{"id":2037,"date":"2026-09-23T13:17:02","date_gmt":"2026-09-23T13:17:02","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=2037"},"modified":"2026-09-23T13:17:05","modified_gmt":"2026-09-23T13:17:05","slug":"thermal-diffusivity","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/thermal-diffusivity\/","title":{"rendered":"Thermal Diffusivity: How Fast Heat Really Moves Through a Material"},"content":{"rendered":"\n
Definition<\/div>

\n\nThermal diffusivity is how quickly a temperature change travels through a material, and it equals the thermal conductivity divided by the density times the specific heat capacity. In symbols it is alpha = k \/ (rho c), measured in square metres per second. Conductivity says how much heat eventually flows; diffusivity says how long the change takes to arrive.\n\n<\/p><\/div>\n\n

Put your hand flat on a stone windowsill on a hot afternoon and it is cool. Come back at midnight, hours after sunset, and the same stone is warm \u2014 the heat of the day has only just finished crossing it.<\/p>\n\n

Nothing about that lag is explained by how well stone conducts. It is set by a second quantity entirely, one that answers a question conductivity cannot touch: not how much heat gets through, but how long it takes.<\/p>\n\n

What Is Thermal Diffusivity?<\/h2>\n\n

Thermal diffusivity is the rate at which a temperature change spreads through a material, defined as its thermal conductivity divided by the heat needed to warm a unit volume of it by one kelvin.<\/p>\n\n

Think of heat arriving at one face of a wall with two jobs to do. Some of it moves on towards the far side; the rest stays behind, warming the material it has just passed through. The second job is a tax on the first.<\/p>\n\n

Conductivity, k, decides how fast heat can move on. Volumetric heat capacity, rho times c, decides how much has to be paid before it can. Diffusivity is simply the ratio \u2014 how much gets through per unit of tax.<\/p>\n\n

That makes a good conductor with a lot of heat to absorb surprisingly slow, and a poor conductor with almost nothing to absorb surprisingly quick. Still air is the standard shock: it is among the worst conductors in any handbook, and yet on the published figures it spreads a temperature change about four and a half times faster than stainless steel.<\/p>\n\n

For the steady state that comes afterwards \u2014 the conductivity table, R-value, U-value and layers in series \u2014 Fourier’s law of thermal conduction<\/a> covers all of it, and this page takes it as read. What follows is the transient that ends in Fourier’s straight line.<\/p>\n\n

The Thermal Diffusivity Formula<\/h2>\n\n

One relation does the whole job, and every symbol in it is something a handbook already lists.<\/p>\n\n

alpha = k \/ (rho c)<\/div>\n\n

Density and specific heat capacity enter only as a product, so they are really one number wearing two hats. That product, rho c, is the volumetric heat capacity: the energy needed to raise a cubic metre of the material by one kelvin. Halve one and double the other and the diffusivity does not move \u2014 which is algebra, not a discovery.<\/p>\n\n

\n\n\n\n
Symbol<\/th>What it is<\/th>SI unit<\/th>Worth knowing<\/th><\/tr>\n<\/thead>\n
alpha<\/td>thermal diffusivity<\/td>m2<\/sup>\/s<\/td>tables almost always print mm2<\/sup>\/s, where ordinary solids run from about 0.08 to 120<\/td><\/tr>\n
k<\/td>thermal conductivity<\/td>W\/(m K)<\/td>the only property here that carries heat forward rather than storing it<\/td><\/tr>\n
rho<\/td>density<\/td>kg\/m3<\/sup><\/td>never appears on its own, only multiplied by c<\/td><\/tr>\n
c<\/td>specific heat capacity<\/td>J\/(kg K)<\/td>per kilogram, not per gram; a factor of 1000 hides here<\/td><\/tr>\n
rho c<\/td>volumetric heat capacity<\/td>J\/(m3<\/sup> K)<\/td>usually quoted in MJ\/(m3<\/sup> K); the whole of what rho and c contribute<\/td><\/tr>\n
L<\/td>layer thickness<\/td>m<\/td>not a material property, and it never changes alpha \u2014 only the clock<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n

The unit falls straight out of the algebra: watts per metre per kelvin divided by joules per cubic metre per kelvin leaves metres squared per second. Those are big units for the job, so almost every table you will meet uses mm2<\/sup>\/s instead, and one mm2<\/sup>\/s is a millionth of an m2<\/sup>\/s.<\/p>\n\n

Any of the four quantities in that relation can be the unknown. If you have a measured diffusivity and want the conductivity behind it, or you know everything except the specific heat capacity, our thermal diffusivity calculator<\/a> rearranges the same relation four ways and prints the settling times beside the answer. If specific heat capacity<\/a> is the term you are least sure of, that guide defines it properly first.<\/p>\n\n

How Fast Does a Temperature Front Cross a Wall?<\/h2>\n\n

A temperature front crosses a wall on a clock set by one dimensionless group, the Fourier number Fo = alpha t \/ L2<\/sup>, and the mid-depth is 99 per cent of the way there when Fo reaches 0.4911.<\/p>\n\n

Picture the cleanest possible experiment. A slab sits at one temperature throughout; at t = 0 one face is stepped to a new temperature and held there, while the far face is pinned at the original one for ever. That is the transient that ends in the straight line Fourier’s law assumes.<\/p>\n\n

Feed that into the heat equation and something useful happens. The elapsed time, the thickness and the diffusivity stop mattering separately and appear only in the combination alpha t \/ L2<\/sup>. Two different materials at the same Fourier number have identical temperature profiles \u2014 they simply took different lengths of time to get there.<\/p>\n\n

Take that literally. A still-air-like gap and a water-like layer of the same thickness, both at Fo = 0.300, sit at exactly the same mid-depth temperature and pass exactly the same fraction of their final flux; the gap took 3.76 min to reach 99 per cent and the layer needs 9.474 h, a factor of about 151.<\/p>\n\n

Notice what is missing, too. The size of the temperature step never appears \u2014 it cancels out of the Fourier number, out of t99<\/sub> and out of every percentage here, so a face stepped by 100 K and one stepped by 5 K trace the same shape. That is algebra rather than a finding.<\/p>\n\n

Three Fourier numbers are worth memorising for this arrangement, and they are pure numbers, the same for brick as for copper:<\/p>\n\n

    \n
  • Fo = 0.0947<\/strong> \u2014 the mid-depth has made half its final rise.<\/li>\n
  • Fo = 0.4911<\/strong> \u2014 the mid-depth has made 99 per cent of it. This is the one that matters, and it drops out of a closed form that is exact here.<\/li>\n
  • Fo = 0.5368<\/strong> \u2014 the far face is finally passing 99 per cent of its steady heat flux. This one is a root solved numerically, and the closed form that nails the second misses it, as problem 8 shows.<\/li>\n<\/ul>\n\n

    Read that third one twice. At the moment the temperature is 99 per cent settled, the flux out of the far side is only 98.43 per cent settled and the slab is holding 99.36 per cent of the energy it will eventually hold. Temperature arrives before flow does, and the gap is real.<\/p>\n\n

    These three constants belong to that boundary condition and to no other.<\/strong> Heat a slab from both faces, insulate its back, or bend it into a cylinder, and the same alpha t \/ L2<\/sup> turns up with different numbers in front of it. Nothing on this page makes them universal.<\/p>\n\n

    Turning the clock into seconds needs only one rearrangement. Setting Fo = 0.4911 gives the time the mid-depth takes to get 99 per cent of the way there, which is the practical answer to “how long until this has warmed through”.<\/p>\n\n

    t99<\/sub> = 0.4911 L2<\/sup> \/ alpha<\/div>\n\n

    Drag the time slider in the lab below and watch the solid curve peel away from the dashed straight line and then settle back onto it. The band between the two is heat that has gone into warming the wall rather than passing through it.<\/p>\n\n

    <\/span>Thermal Diffusivity Lab<\/span><\/div>