Thermal Conductivity Calculator
Calculate conductive heat-transfer rate with Fourier's law (Q = kAΔT/L). Get watts, BTU/hr, heat flux, and R-value from conductivity, area, and thickness.
Thermal Conductivity Calculator
Background.
This calculator applies Fourier's law of heat conduction, Q̇ = kAΔT/L, to compute how fast heat energy flows through a slab of material — a wall, a window pane, an insulation batt, a metal rod — given the material's thermal conductivity, the cross-sectional area it flows through, the temperature difference driving it, and the thickness it must cross. The result, Q̇, is a rate of energy flow measured in watts (joules per second), which is a fundamentally different physical quantity from the energy totals computed by Quanta's specific-heat and latent-heat calculators.
That distinction is worth making explicit because the three calculators are easy to confuse but answer different questions. Specific heat (Q = mcΔT) computes the total energy needed to change a fixed mass of a substance by a given temperature — a one-time energy quantity in joules, with no reference to how long that change takes. Latent heat (Q = mL) computes the total energy needed to drive a phase change, such as melting ice or boiling water, again a one-time energy total. Thermal conductivity, by contrast, computes an ongoing rate: given a temperature difference that is being sustained across a material (a cold winter night outside a warm house, for instance), how many watts of heat continuously leak through per second, for as long as that temperature difference persists? A wall does not have a single, fixed 'heat content' the way a fixed mass of water does at a fixed specific heat — it has a conduction rate that depends on how big a temperature difference is currently pushing heat across it.
The conductive-heat-transfer-rate framing is the backbone of building-science R-value ratings, radiator and heat-sink design, and countless engineering heat-loss calculations. Every material has a characteristic thermal conductivity, k, that describes how readily it conducts heat: metals like copper (k ≈ 401 W/(m·K)) and aluminum (k ≈ 205 W/(m·K)) conduct extremely well, which is why they're used for heat sinks and cookware; common building materials like concrete (k ≈ 1.7 W/(m·K)) and glass (k ≈ 0.96 W/(m·K)) conduct moderately; and purpose-built insulation like fiberglass batts (k ≈ 0.04 W/(m·K)) conducts very poorly — which is exactly the point of insulation. This calculator also reports the R-value (L/k) and total thermal resistance (L/(kA)) directly, because those are the numbers building-science professionals actually specify and compare, and because thermal resistances of stacked material layers add in series (R_total = ΣR_i) the same way electrical resistances do, which is the basis for computing a multi-layer wall's total insulating performance from its individual material layers.
Always enter ΔT as a magnitude — the calculator treats the temperature difference as an unsigned quantity and reports the resulting heat-flow rate as a positive number flowing from the hot side to the cold side; it does not track which direction is 'hot' beyond that convention.
What is thermal conductivity calculator?
Thermal conductivity (k) is a material property that measures how readily a substance conducts heat via molecular vibration and, in metals, free-electron motion, with SI units of watts per meter-kelvin, W/(m·K). It is a property of the material itself, independent of the object's shape — a thick copper bar and a thin copper wire have the identical thermal conductivity k, even though the thick bar conducts far more total heat because of its larger cross-sectional area and, if shorter, its shorter path length.
Fourier's law, first published by Jean-Baptiste Joseph Fourier in his 1822 Théorie analytique de la chaleur, states that the local heat flux (power per unit area) through a material is proportional to the negative of the local temperature gradient: q = −k(dT/dx). For a simple flat slab of uniform material, uniform cross-section, and steady-state (unchanging over time) conditions, that differential relationship integrates directly to the algebraic form this calculator implements: Q̇ = kAΔT/L, where the whole-slab heat-transfer rate Q̇ is proportional to the conductivity, the area, and the temperature difference, and inversely proportional to the thickness. Thicker materials, or materials with lower conductivity, resist heat flow more — exactly the intuition behind building insulation, which pairs a low-conductivity material with meaningful thickness to minimize heat loss.
The R-value commonly printed on insulation packaging is simply L/k — thickness divided by conductivity — expressed per unit area, so a higher R-value means better insulating performance (more resistance to heat flow) for a given thickness of material.
How to use this calculator.
- Enter the material's thermal conductivity (k) — see the hint for common reference values, or look up a specific material's k from a heat-transfer reference.
- Enter the cross-sectional area (A) through which heat is flowing.
- Enter the temperature difference (ΔT) driving the heat flow, as a positive magnitude.
- Enter the thickness (L) — the distance heat must travel through the material.
- Read the heat-transfer rate in watts (and BTU/hr), plus the heat flux and R-value for comparing against insulation specifications.
The formula.
Fourier's law in its local (differential) form states that heat flux is proportional to the negative temperature gradient, q = −k(dT/dx) — heat flows from hot to cold, down the temperature gradient, at a rate set by the material's conductivity. For a flat slab of thickness L with a uniform, steady temperature difference ΔT maintained across it, that gradient is simply ΔT/L (constant across the slab in steady state), so the heat flux becomes q = kΔT/L, in watts per square meter. Multiplying the flux by the total area it acts over gives the whole-slab heat-transfer rate: Q̇ = qA = kAΔT/L, the equation this calculator solves directly.
The calculator also reports two resistance-style intermediates that building-science and heat-transfer engineers use routinely. Thermal resistance, R_th = L/(kA), is the whole object's resistance to heat flow (in K/W) — directly analogous to electrical resistance, since Q̇ = ΔT/R_th mirrors Ohm's law I = V/R. The R-value, R = L/k (in m²·K/W), is the same idea normalized per unit area, which is why it can be compared across products regardless of the exact panel size — and why R-values for stacked material layers (drywall, then insulation, then sheathing) simply add together to give a wall assembly's total R-value, the same way series electrical resistances add.
Finally, the calculator converts the watt result to BTU per hour using the exact relationship 1 BTU_IT = 1055.05585262 joules, so 1 W = 3600/1055.05585262 ≈ 3.4121416331279417 BTU/hr — a conversion many HVAC references in the US still expect.
A worked example.
A 10 m² concrete wall, 0.2 m (20 cm) thick, has a 15°C temperature difference between its warm interior face and cold exterior face — a plausible winter scenario. Concrete's thermal conductivity is about k = 1.7 W/(m·K). Q̇ = 1.7 × 10 × 15 / 0.2 = 255 / 0.2 = 1275 W. That's 1275 joules of heat leaking through the wall every second the temperature difference is sustained — equivalent to about 4350.48 BTU/hr, or roughly the continuous output of twelve 100-watt light bulbs, silently lost through a single concrete wall. The heat flux works out to q = 1.7 × 15 / 0.2 = 127.5 W/m². The wall's total thermal resistance is R_th = 0.2/(1.7×10) ≈ 0.01176 K/W, and its R-value (the number that would appear on an insulation comparison) is R = 0.2/1.7 ≈ 0.1176 m²·K/W — a low figure, which is exactly why bare concrete walls are normally paired with additional insulation layers in any climate with meaningful heating or cooling demand.
Frequently asked questions.
How is thermal conductivity different from specific heat?
How is thermal conductivity different from latent heat?
Can I enter ΔT in Celsius or does it have to be Kelvin?
What is the R-value, and why does a higher R-value mean better insulation?
Why do metals conduct heat so much better than most non-metals?
References& sources.
- [1]ASHRAE. ASHRAE Handbook — Fundamentals. American Society of Heating, Refrigerating and Air-Conditioning Engineers, Atlanta, GA. Fourier's law, R-value = L/(kA), series R-value addition for multi-layer building assemblies.
- [2]University of Illinois Urbana-Champaign, Department of Mechanical Science and Engineering. ME 320 course materials — "Fourier's Law of Heat Conduction."
- [3]National Institute of Standards and Technology (2008). Special Publication 811: Guide for the Use of the International System of Units (SI) — basis for the exact BTU_IT-to-watt conversion.
- [4]Fourier, J.B.J. (1822). Théorie analytique de la chaleur. Firmin Didot, Paris. Original formulation of the law of heat conduction.
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