Blackbody Radiation Calculator

Enter a temperature and emissivity to calculate the spectral radiance of a blackbody using Planck's law. You can also specify a wavelength to find the spectral radiance at that exact point, and the calculator returns the peak wavelength via Wien's displacement law, total radiated power via the Stefan-Boltzmann law, and a full spectral curve across the thermal emission range.

K

Enter the temperature in Kelvin. The Sun's surface is ~5778 K.

1 = perfect blackbody. Real materials are between 0 and 1.

nm

Enter a wavelength in nanometres to get spectral radiance at that specific point.

Results

Peak Wavelength (Wien's Law)

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Spectral Radiance at Given Wavelength

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Peak Spectral Radiance

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Total Radiated Power (Stefan-Boltzmann)

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Temperature (Kelvin)

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Blackbody Spectral Radiance vs. Wavelength

Results Table

Frequently Asked Questions

What is blackbody radiation?

A blackbody is an idealised object that absorbs all incoming electromagnetic radiation and re-emits it purely based on its temperature. The spectrum it emits — described by Planck's law — was the key observation that led to the development of quantum mechanics. Real objects approximate blackbody behaviour with an emissivity factor between 0 and 1.

What is Planck's law for blackbody radiation?

Planck's law gives the spectral radiance B(λ,T) = (2hc²/λ⁵) × 1/(e^(hc/λk_BT) − 1), where h is Planck's constant (6.626×10⁻³⁴ J·s), c is the speed of light, k_B is Boltzmann's constant (1.381×10⁻²³ J/K), λ is wavelength, and T is temperature in Kelvin. This formula quantises radiation and predicts the observed spectrum perfectly.

What is the peak wavelength of a blackbody at 932 °F?

932 °F equals 500 °C, or 773.15 K. Using Wien's displacement law (λ_peak = 2.898×10⁶ nm·K / T), the peak wavelength is approximately 3749 nm, which falls in the mid-infrared region — invisible to the human eye.

How do you calculate the power radiated by a blackbody?

The total power radiated per unit area is given by the Stefan-Boltzmann law: P = ε × σ × T⁴, where σ = 5.67×10⁻⁸ W·m⁻²·K⁻⁴ is the Stefan-Boltzmann constant, ε is the emissivity, and T is temperature in Kelvin. A perfect blackbody (ε = 1) at 5778 K radiates about 63.2 MW/m².

What is Wien's displacement law?

Wien's displacement law states that the peak wavelength of blackbody emission is inversely proportional to temperature: λ_peak = b/T, where b ≈ 2.898×10⁶ nm·K. As temperature increases, the peak wavelength shifts to shorter (bluer) wavelengths, explaining why hotter stars appear blue and cooler stars appear red.

How does the radiation spectrum change as the blackbody temperature increases?

As temperature rises, two things happen: the peak of the emission curve shifts to shorter wavelengths (Wien's law), and the total emitted power increases dramatically (proportional to T⁴). A body at room temperature (~300 K) emits mostly in the far infrared around 9700 nm, while the Sun at 5778 K peaks in the visible green region at ~500 nm.

What is emissivity and how does it affect the calculation?

Emissivity (ε) is a dimensionless factor from 0 to 1 that describes how efficiently a real surface emits thermal radiation compared to a perfect blackbody. A value of 1 means perfect blackbody emission. Real materials have emissivities less than 1 — for example, polished metals can be as low as 0.05, while human skin is about 0.98. The spectral radiance is simply multiplied by ε.

Are black holes perfect blackbodies?

In a quantum mechanical sense, yes — black holes emit Hawking radiation with a near-perfect blackbody spectrum. However, the temperature of a stellar-mass black hole's Hawking radiation is extraordinarily low (nanokelvins), making it completely undetectable in practice. The concept is important theoretically because it links thermodynamics, quantum mechanics, and general relativity.

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