Heat conduction: how thickness and area change the rate
Estimate steady heat flow through a flat layer while distinguishing watts, energy and material conductivity.
Method
For steady conduction through a uniform flat layer, heat-transfer rate is conductivity multiplied by area and the temperature difference, divided by thickness. Use metres for thickness and square metres for area when conductivity is in W/(m·K). A temperature difference has the same numerical size in kelvins and degrees Celsius. The resulting rate is in watts, not an amount of energy. Multiplying a constant rate by elapsed time gives transferred energy in compatible units.
Worked example
The example below uses a stated conductivity and fixed surface temperatures. Increasing thickness reduces the rate; increasing area increases it. Each change is compared separately with the original layer, holding the other inputs fixed. The energy example assumes the original rate persists throughout the interval. Enter the layer properties in Heat conduction and check each unit before calculating. A millimetre thickness entered as metres would create a large error, even though the formula still returns a number.
k = 0.04 W/(m·K); A = 10 m²; ΔT = 20 K; L = 0.1 m
Q̇ = 0.04 × 10 × 20 / 0.1 = 80 W
L = 0.2 m → Q̇ = 40 W; A = 20 m² → Q̇ = 160 W
80 W × 2 h = 160 Wh = 0.16 kWh
Checks and limits
This model omits surface convection, radiation, contact resistance, thermal bridges and changes during warm-up. Conductivity can depend on material condition and temperature, so use a suitable measured or documented value rather than treating the example as a universal material specification. Layers in series require their thermal resistances to be added. A conductivity and a surface heat-transfer coefficient have different units and cannot be substituted for each other. This ideal layer estimate does not predict a complete building's energy bill or replace an engineering assessment.
Related calculators
Source: OpenStax — conduction