Conduction through a slab, and the thermal resistance that makes two layers easy

Q/t = kAΔT/d is one line. The useful form is the thermal resistance d/kA, because resistances add: this page takes two layers, gives the rate of flow, the U-value, and the temperature drop across each layer — which is where the surprise usually is.

Last updated 11 Nov 2026

Rate of heat flow through one or two layers

Thermal resistances in series, and the drop across each

Thermal conductivity of the first layerNine values, all read from one published table on 11 November 2026. Two things about that table are worth knowing. It gives brickwork as 0.6 to 1.0 and concrete as 0.1 to 1.8 depending on density, which are ranges rather than numbers, so they are not offered here. And it links to copper and to carbon steel without printing a value for either — so copper, the material every textbook uses as the example of a good conductor, is absent from this list because this batch could not source a figure for it. Type your question’s value in the last option. The pair worth comparing is aluminium at 215 against still air at 0.0262: a factor of eight thousand, which is the whole of insulation.

Thermal conductivity of the first layer, typed inLocked unless the last option is chosen above. In watts per metre per kelvin. Per °C is the same number. A value in W/cm/K needs multiplying by 100.

Thickness of the first layerIn millimetres, which is how building and glazing thicknesses are quoted. The page converts to metres internally. Rate of flow is inversely proportional to this, so doubling the thickness halves the flow — but only if this layer is what is actually resisting the heat, which the drop rows below will tell you.

Thermal conductivity of the second layerIn watts per metre per kelvin, typed rather than selected because the second layer is usually the gap: still air at 25 °C is 0.0262, mineral wool 0.04, expanded polystyrene 0.03, softwood 0.12. The default is the air in a double-glazed unit. A real sealed unit is filled with argon rather than air and a real cavity also loses heat by convection and radiation across it, neither of which is conduction and neither of which this page models — so the still-air figure is an idealisation of the best case.

Thickness of the second layer (0 for a single layer)In millimetres. Set it to 0 and the second layer disappears, which is how to do a plain single-slab question: the second resistance becomes zero and the whole drop falls across the first layer. Doing the default double-glazed unit and then setting this to 0 is the single most instructive thing on this page.

Area of the slabIn square metres, measured across the flow. Rate is proportional to it, so this just scales everything — which is why the U-value and the flux rows exist: those are per square metre and let you compare build-ups rather than windows.

Temperature difference across the whole slabA difference, in degrees, the same number in Celsius or kelvin. 15 K is about a mild winter day against a heated room. Negative is allowed and simply reverses the direction of flow: conduction is symmetrical, which is why a vacuum flask keeps coffee hot and lemonade cold with the same construction.

36.59WExample

A double-glazed unit: 4 mm of window glass and a 16 mm still-air gap, 1.5 m², with 15 K across it

Write it as a resistance and two layers stop being hard

Q/t = kA ΔT / d · R = d / kA · Q/t = ΔT / (R₁ + R₂) · ΔT₁ = ΔT · R₁/(R₁ + R₂) · U = 1 / (R A)

k: the thermal conductivity, in W m⁻¹ K⁻¹: a property of the material. The range across ordinary materials is enormous — aluminium 215, window glass 0.96, still air 0.0262 — a factor of eight thousand from one end to the other, and that span is the entire reason insulation works
R = d/kA: the thermal resistance of a layer, in K W⁻¹. Writing it this way is the whole trick, because resistances in series ADD, exactly as electrical ones do: ΔT plays the part of voltage and the rate of heat flow the part of current. A two-layer problem then needs no simultaneous equations at all
ΔT share: the drop across each layer is proportional to that layer’s RESISTANCE, not to its thickness and not to its conductivity alone. This is the row most often surprising: in the default double-glazed unit the 4 mm of glass takes 0.1 per cent of the drop and the 16 mm of air takes 99.9 per cent
U and R-value: the per-square-metre forms the building trade uses. U = 1/(RA) in W m⁻² K⁻¹, and the R-value is its reciprocal. They let two build-ups be compared without reference to the size of the window — but a published building U-value includes the surface air films and this page’s does not, so the two are not interchangeable
steady state: the assumption underneath everything here: temperatures have settled and the slab is neither storing nor releasing heat. Before that, the material’s heat capacity matters as well as its conductivity, and the problem becomes a differential equation rather than a division. It is why a stone floor feels cold long after the room has warmed

Worked example

A double-glazed unit: 4 mm of window glass and a 16 mm still-air gap, 1.5 m², with 15 K across it
THE RESISTANCE OF THE GLASS. R = d/kA = 0.004 / (0.96 × 1.5) = 0.004/1.44 = 0.002 778 K W⁻¹. Thickness in METRES, so 4 mm is 0.004 — the single commonest arithmetic slip on this calculation, and a thousandfold one
THE RESISTANCE OF THE AIR GAP. R = 0.016 / (0.0262 × 1.5) = 0.016/0.0393 = 0.407 125 K W⁻¹. Four times the thickness and a conductivity 37 times smaller, so about 147 times the resistance
ADD THEM, because resistances in series add. R_total = 0.002 778 + 0.407 125 = 0.409 902 K W⁻¹. No simultaneous equations and no interface temperature needed to get here, which is the whole reason for using the resistance form
THE RATE. Q/t = ΔT/R = 15 / 0.409 902 = 36.59 W, which is 24.4 W m⁻² and a conduction-only U-value of 1.63 W m⁻² K⁻¹. Over a day that is 0.878 kWh
AND NOW THE PART WORTH THE PAGE. The drop across each layer is proportional to its resistance: the glass takes 15 × 0.002 778/0.409 902 = 0.102 K and the air takes 14.898 K. The glass carries 0.68 per cent of the resistance. Remove the gap entirely and the same window conducts 15 × 0.96 × 1.5/0.004 = 5400 W — 148 times as much. Glass is not what keeps heat in a window; the air is, and the glass is there to hold the air still

Where the resistance actually is: 1.5 m² with 15 K across it

The second row is the page’s default and the first row is the same window with the gap removed: 5400 W against 36.59 W, a factor of 148, from 16 mm of air. The final column is the one that reorganises how you think about this: in the glazed unit the glass takes 0.1 of a degree out of 15 and the air takes the other 14.9, so the glass is very nearly not there. That is why specifying a thicker pane barely helps a window and widening the cavity does, and why the aluminium row is a reminder rather than a design — 1.2 MW through a window-sized sheet is what “good conductor” actually means in numbers. All five rows are conduction only. A real window’s surface air films add resistance of their own and a real cavity loses heat by convection and radiation across it as well, so the honest reading of the second row is a best case for the gap and an underestimate of the whole assembly’s resistance.
Build-up	R of layer 1	R of layer 2	Total R	Rate of flow	Drop across layer 1
4 mm glass alone	0.002 778 K/W	—	0.002 778 K/W	5400.00 W	15.000 K
4 mm glass + 16 mm still air	0.002 778	0.407 125	0.409 902	36.59 W	0.102 K
4 mm glass + 16 mm mineral wool	0.002 778	0.266 667	0.269 444	55.67 W	0.155 K
4 mm glass + 100 mm polystyrene	0.002 778	2.222 222	2.225 000	6.74 W	0.019 K
4 mm aluminium alone	0.000 012	—	0.000 012	1 209 375 W	15.000 K

Resistances add, and that one fact does most of the work

Q/t = kAΔT/d is the formula every syllabus prints, and the version worth learning is R = d/kA. Write a layer as a resistance and the analogy with electricity becomes exact: the temperature difference plays the part of voltage, the rate of heat flow the part of current, and resistances in series simply ADD. A two-layer wall then needs no simultaneous equations and no guess at the interface temperature — add the resistances, divide the temperature difference by the total, and you have the rate.

The row that reorganises people’s thinking is the temperature drop across each layer. It is proportional to that layer’s RESISTANCE, not to its thickness and not to its conductivity. In a double-glazed unit of 4 mm glass and a 16 mm air gap, the glass carries 0.68 per cent of the resistance and takes 0.1 of a degree out of 15; the air takes the other 14.9. The glass is very nearly not there. That is why ordering a thicker pane barely improves a window while widening the cavity does, and why the glass in a sealed unit is best thought of as a device for holding air still rather than as insulation.

Where this goes wrong, in order. Thickness in millimetres instead of metres, which is a factor of a thousand and the commonest slip by a distance. Then using the area of the wrong face: the area is measured ACROSS the flow, so for a pipe or a rod it is the cross-section and not the surface. Then treating the formula as though it applied before the temperatures had settled, which it does not: everything here is steady state, and before that the material’s heat capacity matters as much as its conductivity, which is why a stone floor stays cold long after the room is warm. And finally, mistaking this page’s U-value for a building U-value. A published U-value includes the still-air films clinging to each surface, which for a window are a large part of the total; this page is conduction only, by design, because that is what the syllabus formula describes.

What the numbers are, and what could not be sourced. The nine conductivities in the dropdown came from one published table, read on 11 November 2026. The table gives brickwork as 0.6 to 1.0 and concrete as 0.1 to 1.8 by density, which are ranges and so are not offered as single values. More awkwardly, it links to copper and to carbon steel without printing a figure for either — so copper, the material every textbook names as the example of a good conductor, is not in this list, because no source for it was established for this batch. Aluminium at 215 is there, and the comparison that matters is aluminium against still air at 0.0262: a factor of eight thousand. Insulation is not a clever material, it is trapped air with something to stop it moving.

Which also explains what this page cannot see. Conduction is one of three ways heat moves, and across a 16 mm cavity the other two are not negligible: air in a gap convects, carrying heat by bulk movement, and the two surfaces exchange heat by radiation regardless of the air. A real sealed unit is filled with argon to suppress the first and coated to suppress the second. The still-air conduction figure this page computes is therefore the best case for the gap, and a real cavity performs worse than it. The rest of the chapter — convection, Newton’s law of cooling, Stefan’s law and Wien’s law — is where those belong, and the arithmetic there is a different shape.

Frequently asked questions

Why use thermal resistance instead of the formula in my textbook?

Because it makes a multi-layer problem trivial. Q/t = kAΔT/d applies to one uniform slab with a known temperature difference across it; the moment there are two layers you do not know the interface temperature, and solving for it is fiddly. Rewrite each layer as R = d/kA and the resistances add, so the rate is just ΔT divided by the total. It is the same physics and it also hands you the temperature drop across each layer for free, which is the number that usually matters. Note that resistances add in SERIES, with heat passing through one layer and then the next; layers side by side are a parallel problem and their conductances add instead.

Why does 4 mm of glass conduct 5400 W when a real window loses far less?

Because glass is not what resists heat in a window, and the bare conduction figure is not a window. The 5400 W assumes a full 15 K difference between the two faces of the glass, and in reality the still air clinging to each surface, the cavity in a sealed unit, and the air movement in the room all stand in the way first, so the glass itself never sees more than a fraction of a degree across it. The calculation is correct and the inputs are unphysical: in a real window the glass is at almost one uniform temperature. That is exactly why a published U-value includes surface films and why this page says plainly that its own U-value does not.

Is the temperature drop bigger across the thicker layer?

Only if that layer also has the greater resistance, and resistance is thickness DIVIDED by conductivity. In the default build-up the air gap is four times thicker than the glass and has a conductivity thirty-seven times smaller, so it has about 147 times the resistance and takes 99.3 per cent of the drop. Make the second layer 4 mm of aluminium instead and the thickness is equal while the drop across it is essentially zero. Thickness alone tells you nothing; the ratio d/k is what to compare, and this page prints both shares so you can see which layer is actually doing the work.

What is the difference between a U-value and an R-value?

They are reciprocals of each other, both expressed per square metre so that build-ups can be compared without reference to size. U is in W m⁻² K⁻¹ and lower is better; R is in m² K W⁻¹ and higher is better. The practical advantage of the R form is that it adds: the R-values of the layers of a wall sum to the wall’s R-value, whereas U-values do not add at all. Both are printed on this page, and both are conduction only — a building U-value includes internal and external surface resistances and will be lower than the figure here.

Does this work for a rod or a pipe as well as a flat slab?

For a rod of uniform cross-section, yes: the area is the cross-section measured across the flow, the length is d, and the formula is unchanged — that is exactly the standard experiment for measuring a metal’s conductivity. For a pipe WALL, no, not strictly: heat spreading outward through a cylindrical shell passes through an area that grows with radius, so the correct resistance involves a logarithm of the radius ratio rather than a simple thickness over area. For a thin wall on a large pipe the difference is small and using the mean area is a good approximation; for lagging several times thicker than the pipe it is not.

Related calculators

References

Thermal Conductivity of Common Materials: Solids, Liquids and Gases, The Engineering ToolBox (2003) (read 11 November 2026). The source of every conductivity in this page’s dropdown: window glass 0.96, ordinary glass 1.05, liquid water 0.606, still air 0.0262 at 25 °C, softwood (fir, pine) 0.12, wool blanket 0.04, expanded polystyrene 0.03 and aluminium 215, all in W m⁻¹ K⁻¹. Two limitations are recorded rather than hidden. The table gives brickwork as 0.6–1.0 and concrete as 0.1–1.8 by density, which are ranges and are therefore not offered as single options here. And it links to copper and to carbon steel while printing no value for either, which is why copper — the textbook example of a good conductor — does not appear in this page’s list at all. The page names no primary source for its own figures and cites only itself; that is stated here because it is the reason the typed field exists and the reason the aluminium figure, 215, is taken from a worked example on the page rather than from its main table.
NCERT / CIET, the Physics-02 module series for Unit 7, “Properties of Bulk Matter” (read 11 November 2026). Cited only to place this page in the CBSE class XI syllabus, not for a number. The modules that could be read identify thermal expansion, calorimetry, change of state, conduction, convection and radiation as chapter 11, “Thermal Properties of Matter”, of Physics Part II. The module whose file is keph_201106_eContent2019.pdf — its own heading reads Keph_201105, so the file name and the heading disagree by one — is Module 17, “Radiation (Black body)”, and lists “coefficient of thermal conductivity” as a Module 16 topic while giving no numerical value for it. That module also gives the Stefan constant as 5.6704 × 10⁻⁸ W m⁻² K⁻⁴, Wien’s constant as 2.88 × 10⁻³ m K, and Newton’s law of cooling as −dQ/dt = k(T₂ − T₁), valid only for small temperature differences.
COULD NOT BE READ, and nothing was substituted for any of it. The NCERT textbook pages on ncert.nic.in, and four CIET modules in the same series as the ones cited here — Keph_201101 (temperature and its measurement), Keph_201102 (thermal expansion), Keph_201103 (specific heat capacity and calorimetry) and Keph_201105 (conduction and convection) — were all refused by the server’s robots policy on 11 November 2026. That is why this batch quotes no NCERT table number, and why the coefficients of expansion and the thermal conductivities are attributed to the reference actually read rather than to the textbook table you are more likely to be holding. If your table disagrees, use your table: these pages take the constant from you for exactly that reason.
The International System of Units (SI), 9th edition, BIPM, published May 2019; the version read for this batch is 4.01, revised June 2026 (read 11 November 2026). The source of the kelvin’s CURRENT definition: the kelvin is defined by fixing the Boltzmann constant at exactly k = 1.380 649 × 10⁻²³ kg m² s⁻² K⁻¹, which the brochure states as an exact relation. The degree Celsius is defined equal in magnitude to the kelvin, with T₀ = 273.15 K and t/°C = T/K − 273.15. Most school textbooks still define the kelvin as 1/273.16 of the triple point of water. The brochure records that as the PREVIOUS definition, and says the triple-point temperature must now be determined experimentally — it stood at 273.16 K with a relative standard uncertainty of 3.7 × 10⁻⁷ at the redefinition, and is a measured quantity from then on.