure, and an infrared camera is set up on the other side to monitor the temperature progression. This method works by tracking the heat flow for any anomaly in the temperature field to locate defects or damage. The diffusivity is one of the parameters used to estimate the heat flow across the thickness of the sample.


How do I define thermal diffusivity?

Thermal diffusivity is defined as the rate of heat transfer in material from one point to another. In other words, it is the ratio of thermal conductivity and volumetric heat capacity.

What are the units of thermal diffusivity?

Thermal diffusivity is measured in the area per unit time, i.e., meters square per second or feet square per second. Considering, the units of thermal conductivity as W/m⋅K, density being kg/m³, and specific heat as J/kg⋅K, we get: (W/m K) / ((kg/m³) × (J/kg⋅K)) = m²/s.

How do you calculate thermal diffusivity?

To calculate thermal diffusivity of a material:

  1. Multiply the density of material by its specific heat capacity.
  2. Divide the thermal conductivity by the product in the previous step to obtain the thermal diffusivity.

What is thermal diffusivity of air at 0 °C?

Thermal diffusivity of air is 18.46 mm2/s. Considering density of air as 1.2922 kg/m3, thermal conductivity of 0.024 W/m⋅K and specific heat of 1006 J/kg⋅K, the thermal diffusivity is calculated as α = 0.024 / (1.2922 × 1006) = 18.46 mm²/s.

What is thermal diffusivity of water?

Thermal diffusivity of water is 0.14558 mm2/s. Considering density of air as 997 kg/m3, thermal conductivity of 0.607 W/m⋅K and specific heat of 4182 J/kg⋅K, the thermal diffusivity is calculated as α = 0.607 / (4182 × 997) = 0.14558 mm²/s.

What are the factors affecting thermal diffusivity?

Thermal diffusivity is a function of:

  • Density;
  • Specific heat capacity; and
  • Thermal conductivity of the material.

The thermal diffusivity increases with an increase in thermal conductivity and decreases when the density or specific heat of the material increases.

Rahul Dhari
Fluid properties
Thermal conductivity (k)
Density (ρ)
lb/cu ft
Specific heat (Cₚ)
Thermal diffusivity (α)
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