If the Planck distribution function spectral irradiance is integrated over all wavelengths, then the total irradiance emitted into a hemisphere is given by the **Stefan-Boltzmann Law**:

$${F}_{s}=\sigma \text{\hspace{0.05em}}{T}^{4}$$

where σ is called the Stefan-Boltzmann constant (5.67x10^{-8} W m^{-2} K^{-4}). F_{s} has units of W m^{-2}, where the m^{-2} refers to the surface area of the object that is radiating.

The Stefan-Boltzmann law (total) irradiance applies to an object that radiates according to the Planck distribution function spectral irradiance. If we look at the figure below, we see that the solar spectrum at the top of the atmosphere is similar to the Planck distribution function but does not follow it perfectly. However, the Planck distribution function with the same total irradiance as the sun has a temperature of 5777K, as in the second figure.

#### Check Your Understanding

Clouds radiate. Assume two spherical clouds, one with a radius of 100 meters and a temperature of 275K and a second with a radius of 100 meters and a temperature of 230K. Assuming that they both radiate according to the Planck distribution function, which one is radiating more total energy?

**Click for answer.**

ANSWER:

Cloud T (K) | Cloud radius (m) | F_{s} (W m^{-2}) |
F_{s} x 4πR_{c}^{2} (W) |
---|---|---|---|

275 | 100 | 324 | 4.1x10^{7} |

230 | 100 | 159 | 2.0x10^{7} |

These little clouds are radiating quite a lot of energy in all directions, but some of it is down toward Earth’s surface. If we make the simple assumption that half the radiation goes up and the other half goes down, the amount of energy radiated toward Earth’s surface per second is approximately 10 million Watts. If the clouds are not too far from the surface, this downward radiation could contribute a few hundred W m^{-2} of heating at Earth’s surface. Thus clouds can act like additional heat sources for Earth’s surface, keeping its temperature higher than it would be on a clear night.