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Library / Heat Transfer / HTR-09

Heat Transfer

Blackbody emissive power

Radiation emitted per unit area by a perfect emitter.

Ref
HTR-09
Rev
v1
Field
Heat Transfer
Result
E [W/m²]

T is the absolute temperature of the surface in kelvin and σ the Stefan–Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴). E is the power emitted per unit area into the half-space in front of the surface. A real surface emits ε times this, where the emissivity ε is between 0 and 1.

Radiation emitted per unit area by a perfect emitter.

Calculator

Solve for
W/m²
K
1005,000
  • Emissive power at twice the temperature56,700 W/m²
  • Wavelength of peak emission (Wien), 2898/T5.796 µm

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Emissive power (W/m²) current value 3,544 W/m²
TTemperatureKey driver
5.673.54e7
  • A surface at 500 K (227 °C) can emit at most 3.5 kW/m². The Sun’s surface at about 5778 K emits about 63 MW/m², 18 000 times more.
  • The peak wavelength is 5.8 µm at 500 K, in the infrared, which is why thermal imagers see hot engines and exhausts.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (W/m²)Local stepChange in result
T Temperature100 to 5,000 K5.673.544e73.544e7+10%+46.4 %

Sweep of Temperature

Temperature (K)Emissive power (W/m²)
1005.67
712.514,610
1,325174,800
1,938799,100
2,5502,398,000
3,1635,672,000
3,7751.152e7
4,3882.101e7
5,0003.544e7

Sweep one variable

Emissive power (W/m²)Hover, touch, or use the arrow keys on the chart
010M20M30M 10002000300040005000 Temperature (K) now 3,544

Variables and how each one contributes

E

Emissive power

Result · W/m²

What the equation gives you. Everything else feeds it.

T

Temperature

K · typical 100 to 5,000

The absolute temperature of the surface, in kelvin. Fourth power: doubling the temperature increases the emission 16 times. A 10 % rise in T raises E by 46 %.

Worked example with the values above

  1. Fourth power: T⁴ = (500)⁴ = 6.25×1010 K⁴.
  2. Multiply by σ = 5.67×10−8.
  3. E = 3,544 W/m².

Assumptions and limits

  • A perfect emitter (blackbody) with emissivity 1.
  • Uniform surface temperature and emission only; absorbed radiation from the surroundings is a separate calculation.

Common mistakes

  • Using °C. The formula needs kelvin.
  • Forgetting the emissivity of a real surface, which can be 0.05 for polished metal and 0.95 for paint.

Statistics and uncertainty

E ∝ T⁴, so a 1 % error in temperature gives a 4 % error in the emitted power. This is the most temperature-sensitive formula in heat transfer, and the reason a thermal camera must be calibrated for the surface’s emissivity as well as its temperature.

Where it is used

  • The upper limit of thermal emission, used to compare real surfaces.
  • Estimating radiant heating of equipment from a hot surface.

Maths you need

  • Absolute temperature: kelvin
  • Powers: fourth power

Related formulas

Practise this formula