Radiation Heat Loss Calculator (Stefan Boltzmann)
Calculate heat lost by thermal radiation from a hot surface to its surroundings, using the Stefan Boltzmann law and the surface emissivity, and add natural convection for the total.
- qrad = εσA(Ts⁴ − Ta⁴) = 0.9 × 5.670 × 10⁻⁸ × 2 × (473.15⁴ − 298.15⁴) = 4,309 W
- qconv = h × A × ΔT = 2,800 W
Stefan Boltzmann Law
qconv = h × A × (Ts − Ta)
σ = 5.670374 × 10⁻⁸ W/m²·K⁴
ε is the surface emissivity, σ the Stefan Boltzmann constant, A the area and T absolute temperatures in kelvin. The formula assumes a small surface radiating to large surroundings at Ta. Emissivity values are typical; measured values vary with surface condition.
Worked example
A 2 m² painted steel surface (ε 0.9) at 200 °C in a 25 °C room radiates about 4.31 kW and loses another 2.8 kW by convection (h = 8), so radiation is about 61 % of the loss.
Frequently Asked Questions
What is emissivity?
How well a surface radiates compared with a perfect black body, from 0 to 1. Painted surfaces are about 0.9; polished metals can be below 0.1.
Why must temperature be in kelvin?
Radiation depends on absolute temperature to the fourth power. Using °C would give wrong results.
Does emissivity matter for IR thermometers too?
Yes. Pyrometers need the correct emissivity setting; see the pyrometer emissivity calculator.
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