Thermal diffusivity is how fast a temperature change travels through a material, as opposed to how much heat finally flows through it. This free thermal diffusivity calculator solves alpha = k / (rho × c) four ways — for the diffusivity itself, or for the conductivity, the density or the specific heat capacity behind it. Beside the answer it prints the thermal effusivity, the volumetric heat capacity, the time the layer needs to reach 99 per cent of its steady state, the time the mid-depth is half way there, the Fourier number after one hour and the steady flux the layer passes per kelvin.
Each button puts the Solve for menu on the unknown that case is asking about, sets every unit menu the case names, and fills the remaining boxes. Whatever the widget then works out is read back into the line underneath, so nothing there is stored text. The six triples are idealised values that land near a published diffusivity rather than measurements of any real brick, bar or board, which is why every name carries a "-like"; the three material figures are typed to six significant figures, and the diffusivity to seven, so that a case run backwards returns exactly the figures it started from.
Pick a case above, or type your own numbers.

The thermal diffusivity calculator is a free online tool for how fast a temperature change travels through a material, rather than how much heat finally flows through it. Enter the thermal conductivity, the density and the specific heat capacity and it returns alpha = k / (rho × c) in square millimetres per second; move the Solve for menu and the same relation runs backwards for whichever of those three you are missing. Beside the answer it prints the thermal effusivity e = sqrt(k × rho × c), the volumetric heat capacity, the Fourier number after one hour, the steady flux the layer passes per kelvin, and the times the layer takes to reach half and 99 per cent of its steady state for a slab with one face stepped and the far face held.
| Symbol | Quantity | Default unit | Also accepts | Example value |
|---|---|---|---|---|
| k | Thermal conductivity | W/(m K) | mW/(m K) | 1 |
| rho | Density | kg/m3 | g/cm3 | 1995.26 |
| c | Specific heat capacity | J/(kg K) | kJ/(kg K) | 900 |
| alpha | Thermal diffusivity | mm2/s | cm2/s, m2/s | 0.5568754 |
| L | Layer thickness | mm | m | 100 |
This page starts where a steady-state tool stops. Once nothing is changing any more, the heat flow through a slab is Fourier's law and the thermal conduction calculator is the right tool, with its area, its temperature difference and its thickness. What it cannot tell you is how long the wall took to get there, because it has nowhere to put a density or a specific heat.
For the underlying law, the conductivity table and the way resistances add in series, the guide to Fourier's law of thermal conduction covers all of it, and this page assumes it. The one quantity it names and then hands on is this one. If the specific heat capacity in the denominator is the part you are unsure of, the guide to specific heat capacity defines it properly and the specific heat calculator works it out from an energy and a temperature rise.
Two mistakes account for most wrong answers here, and both are about what was typed. The first is a specific heat capacity entered per gram rather than per kilogram, which makes the material a thousand times slower than it is; the kJ/(kg K) option is there because that is how many tables print it. The second is a unit menu left on the wrong option after a preset, which the second line of Show working will always catch.
1.79573 MJ/(m3 K) chip and the 0.556875 mm2/s headline are the same arithmetic printed twice — the answer is the conductivity divided by that chip — so their agreeing confirms nothing at all.The table starts at the defaults and moves one thing at a time: which quantity is the unknown, then the material, then the thickness, then the unit the figures are typed in. Every Result cell was read out of the running widget rather than worked out by hand, so where a cell and the tool ever part company, believe the tool. The input column is simply what you type, in the order the boxes appear.
| Step | Solving for | Case | What you type | Result | Time to 99 per cent |
|---|---|---|---|---|---|
| The page as it opens | Thermal diffusivity | Brick-like wall | 1 W/(m K), 1995.26 kg/m3, 900 J/(kg K), 100 mm | 0.556875 mm2/s | 2.450 h |
| Feed that answer back | Thermal conductivity | Brick-like wall | 0.5568754 mm2/s, 1995.26 kg/m3, 900 J/(kg K), 100 mm | 1 W/(m K) | 2.450 h |
| and again, for the density | Density | Brick-like wall | 1 W/(m K), 900 J/(kg K), 0.5568754 mm2/s, 100 mm | 1995.26 kg/m3 | 2.450 h |
| and the specific heat | Specific heat capacity | Brick-like wall | 1 W/(m K), 1995.26 kg/m3, 0.5568754 mm2/s, 100 mm | 900 J/(kg K) | 2.450 h |
| Same product, other split | Thermal diffusivity | Brick-like wall | 1 W/(m K), 997.63 kg/m3, 1800 J/(kg K), 100 mm | 0.556875 mm2/s | 2.450 h |
| A gap of still air | Thermal diffusivity | Still-air-like gap | 0.0263027 W/(m K), 1.20226 kg/m3, 1005 J/(kg K), 100 mm | 21.7689 mm2/s | 3.76 min |
| Stainless steel instead | Thermal diffusivity | Stainless-like plate | 15.1356 W/(m K), 7943.28 kg/m3, 500 J/(kg K), 100 mm | 3.81092 mm2/s | 21.48 min |
| A bar of copper | Thermal diffusivity | Copper-like bar | 398.107 W/(m K), 9120.11 kg/m3, 385 J/(kg K), 100 mm | 113.381 mm2/s | 43.3 s |
| A layer of water | Thermal diffusivity | Water-like layer | 0.60256 W/(m K), 1000 kg/m3, 4185 J/(kg K), 100 mm | 0.143981 mm2/s | 9.474 h |
| A pine board | Thermal diffusivity | Pine-like board | 0.1 W/(m K), 630.957 kg/m3, 1900 J/(kg K), 100 mm | 0.0834155 mm2/s | 16.353 h |
| The same brick, 25 mm thin | Thermal diffusivity | Brick-like wall | 1 W/(m K), 1995.26 kg/m3, 900 J/(kg K), 25 mm | 0.556875 mm2/s | 9.19 min |
| The same brick, 300 mm thick | Thermal diffusivity | Brick-like wall | 1 W/(m K), 1995.26 kg/m3, 900 J/(kg K), 300 mm | 0.556875 mm2/s | 22.046 h |
| Density typed in g/cm3 | Thermal diffusivity | Brick-like wall | 1 W/(m K), 1.99526 g/cm3, 900 J/(kg K), 100 mm | 0.556875 mm2/s | 2.450 h |
| A density of zero | Thermal diffusivity | nothing at all | 1 W/(m K), 0 kg/m3, 900 J/(kg K), 100 mm | no answer | — |
Rows 1 to 4 are one state read four different ways. Each answer returns the figure the row above it started from, so the diffusivity gives back the conductivity, then the density, then the specific heat capacity. Four unknowns, one relation, and no arrangement more fundamental than another.
That round trip is the genuine check on this page, and it is worth saying why the volumetric heat capacity chip is not. In row 1 the answer is the conductivity divided by that chip, so the two agreeing is the definition restated. Rows 2 to 4 do different arithmetic in the other direction and still land on 1 W/(m K), 1995.26 kg/m3 and 900 J/(kg K), which is a claim that can fail.
Row 5 halves the density and doubles the specific heat capacity, and the diffusivity does not move. That is algebra rather than a discovery: the two appear only as their product, and the tool is restating its own definition rather than testing it. It is still worth seeing, because it is the reason a light foam and a dense solid can share a diffusivity.
Rows 6 and 7 are the demonstration this page exists for. The still-air-like gap conducts 575 times worse than the stainless-like plate and diffuses a temperature change 5.71 times faster, so the two quantities put the same pair of materials in opposite orders. A steady-state conduction tool has no density and no specific heat capacity, so it can never show that.
Rows 8 to 10 run from the quickest of the six presets to the slowest, with the water-like layer sitting between them: a copper-like bar at 113.381 mm2/s against a pine-like board at 0.0834155, a factor of about 1359 across that pair. The times beside them say the same thing more usefully: 43.3 s against 16.353 h for the same 100 mm.
Rows 11 and 12 change only the thickness, and the headline does not move, because thickness is not a material property. The time does: 25 mm takes 9.19 min and 300 mm takes 22.046 h, which are the same 2.450 h multiplied by a sixteenth and by nine. Rows 13 and 14 are the edges — the same brick typed as 1.99526 g/cm3 returns the identical answer, and a density of zero is refused rather than answered with infinity.
The calculator uses one relation in four arrangements. The diffusivity is alpha = k / (rho × c), so the conductivity is k = alpha × rho × c, the density is rho = k / (alpha × c) and the specific heat capacity is c = k / (alpha × rho). The effusivity chip beside them is e = sqrt(k × rho × c), the same three properties arranged the other way round.
Those two carry everything the three do, for conduction: k = e × sqrt(alpha) and rho × c = e / sqrt(alpha). It is worth knowing because the two say opposite things about the same material — of the six presets only the copper-like bar diffuses faster than the still-air-like gap, and not one of them has an effusivity anywhere near as low as that gap's.
| Symbol | Meaning | SI unit | Values used on this page |
|---|---|---|---|
| alpha | Thermal diffusivity: how fast a temperature change travels, and the quantity this page opens on. Written with a Greek alpha in textbooks and spelled out here, because the labels beside it are uppercased by the stylesheet | square metre per second, m2/s | Boxes take mm2/s, cm2/s or m2/s: 0.556875 mm2/s on the opening brick-like wall, 113.381 for the copper-like bar, 0.0834155 for the pine-like board. |
| k | Thermal conductivity: how much heat crosses a metre of the material per kelvin of temperature difference, once nothing is changing any more. The numerator of the diffusivity | watt per metre per kelvin, W/(m K) | Boxes take W/(m K) or mW/(m K): 1 for the brick-like wall, 0.0263027 for the still-air-like gap, 398.107 for the copper-like bar. |
| rho | Density. It never acts alone here: it enters only multiplied by the specific heat capacity, and a Greek rho is what most textbooks print | kilogram per cubic metre, kg/m3 | Boxes take kg/m3 or g/cm3: 1995.26 for the brick-like wall, 1.20226 for the still-air-like gap, 1000 for the water-like layer. |
| c | Specific heat capacity: the energy it takes to raise a kilogram of the material by one kelvin. Together with the density it makes the volumetric heat capacity in the denominator | joule per kilogram per kelvin, J/(kg K) | Boxes take J/(kg K) or kJ/(kg K): 900 for the brick-like wall, 4185 for the water-like layer, 1900 for the pine-like board. |
| L | Layer thickness. The one box that is never the unknown, because it changes no material property - it sets the clock the diffusivity is measured against | metre, m | Boxes take mm or m: 100 mm throughout the presets, 25 mm and 300 mm in the two rows that show the thickness-squared law. |
| e | Thermal effusivity, the square root of k times rho times c. It decides the contact temperature at the first instant two surfaces touch, and it is printed as a chip rather than typed | J/(m2 K sqrt(s)) | Computed, never typed: 1340.05 on the opening state, 37387.8 for the copper-like bar, 346.24 for the pine-like board. |
Heat arriving at one face of a slab does two things at once: some of it moves on through the material, and some of it stays behind to warm the material it has passed. Conductivity governs the first and the volumetric heat capacity governs the second, and the speed of the front is the ratio of the two. That ratio is the diffusivity, and it is why a good conductor with a great deal of heat to absorb can be slow.
The model behind the two times on the chips is the simplest one that ends in Fourier's law. A slab starts at a single temperature; one face is stepped to a new one and the far face is held at the old one for ever. The temperature profile then climbs from flat to the straight line that the steady-state law assumes from the outset, and the whole of the transient is carried by one dimensionless group.
That group is the Fourier number, Fo = alpha × t / L2. Two different materials at the same Fourier number have identical temperature profiles, differing only in how long they took to get there, which is why the chips print one after an hour: it says where a given layer is in its own story. The thickness enters squared, so it is the most powerful thing on the page.
Mid-depth reaches 99 per cent of its final rise at Fo = 0.4911, and that number is exact algebra rather than a fit. The mid-depth series contains only odd terms, so the one-term form is wrong by the third term, a matter of 4e-20. The companion figure for the far face is a solved root rather than a closed form, and it is larger: the far face is still delivering only 98.43 per cent of its final flux when the mid-depth is 99 per cent of the way there, so temperature settles before flux does.
Both constants belong to that boundary condition and to no other. Heat the slab from both faces, insulate the back of it, or make it a cylinder, and the transient has different numbers in front of the same L2/alpha. What survives every version is the L2 itself, which is why the 25 mm and 300 mm rows of the table differ by exactly a factor of 144.
The effusivity chip answers a different question again: not how fast the front moves, but what happens in the instant two surfaces touch. Two bodies brought into contact settle their interface immediately at a temperature weighted by their effusivities, so equal effusivities meet exactly in the middle and a much more effusive body keeps the interface close to its own temperature. The copper-like bar's chip reads 37387.8 against the pine-like board's 346.24, a factor of 108, which is the honest version of the familiar observation that metal feels colder than wood at the same temperature.
None of this brings in a property the material did not already have. The same three figures make the diffusivity and the effusivity, the temperature difference cancels out of every percentage and every time on this page, and the Fourier's law simulator shows what the same slab is doing once all of it has finished happening.
5.63745 against 7753.27, is lower by a factor of 1375.The arithmetic is short and hard to get wrong. What fails is the relation being pressed onto a situation it does not describe, or a constant being carried somewhere it does not belong.
L2/alpha carries a different number in front of it. The diffusivity itself is unaffected; only the schedule changes.
398.107 it came from; rounded to the six the headline carries, 113.381, it returns 398.108 instead.For the method in full, with worked problems and the diagrams that go with them, read Thermal Diffusivity: How Fast Heat Moves. For the steady state this transient ends in — the conductivity table, the R-value and how layers add up — Fourier's Law of Thermal Conduction is the place to go, with the thermal conduction calculator beside it. The specific heat calculator handles the denominator on its own, and the whole physics lab library is open if you would rather watch a temperature front than type one.
It works out how fast a temperature change travels through a material rather than how much heat flows once things have settled. Enter the conductivity, the density and the specific heat capacity and it returns alpha = k/(rho c) in square millimetres per second, along with the effusivity, the volumetric heat capacity and the time the layer takes to reach 99 per cent of its steady state. Move the Solve for menu and the same relation runs backwards, recovering whichever one of the three material figures you are missing.
Conductivity answers how much heat flows once the temperatures have stopped changing; diffusivity answers how quickly the change gets there in the first place. They can put two materials in opposite orders, which is not a contradiction: conductivity is the numerator of the diffusivity, and the volumetric heat capacity underneath it can be larger still. Still air is the standard demonstration, since it conducts hundreds of times worse than stainless steel and diffuses a temperature change several times faster.
The SI unit is the square metre per second, but published tables almost always use square millimetres per second, because ordinary solids land between roughly 0.08 and 120 in those units. One square millimetre per second is a millionth of a square metre per second. The menu on this page carries all three of mm2/s, cm2/s and m2/s, and the working lines print the SI figure beside the one you asked for.
Because the relation contains them only that way. Their product is the volumetric heat capacity, the energy it takes to raise a cubic metre of the material by one kelvin, and that is the whole of what they contribute. Halve the density and double the specific heat and the diffusivity does not move at all; this is algebra rather than a property of any particular material, and the tool is restating the definition rather than discovering anything.
For a layer that starts at one temperature, with one face stepped to a new one and the far face held, the mid-depth is 99 per cent of the way to its final value at t99 = 0.4911 L squared divided by alpha. That constant belongs to that boundary condition and to no other, so a wall warmed on both faces, or one backed by insulation, is a different problem with a different constant. The thickness squared is the part that always holds: three times the thickness is nine times the wait.
Effusivity is the square root of k times rho times c, and it decides how a surface behaves in the first instant of contact rather than how a slab behaves over time. Two bodies pressed together settle their interface immediately at a temperature weighted by their effusivities, so the more effusive one wins the interface. Diffusivity has the same three properties arranged the other way round, which is why the two quantities can disagree completely about which of two materials is the extreme one.
Because thickness is not a material property and does not appear in alpha = k/(rho c) at all. What it changes is the clock the diffusivity is measured against: the Fourier number is alpha times the time divided by the thickness squared, so the times on the chips move as the square while the headline stays put. That is exactly why thickness is an input here and never an answer.
No, and each name carries a "-like" for that reason. Each preset is an idealised triple chosen to land near a published diffusivity, and the page states how far it lands from it, which for the air-like gap is 14.6 per cent high. Use them to see the shape of the comparison, then type your own figures from a source you trust before you rely on a number.
It does change, unless you also change the number. The menus convert whatever you type into SI base units before any arithmetic happens, so 2.0 g/cm3 and 2000 kg/m3 are the same density and give the same diffusivity, while 2000 g/cm3 is a thousand times denser than most solids and shows it. The second line of Show working always restates your figures in SI, which is the quickest way to catch a menu left on the wrong option.
Not directly: a layered wall has no single diffusivity, because each layer has its own and the transient depends on the order they are in as well as on their thicknesses. Running one layer at a time gives you the individual timescales, and the slowest of them usually dominates the answer. For the steady heat flow through the finished wall, the resistances in series are the right tool rather than this one.