Heat Transfer Calculator

Heat moves from warmer objects to cooler ones through three mechanisms: conduction, convection, and radiation. This calculator handles the two most common analytical cases, conduction through a solid wall or slab using Fourier's law, and convective heat exchange between a surface and a fluid using Newton's law of cooling. Knowing the rate of heat transfer in watts is essential for sizing insulation, selecting heating or cooling equipment, assessing energy loss through building elements, and designing heat exchangers. For conduction, Fourier's law says that heat flow equals the thermal conductivity of the material multiplied by the cross-sectional area and the temperature difference across the material, divided by its thickness. High conductivity materials like copper and steel pass heat quickly; low conductivity materials like glass wool and timber resist it. For convection, Newton's law of cooling replaces the conductivity and thickness terms with a single convective heat transfer coefficient, which combines the effects of fluid properties and flow conditions into one practical number. You select the mode you want, enter the relevant dimensions and material properties, and the calculator returns the heat transfer rate in watts and kilowatts, along with supporting values. Use the results as a starting point for sizing and design work rather than as a final engineering specification.

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W/m·K
°C
m
200.0 W
heat transfer rate
In kilowatts0.200 kW
Thermal resistance0.100 K/W
Heat flux100.0 W/m²

Results are theoretical; real-world values depend on surface finish, contact resistance, and fluid flow conditions. For engineering design, consult a qualified professional.

How it works

For conduction, Fourier's law gives Q = k × A × ΔT / d, where k is thermal conductivity in W/m·K, A is the cross-sectional area in m², ΔT is the temperature difference in kelvin or degrees Celsius, and d is the thickness in metres. Thermal resistance R = d / (k × A) in K/W, and heat flux is Q / A in W/m². For convection, Newton's law of cooling gives Q = h × A × ΔT, where h is the convective heat transfer coefficient in W/m²·K. The thermal resistance in this case is 1 / (h × A). Heat flux is Q / A in both modes.

Worked example

A concrete panel has thermal conductivity k = 0.5 W/m·K, area A = 2 m², a temperature difference of 20 °C across it, and a thickness of 0.1 m. Fourier's law gives Q = 0.5 × 2 × 20 / 0.1 = 200 W (0.200 kW). Thermal resistance is 0.1 / (0.5 × 2) = 0.100 K/W and heat flux is 200 / 2 = 100 W/m². These match the default values pre-filled in the calculator above.

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