Thermal Conductivity Converter

Thermal Conductivity (symbolized by k or λ) measures an intrinsic material’s physical capacity to conduct heat energy through molecular vibration and electron transport (governed by Fourier’s Heat Conduction Law: q = -k · A · (dT ÷ dx) or k = (Q · L) ÷ (A · ΔT)). Across building envelope insulation k-factor ratings, electronics copper heat spreader sizing, aerospace thermal protection shields, metallurgy heat treating, and geothermal well design, thermal conductivity is categorized across three major engineering unit systems: International System of Units (SI metric fundamental: Watt per Meter Kelvin / W/(m·K), Watt per Meter Degree Celsius / W/(m·°C), Watt per Centimeter Degree Celsius / W/(cm·°C), Kilowatt per Meter Kelvin / kW/(m·K)), Imperial / US Customary standards (Btu Foot per Hour Sq. Foot °F / Btu·ft/(h·ft²·°F), Btu Inch per Hour Sq. Foot °F / Btu·in/(h·ft²·°F), Btu Inch per Second Sq. Foot °F / Btu·in/(s·ft²·°F)), and metric laboratory standards (Calorie per Second Centimeter °C / cal/(s·cm·°C), Kilocalorie per Hour Meter °C / kcal/(h·m·°C)).

Our free online Thermal Conductivity Converter provides instant, high-precision conversions across all SI metric, Imperial insulation, metallurgical, and physical science units:

• Btu Foot per Hour Sq. Foot °F to W/(m·K): Multiply Btu·ft/(h·ft²·°F) by 1.73073467 (1 Btu·ft/(h·ft²·°F) = 1.73073 W/(m·K) = 1.48816 kcal/(h·m·°C)).
• Btu Inch per Hour Sq. Foot °F to W/(m·K): Multiply Btu·in/(h·ft²·°F) by 0.14422789 (1 W/(m·K) = 6.93347 Btu·in/(h·ft²·°F)).
• Watt per Meter Kelvin to Btu Foot per Hour Sq. Foot °F: Multiply W/(m·K) by 0.5777888 (1 W/(m·K) = 0.57779 Btu·ft/(h·ft²·°F) = 0.85985 kcal/(h·m·°C)).
• Watt per Centimeter °C [W/(cm·°C)] to W/(m·K): Multiply W/(cm·°C) by 100.0 (1 W/(cm·°C) = 100.0 W/(m·K) = 57.779 Btu·ft/(h·ft²·°F)).
• Calorie per Second Centimeter °C to W/(m·K): Multiply by 418.68 (1 cal/(s·cm·°C) = 418.68 W/(m·K) = 241.91 Btu·ft/(h·ft²·°F)).
• Kilocalorie per Hour Meter °C [kcal/(h·m·°C)] to W/(m·K): Multiply by 1.163 (1 kcal/(h·m·°C) = 1.163 W/(m·K) = 0.67197 Btu·ft/(h·ft²·°F)).

Master Material Thermal Conductivity Conversion Table

The table below displays exact mathematical conversion relationships, SI W/(m·K) multipliers, and imperial Btu equivalents relative to 1 Watt per Meter Kelvin (1 W/(m·K) = 1 W/(m·°C)):

Step-by-Step Copper Heat Spreader Conductivity Calculation Example

To convert pure copper (C11000 ETP) with a thermal conductivity of 390.0 Watts per Meter Kelvin (390 W/(m·K)) into Imperial Btu·ft/(h·ft2·°F) and calculate heat flow (Q) through a 0.01m thick, 0.05 m² plate across a Δ10°C temperature difference:

Step 1 (Imperial Conversion): k = 390.0 × 0.5777888 = 225.3376 Btu·ft/(h·ft2·°F) (225.34 Btu·ft/(h·ft2·°F))

Step 2 (Heat Flow Rate Q): Q = k · A · (ΔT ÷ L) = 390.0 × 0.05 m2 × (10°C ÷ 0.01 m) = 19,500.0 Watts (19.5 kW)

Step 3 (Building Insulation k-Factor Equivalent): k = 390.0 × 6.933472 = 2,704.05 Btu·in/(h·ft2·°F)

Thus, pure copper (390 W/(m·K)) equals 225.34 Btu·ft/(h·ft²·°F), dissipating 19,500 Watts across a 10°C gradient.

Real-World Engineering Material Thermal Conductivity Benchmarks

Below is a comparative reference chart showing thermal conductivity values (k) across air, fiberglass, steel, copper, silver, and synthetic diamond:

History & Physics: 1822 Fourier Law vs 1853 Wiedemann-Franz Law

1822 Joseph Fourier & the Law of Heat Conduction

In 1822, French mathematician Joseph Fourier formulated Fourier’s Law of Heat Conduction in Théorie Analytique de la Chaleur: q = -k · ∇T. Fourier established thermal conductivity (k) as the fundamental material constant governing conductive heat transport across temperature gradients, establishing the unit W/(m·K).

1853 Wiedemann-Franz Law (k ÷ σ = L · T)

In 1853, German physicists Gustav Wiedemann and Rudolf Franz discovered that for metals, the ratio of thermal conductivity (k) to electrical conductivity (σ) is directly proportional to temperature (k ÷ σ = L · T, where L is the Lorenz number). This proved that free electrons conduct both electrical current and heat energy in metals, explaining why silver (429 W/(m·K)) and copper (390 W/(m·K)) excel as electrical and thermal conductors.

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Frequently Asked Questions (FAQ)

How do you convert W/(m·K) to Btu·ft/(h·ft²·°F)?

To convert Thermal Conductivity from W/(m·K) to Btu·ft/(h·ft²·°F), multiply by 0.577789 (or divide by 1.730735). For example, copper’s 390 W/(m·K) × 0.577789 = 225.34 Btu·ft/(h·ft²·°F).

How do you convert W/(m·K) to insulation k-factor in Btu·in/(h·ft²·°F)?

To convert W/(m·K) to insulation k-factor in Btu·in/(h·ft²·°F), multiply W/(m·K) by 6.933472. For example, fiberglass insulation with k = 0.040 W/(m·K) × 6.933472 = 0.277 Btu·in/(h·ft²·°F).

What material has the highest thermal conductivity in the world?

Single-crystal synthetic diamond has the highest known thermal conductivity of any bulk material, reaching up to 2,200 W/(m·K)—over five times higher than pure copper (390 W/(m·K))—due to strong covalent carbon bonds that transfer heat via high-velocity acoustic phonons.

What is the difference between Thermal Conductivity and Thermal Resistance?

Thermal Conductivity (k in W/(m·K)) is an intensive material property (independent of size or shape). Thermal Resistance (R_th in K/W) is an extensive geometry-dependent property calculated as Rth = L ÷ (k · A).