Thermal Conductivity Calculator

Fourier's law gives the rate heat conducts through a slab of material.

How the Math Works

The Thermal Conductivity Calculator uses the formula Q/t = k·A·ΔT ÷ d to determine the rate of heat conduction through a material. Here, Q/t represents the heat transfer rate (energy per second), k is the material's thermal conductivity (a measure of its ability to conduct heat), A is the cross-sectional area through which heat flows, ΔT is the temperature difference across the material, and d is the thickness of the material. The formula shows that heat transfer increases with higher thermal conductivity, larger area, or greater temperature difference, but decreases with greater thickness. By rearranging the equation, you can solve for any single variable if the others are known.

Practical Applications

This calculation is essential in engineering and material science for designing systems that manage heat transfer, such as heat sinks in electronics, insulation in buildings, or thermal barriers in manufacturing. For example, engineers can determine the required thickness of insulation for a pipe by inputting the material's thermal conductivity, the desired heat loss limit, and the temperature difference between the pipe and surroundings. It also helps in selecting appropriate materials for applications like cooking utensils, where minimizing heat retention is key, or in aerospace design, where maximizing heat dissipation is critical for component safety.

Day-to-Day Use

Understanding heat conduction helps explain everyday phenomena, such as why metal feels colder than wood at the same temperature (metal conducts heat away from your hand faster) or why double-pane windows use air gaps to reduce heat loss. Homeowners can use this concept to choose better insulators for their attics or walls, improving energy efficiency and lowering heating/cooling costs. It also applies to cooking—selecting a pan with high thermal conductivity ensures even heat distribution for better cooking results, while low-conductivity materials can help retain heat in ovens.

Worked example

Insulation (k 0.04), 10 m², ΔT 20 K, 0.1 m → 80 W.

FAQ