Stefan-Boltzmann Radiation Calculator
Calculate thermal radiation power emitted by an object using the Stefan-Boltzmann law.
Find heat radiated by temperature and surface area.
The Stefan-Boltzmann Law describes how much thermal radiation (heat in the form of electromagnetic waves) an object emits based on its temperature. It was first proposed by Slovenian physicist Josef Stefan in 1879 and later derived theoretically by Ludwig Boltzmann in 1884.
The Formula
For a perfect blackbody: P = σ × A × T⁴
For a real object (with emissivity): P = ε × σ × A × T⁴
Where:
- P = Power radiated (Watts)
- ε = Emissivity (0 to 1; 1 = perfect blackbody)
- σ = Stefan-Boltzmann constant = 5.670 × 10⁻⁸ W·m⁻²·K⁻⁴
- A = Surface area (m²)
- T = Absolute temperature (Kelvin)
The T⁴ Relationship
The most important feature of this law is the fourth-power dependence on temperature. Double the absolute temperature and the radiated power increases by 2⁴ = 16 times. This is why hotter objects radiate so much more intensely — a star at 6,000 K radiates enormously more than one at 3,000 K.
Emissivity Reference
| Material | Emissivity (ε) |
|---|---|
| Perfect blackbody | 1.00 |
| Human skin | 0.95–0.98 |
| Snow | 0.95–0.99 |
| Concrete | 0.88–0.93 |
| Wood | 0.80–0.90 |
| Glass | 0.85–0.95 |
| Polished steel | 0.07–0.17 |
| Polished aluminum | 0.05–0.09 |
| Gold (polished) | 0.02–0.04 |
Low-emissivity (low-e) coatings on windows exploit this — by reducing emissivity, they reduce heat radiation loss in winter.
Real-World Examples
- The Sun (surface ~5,778 K) radiates about 3.85 × 10²⁶ watts
- A human body at 37°C (310 K) radiates about 500 W from a surface area of ~1.7 m² — but also absorbs radiation from surroundings, so the net loss is much less
- Incandescent bulbs at 2,700 K radiate mostly infrared (heat), not visible light — which is why they are inefficient
- Thermal cameras detect the infrared radiation predicted by this law