What is the Blackbody Radiation Calculator?
A blackbody is an idealized object that absorbs all incident radiation and re-emits it purely as a function of its temperature. This calculator takes a single temperature in kelvin and returns two key quantities: the total radiant exitance (energy radiated per second per square metre) from the Stefan-Boltzmann law, and the wavelength at which emission peaks from Wien's displacement law.
How to use it
Enter the absolute temperature in kelvin (K). For reference, ice melts at 273 K, the Sun's photosphere is about 5778 K, and a tungsten filament glows near 3000 K. Press calculate to see the radiant exitance in W/mĀ² and the peak wavelength in both nanometres and metres.
The formulas explained
The Stefan-Boltzmann law, \(j = \sigma T^{4}\), shows that total emitted power scales with the fourth power of temperature ā doubling the temperature increases output sixteenfold. Wien's law, \(\lambda_{\max} = b/T\), shows the peak wavelength shifts inversely with temperature, so hotter objects glow bluer and cooler ones redder.
$$j = \sigma\, \text{Temperature (K)}^{4} \qquad \lambda_{\max} = \frac{b}{\text{Temperature (K)}}$$ $$\text{where}\quad \left\{ \begin{aligned} \sigma &= 5.670374419 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4} \\ b &= 2.897771955 \times 10^{-3}\ \text{m K} \end{aligned} \right.$$
Worked example
For the Sun's photosphere at 5778 K: $$j = 5.670374419 \times 10^{-8} \times 5778^{4} \approx 6.32 \times 10^{7}\ \text{W/m}^2.$$ The peak wavelength is $$\lambda_{\max} = \frac{2.897771955 \times 10^{-3}}{5778} \approx 5.015 \times 10^{-7}\ \text{m} \approx 501.5\ \text{nm},$$ in the green part of the visible spectrum.
Key Terms & Variables
- Blackbody
- An idealized object that absorbs all incident electromagnetic radiation at every wavelength and re-emits energy purely as a function of its temperature. It is the perfect thermal emitter against which real surfaces are compared.
- Radiant exitance \(j\)
- The total radiant power emitted per unit surface area, expressed in watts per square meter (\(\text{W m}^{-2}\)). For a blackbody it follows the StefanāBoltzmann law \(j = \sigma T^4\).
- StefanāBoltzmann constant \(\sigma\)
- The proportionality constant linking exitance to the fourth power of temperature: \(\sigma = 5.670374419\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}\).
- Wien displacement constant \(b\)
- The constant in Wien's displacement law relating peak wavelength to temperature: \(b = 2.897771955\times10^{-3}\ \text{m K}\).
- Peak wavelength \(\lambda_{\max}\)
- The wavelength at which the spectral radiance of a blackbody is greatest, given by \(\lambda_{\max} = b/T\). Higher temperatures shift the peak toward shorter wavelengths.
- Emissivity \(\varepsilon\)
- A dimensionless factor between 0 and 1 describing how efficiently a real surface emits compared with an ideal blackbody. The real exitance is \(j = \varepsilon\sigma T^4\); a perfect blackbody has \(\varepsilon = 1\).
- Absolute temperature \(T\)
- Temperature measured on the Kelvin scale, where 0 K is absolute zero. Both blackbody laws require Kelvin, since they are defined relative to absolute thermal energy.
FAQ
Why does the Sun appear yellow-white if its peak is green? The eye integrates across the whole broad spectrum; the mix of all visible wavelengths appears white-yellow.
Does this assume emissivity of 1? Yes ā it models an ideal blackbody. A real (grey) body emits \(j = \varepsilon \sigma T^{4}\) with emissivity \(\varepsilon < 1\).
What units are used? Temperature in kelvin, exitance in W/mĀ², wavelength in nanometres and metres.
