Engineering
Advanced

Conductive Heat Transfer

Heat conducted through a material.

Formula

Q=kAΔTdQ = \frac{kA\Delta T}{d}

Variables

kConductivity
AArea
\Delta TTemp difference
dThickness

Example

Insulation reduces Q

How it works

This equation gives the rate at which heat flows through a solid when one side is hotter than the other. It comes from Fourier’s law: thermal energy moves from high temperature to low temperature, and the amount transferred is larger for materials with better conductivity, larger cross-sectional area, and a bigger temperature difference, but smaller for thicker barriers. In steady state, it assumes the temperature profile through the material is essentially linear and the material properties are uniform.

When to use it

Use this when estimating heat loss or heat gain through walls, windows, pipes, plates, or any solid layer where conduction dominates. It is especially useful in thermal design, insulation analysis, and engineering calculations for steady-state heat transfer.

Worked example, step by step

A steel plate has thermal conductivity k = 45 W/(m·K), area A = 0.80 m², thickness d = 0.020 m, and a temperature difference ΔT = 30 K. Q = kAΔT/d Q = (45)(0.80)(30)/(0.020) Q = 1080/0.020 Q = 54,000 W Final answer: Q = 5.4 × 10^4 W, so the plate conducts 54 kW of heat.

Common mistakes to avoid

  • Using the wrong thickness unit, such as mm instead of m, which changes the result by orders of magnitude
  • Treating Q as total energy instead of heat transfer rate; this formula gives watts, not joules
  • Using the formula when convection or radiation dominates instead of conduction through a solid

Did You Know?

Heat transfer rate rises with area and temperature difference, the logic behind every radiator's fins.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026