Net Radiation Exchange Between Surfaces

Q˙=εσA(T14−T24)\dot{Q} = \varepsilon \sigma A (T_1^4 - T_2^4)
sig figs

Worked example: eps 0.85, 1.2 m2, 500 K to 300 K → 3146 W. Press Try an example to run it live, then adjust anything.

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See also

Combined Convection and Radiation Coefficient: Both use emissivity and surface temperature.Stefan-Boltzmann Law: Both use emissivity and surface area.Net Radiative Cooling to the Sky: Both use emissivity and surface temperature.Newton's Law of Cooling (Q = hAΔT): Both use net radiant heat rate and surface area.Conduction Through a Pipe Wall: Both use net radiant heat rate.Fin Heat Transfer Rate: Both use net radiant heat rate.
6 more relatedGeothermal Gradient from a Bottom-Hole Temperature: Both use surface temperature.Heat Flow from Thermal Resistance: Both use net radiant heat rate.Heat Loss Through an Assembly (Q = A·ΔT/R): Both use net radiant heat rate.Surface Temperature Depression Under a Clear Sky: Both use surface temperature.Glycol Loop Heat Transfer (Capacity Derate): Both use net radiant heat rate.Hydronic Heat Transfer (Water): Both use net radiant heat rate.
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Net Radiation Exchange Between Surfaces explained

T1T2QεA

Every surface both emits and absorbs radiation, and the net exchange with large surroundings goes as the difference of fourth powers, not as a simple ΔT. That non-linearity is why radiation is negligible in a chilled-water pipe and dominant in a furnace. A 1.2 m² oxidised steel panel (ε = 0.85) at 500 K facing a 300 K room radiates 0.85 × 5.670 × 10⁻⁸ × 1.2 × (6.25 × 10¹⁰ − 8.1 × 10⁹) = 3146 W: with a convective coefficient of 10 W/(m²·K) it would shed only about 2400 W by convection, so more than half the heat leaves as light you cannot see.

Emissivity is where field work goes wrong. Polished aluminium is 0.04, mill-finish steel 0.2–0.3, the same steel after a summer outdoors 0.7–0.85, and almost every paint, oxide, brick and organic surface is 0.85–0.95 regardless of colour: white paint is as good a radiator in the infrared as black. That is the trap in infrared thermography: point a camera set for ε = 0.95 at a shiny bus bar and it reports the temperature of whatever the bar is reflecting, which is usually you. Tape a square of matte tape on the target and read that instead. The formula assumes the surroundings are large enough to behave as a black enclosure; two comparable surfaces facing each other need view factors and a full radiosity network.

Net Radiation Exchange Between Surfaces formula

Q˙=εσA(T14−T24)\dot{Q} = \varepsilon \sigma A (T_1^4 - T_2^4)
Where
  • Q˙\dot{Q}= Net radiant heat rate (W)
  • ε\varepsilon= Emissivity
  • AA= Surface area (m²)
  • T1T_1= Surface temperature (°C)
  • T2T_2= Surroundings temperature (°C)
Revised 16 October 2026
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formula.expert. (2026, October 16). Net Radiation Exchange Between Surfaces Formula & Calculator. https://formula.expert/solvers/radiation-net-exchange