Blackbody Radiation Calculator
Calculates total radiated power, peak emission wavelength, spectral intensity, or energy density for a blackbody or grey body at a given temperature. Use it in astrophysics, thermal engineering, or infrared sensor design.
About this calculator
A blackbody emits radiation across all wavelengths. The total radiated power is given by the Stefan–Boltzmann law: P = ε · σ · A · T⁴, where σ = 5.67 × 10⁻⁸ W/(m²·K⁴) is the Stefan–Boltzmann constant, ε is emissivity (1 for a perfect blackbody), A is surface area, and T is temperature in kelvin. The peak emission wavelength is found from Wien's displacement law: λ_max = b / T, where b = 2.898 × 10⁻³ m·K. The spectral radiance at a specific wavelength follows Planck's law: B(λ,T) = (2hc²/λ⁵) / (exp(hc/λkT) − 1), using Planck's constant h = 6.626 × 10⁻³⁴ J·s, speed of light c = 3 × 10⁸ m/s, and Boltzmann's constant k = 1.381 × 10⁻²³ J/K. Together these laws explain why hotter objects glow blue-white while cooler ones appear red.
How to use
Find the peak wavelength and total power radiated by the Sun, modelled as a blackbody at T = 5778 K with surface area A = 6.08 × 10¹⁸ m² and emissivity ε = 1. Peak wavelength: λ_max = 2.898 × 10⁻³ / 5778 ≈ 5.015 × 10⁻⁷ m = 501.5 nm — green light, consistent with solar spectrum data. Total power: P = 1 × 5.67 × 10⁻⁸ × 6.08 × 10¹⁸ × 5778⁴ ≈ 3.85 × 10²⁶ W, matching the accepted solar luminosity of 3.83 × 10²⁶ W.
Frequently asked questions
What is emissivity and how does it affect radiated power?
Emissivity ε is a dimensionless number between 0 and 1 that describes how efficiently a surface emits thermal radiation compared to a perfect blackbody (ε = 1). A polished metal mirror might have ε ≈ 0.05, meaning it emits only 5% of the power a blackbody at the same temperature would. Matte black paint has ε ≈ 0.97. The Stefan–Boltzmann law multiplies the blackbody power by ε, so knowing the emissivity of a material is essential for accurate heat-loss calculations in building insulation, spacecraft thermal control, and industrial furnaces.
Why does Wien's law predict that hotter stars appear blue and cooler stars appear red?
Wien's displacement law states that the peak emission wavelength is inversely proportional to temperature: λ_max = 2.898 × 10⁻³ / T. A hot star at 20,000 K peaks at about 145 nm (ultraviolet), but its visible emission is strongest toward the blue end. A cool star at 3,000 K peaks at about 966 nm (near-infrared), making its visible output predominantly red. Our Sun at ~5778 K peaks near 500 nm (green-yellow), but the combined spectrum across all wavelengths appears white to our eyes.
How is the Stefan–Boltzmann law used in real-world engineering?
Engineers use P = ε · σ · A · T⁴ whenever radiative heat transfer dominates — in spacecraft thermal design, industrial furnace efficiency, and passive cooling of electronics. Because power scales with T⁴, doubling the temperature increases radiated power 16-fold, making temperature control critical. In space, where conduction and convection are absent, radiation is the only cooling mechanism, so satellite designers carefully select surface coatings with the right emissivity to maintain safe operating temperatures.