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Thermal Conductivity Calculator

Heat transfer rate equals conductivity times temperature differential divided by distance

Solution

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How It Works

Fourier's law of heat conduction states that heat flows from hot to cold at a rate proportional to the temperature difference and the material's conductivity: q = K × ΔT / x. Knowing any three of these variables lets you solve for the fourth. High-conductivity materials like copper (K ≈ 385 W/(m·K)) transfer heat quickly, while insulators like fiberglass (K ≈ 0.04) resist heat flow. This equation is the foundation of building insulation design, electronics cooling, and industrial heat exchanger sizing.

Example Problem

A 0.15 m thick brick wall (K = 0.72 W/(m·K)) separates a 22 °C interior from −5 °C outside. What is the heat loss per square meter?

  1. Identify the known values: thermal conductivity K = 0.72 W/(m·K), inside temperature = 22 °C, outside temperature = −5 °C, wall thickness x = 0.15 m.
  2. Determine what we are solving for: the heat transfer rate q (heat flux) through the wall in watts per square meter.
  3. Calculate the temperature differential: ΔT = 22 − (−5) = 27 °C. Since Fourier's law uses the magnitude of the temperature difference, both °C and K give the same ΔT.
  4. Write Fourier's law of heat conduction: q = K × ΔT / x.
  5. Substitute the known values: q = 0.72 W/(m·K) × 27 °C / 0.15 m = 19.44 / 0.15.
  6. Compute the result: q = 129.6 W/m². This means every square meter of wall loses about 130 watts — roughly equivalent to leaving a bright incandescent bulb on per square meter of wall area.

Adding 0.10 m of fiberglass insulation (K = 0.04) would dramatically reduce this to about 10 W/m², showing why insulation matters.

When to Use Each Variable

  • Solve for Heat Transfer Rate (q) — when you know the material's conductivity, temperature difference, and wall thickness, e.g., estimating heat loss through a building wall.
  • Solve for Thermal Conductivity (K) — when you have measured heat flux, thickness, and temperature difference and need to determine the material's conductivity.
  • Solve for Temperature Differential (ΔT) — when you know the heat transfer rate and material properties and need to find the resulting temperature difference across a wall or layer.
  • Solve for Distance / Thickness (x) — when you need to determine the insulation thickness required to limit heat loss to a target rate.

Key Concepts

Fourier's law of conduction (q = K × ΔT / x) is the basis of steady-state heat transfer through solids. Thermal conductivity K is a material property measured in W/(m·K). The equation assumes one-dimensional heat flow through a flat slab — for cylinders or spheres, logarithmic or spherical geometry corrections apply. For composite walls, resistances add in series: R_total = Σ(x_i / K_i).

Applications

  • Building energy analysis: calculating heat loss through walls, roofs, and windows to size HVAC systems and meet energy codes
  • Electronics cooling: determining whether a heat sink can dissipate enough power to keep a chip below its maximum junction temperature
  • Industrial insulation: specifying pipe insulation thickness to prevent heat loss in steam lines or condensation on chilled water pipes
  • Geothermal engineering: estimating heat flow through subsurface rock layers for ground-source heat pump design

Common Mistakes

  • Using the wrong units — mixing imperial (BTU·in/(h·ft²·°F)) and SI (W/(m·K)) without converting leads to orders-of-magnitude errors
  • Ignoring convective resistance at surfaces — real walls have air films on each side that add thermal resistance beyond conduction alone
  • Applying the flat-wall formula to cylindrical pipes — pipe insulation requires the logarithmic mean radius formula for accurate results
  • Assuming conductivity is constant across all temperatures — K varies with temperature, especially for metals at cryogenic or high temperatures

Frequently Asked Questions

What is thermal conductivity and what does it measure?

Thermal conductivity (K) measures how well a material conducts heat. It is defined as the amount of heat (in watts) that passes through a 1-meter thickness of material per square meter of area for each degree of temperature difference. Higher values mean the material transfers heat more readily.

How do you calculate heat transfer through a wall?

Use Fourier's law: q = K × ΔT / x. Multiply the material's thermal conductivity (K in W/(m·K)) by the temperature difference (ΔT in °C or K), then divide by the wall thickness (x in meters). The result is the heat flux in watts per square meter (W/m²).

What is the thermal conductivity of common building materials?

Concrete has K ≈ 1.0 W/(m·K), brick about 0.72, wood around 0.12–0.17, fiberglass insulation 0.04, and still air 0.026. Engineers use these values to design walls, roofs, and insulation layers that meet energy codes.

How does thickness affect heat transfer through a wall?

Heat transfer rate is inversely proportional to thickness. Doubling the wall thickness cuts the heat flow in half, which is why adding insulation is one of the most effective ways to reduce energy loss in buildings.

What is the difference between thermal conductivity and thermal diffusivity?

Conductivity (K) measures how much heat a material transfers per unit time, while diffusivity (α = K/(ρ·c_p)) measures how fast temperature changes propagate through it. A material can conduct a lot of heat but change temperature slowly if it has high density and heat capacity.

What is R-value and how does it relate to thermal conductivity?

R-value is the thermal resistance of a material layer: R = x / K, where x is thickness and K is thermal conductivity. Higher R-values mean better insulation. In the US, R-value is measured in ft²·°F·h/BTU; in SI units it is m²·K/W. Building codes specify minimum R-values for walls, roofs, and floors.

Can you use Fourier's law for cylindrical pipes?

Fourier's law in the form q = K × ΔT / x applies to flat slabs. For cylindrical pipes, the heat transfer equation uses the logarithmic mean radius: q = 2πKL × ΔT / ln(r₂/r₁), where r₁ and r₂ are the inner and outer radii. This calculator uses the flat-slab formula, which is a good approximation when the insulation thickness is small relative to the pipe diameter.

Reference: Incropera, Frank P. & DeWitt, David P. 2002. Fundamentals of Heat and Mass Transfer. John Wiley & Sons. 5th ed. Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th ed.

Thermal Conductivity Formula

Fourier's law of heat conduction describes steady-state heat transfer through a solid slab:

q = K × ΔT / x

Where:

  • q — heat transfer rate per unit area, measured in watts per square meter (W/m²)
  • K — thermal conductivity, measured in watts per meter-kelvin (W/(m·K))
  • ΔT — temperature difference across the material, measured in degrees Celsius (°C) or kelvin (K)
  • x — thickness or distance of the material, measured in meters (m)

The formula assumes one-dimensional steady-state heat conduction through a flat slab of uniform material. For composite walls, thermal resistances add in series: Rtotal = Σ(xi / Ki). For cylindrical or spherical geometries, logarithmic and radial corrections are needed.

Worked Examples

Building Insulation

How much heat is lost through a brick wall in winter?

A 0.25 m thick brick wall (K = 0.72 W/(m·K)) separates a 21 °C heated interior from −3 °C outside air. Find the heat loss per square meter.

  • Temperature difference: ΔT = 21 − (−3) = 24 °C
  • q = K × ΔT / x = 0.72 × 24 / 0.25
  • q = 69.12 W/m²

Adding 100 mm of fiberglass insulation (K = 0.04) would reduce this by roughly 90%, which is why building codes require minimum R-values for exterior walls.

Electronics Cooling

What thermal conductivity does a heat sink need to keep a chip cool?

A processor dissipates 95 W through a heat sink base that is 0.005 m thick with 0.003 m² contact area. The temperature rise across the base must stay below 5 °C. What minimum thermal conductivity is needed?

  • Heat flux: q = 95 / 0.003 = 31,667 W/m²
  • K = q × x / ΔT = 31667 × 0.005 / 5
  • K = 31.667 W/(m·K)

Aluminum (K ≈ 205) and copper (K ≈ 385) both exceed this requirement. The actual design choice depends on weight, cost, and manufacturing constraints.

Industrial Process

How thick should pipe insulation be to limit heat loss?

A steam pipe at 180 °C runs through a 25 °C facility. The insulation (mineral wool, K = 0.04 W/(m·K)) must limit heat loss to 50 W/m². What minimum insulation thickness is required?

  • Temperature difference: ΔT = 180 − 25 = 155 °C
  • x = K × ΔT / q = 0.04 × 155 / 50
  • x = 0.124 m (124 mm)

In practice, pipe insulation uses a cylindrical geometry correction (log-mean radius), which would slightly increase the required thickness compared to this flat-slab estimate.

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