The **total** amount of energy radiated per second by a perfect
blackbody depends only on its temperature **T** and
area **A**:

4 energy per second = sigma * T * Awhere

-8 Joules sigma is the Stefan-Boltzmann constant = 5.67 x 10 --------- s*m^2*K^4 T is the temperature of the object, in Kelvin A is the surface area of the object, in square meters

This relationship is called the Stefan-Boltzmann Law.

If we consider a patch of area exactly one square meter, then the energy radiated from it per second is

4 energy per second = sigma * T per square meter

A blackbody doesn't emit equal amounts of radiation at all wavelengths; instead, most of the energy is radiated within a relatively narrow band of wavelengths. The location of that band varies with the body's temperature; for example,

object T (Kelvin) radiates mostly ------------------------------------------------------------------------ very cold gas in insterstellar space 20 radio a live human being 310 infrared the Sun 5,600 visible interior of nuclear explosion 3,000,000 X-rays

The exact amount of energy emitted at a particular wavelength
**lambda** is given by the Planck function:

(2*h*c^2) / lambda^5 B (T) = ---------------------- lambda h*nu/k*T e - 1

Or, in beautiful typeset format,

In this equation,

B (T) is the energy (Joules) emitted per second per unit lambda wavelength per steradian from one square meter of a perfect blackbody at temperature T T is the temperature of the blackbody h is Planck's constant = 6.63 x 10^(-34) J*s c is the speed of light = 3.00 x 10^(8) m/s lambda is the wavelength k is Boltzmann's constant = 1.38 x 10^(-23) J/K

The Planck function has a distinctive shape: it rises very sharply at short wavelengths (due to the exponential), reaches a peak at some wavelength, then falls gradually at longer wavelengths.

In order to find the total energy emitted per second in some wavelength range

lambda 1 < lambda < lambda 2from one square meter of a perfect blackbody, one can integrate the Planck function

lambda 1 / (2*h*c^2) / lambda^5 energy emitted = | ---------------------- d(lambda) per second / h*nu/k*T per square meter lambda 2 e - 1 per steradian

and then multiply by (pi) steradians (we are interested in the energy emitted at all angles from one face of a flat plate, so we integrate over solid angle to cover half the sky; that ends up being the same as multiplying by a factor of (pi) = 6.28...)

Or, in nicely typeset form,

Obviously, if one integrates from the shortest possible wavelength (lambda = 0) to the longest possible wavelength (lambda = infinity), and multiplies by (pi), one ought to end up with the same total energy emitted per second as given by the Stefan-Boltzmann Law,

4 energy per second = sigma * T per square meter

* Last modified 4/06/2007 by MWR.*

Copyright © Michael Richmond. This work is licensed under a Creative Commons License.