Physics & Mechanics

Stefan-Boltzmann Law Calculator

Calculate the thermal power radiated by a black body based on its absolute temperature and surface area.

K
Radiated Power
126,401,399.47

Calculated locally in your browser. Fast, secure, and private.

The Power of Thermal Radiation

Unlike conduction or convection, which require physical matter (like metal or air) to transfer heat, Thermal Radiation can travel through the absolute vacuum of space. Every single object in the universe that is above absolute zero constantly radiates electromagnetic energy (light).

The Stefan-Boltzmann Law states that the total thermal power radiated by a black body is proportional to the fourth power of its absolute temperature. This fourth-power relationship is staggeringly powerful: if you simply double the temperature of an object, it will radiate $2^4$ (which is $16$) times more energy!

The Glowing Universe

  • The Sun: The Sun is floating in a perfect vacuum. The only way it can transfer heat to Earth is via thermal radiation. By measuring the total power of sunlight hitting the Earth, and knowing the distance to the Sun, astrophysicists used this exact law to accurately calculate the surface temperature of the Sun (roughly $5778 , \text{K}$) without ever having to go there.
  • Night Vision Goggles: You are constantly glowing. Because human body temperature is relatively low (around $310 , \text{K}$), the radiation we emit is not visible light; it is long-wave infrared light. Thermal cameras use the Stefan-Boltzmann law to detect this infrared glow and convert it into a visible image, allowing soldiers to see in total darkness.
  • Incandescent Lightbulbs: A traditional lightbulb passes electricity through a thin tungsten filament until it reaches roughly $3000 , \text{K}$. According to this law, it radiates a massive amount of power. However, because it's not hot enough, 95% of that radiated power is invisible infrared heat, making these bulbs terribly inefficient for actually producing visible light.

The Formula

P=εσAT4\begin{aligned} P = \varepsilon \cdot \sigma \cdot A \cdot T^4 \end{aligned}

Where:
P=
Radiated Power (Watts, W)
ε\varepsilon=
Emissivity of the material (0.0 to 1.0)
σ\sigma=
Stefan-Boltzmann Constant (5.670374 × 10⁻⁸ W/(m²·K⁴))
A=
Surface Area (m²)
T=
Absolute Temperature (K)

Example Calculation

Let's calculate the radiated power of a standard human being. The average human has a surface area of about $1.8 , \text{m}^2$ and a skin temperature of roughly $306 , \text{K}$ ($33^\circ\text{C}$). Human skin, surprisingly, is nearly a perfect "black body" in the infrared spectrum, so its emissivity is $e = 0.98$.

  1. Calculate Temperature to the 4th Power ($T^4$): $306^4 = 8,767,522,096$.
  2. Multiply by Stefan-Boltzmann Constant ($\sigma$): $8,767,522,096 \cdot (5.67 \times 10^{-8}) \approx 497.11$.
  3. Multiply by Area and Emissivity: $497.11 \cdot 1.8 \cdot 0.98 = 876.9 , \text{Watts}$.

You are constantly blasting almost $900 , \text{Watts}$ of raw infrared energy into the room! (Note: The walls of the room are also radiating heat back at you, which is why you don't freeze to death instantly. Your net heat loss is much lower, typically around 100 Watts).

Frequently Asked Questions

In physics, a black body is an idealized theoretical object that perfectly absorbs all electromagnetic radiation that hits it, and perfectly emits it back out. A true black body has an emissivity ($\varepsilon$) of exactly 1.0. In the real world, things like soot or Vantablack get very close (0.99), while shiny silver mirrors are terrible emitters (0.02).

Because the law relies on raising the temperature to the 4th power ($T^4$). If you used Celsius and plugged in $0^\circ\text{C}$, the math would incorrectly say the object radiates zero power. Only absolute zero ($0 , \text{K}$) radiates zero power.

Because of Net Radiation. You are radiating 900W out, but the walls of a cold room are only radiating 700W back at you. You are operating at a net loss of 200W, which drains your body heat. In a 30°C room, the walls radiate 850W back at you, so your net loss is only 50W, making you feel warm.