The Stefan-Boltzmann Law was first discovered experimentally in 1879 by Stefan, then derived theoretically in 1884 by his student Boltzmann. It says:

Note the strong dependence on temperature: -- as the fourth
power of the temperature. I.e., double the temperature, and the radiated power
will increase by a factor 2^{4} = 16. Triple the temperature, and the
power will increase by a factor 3^{4 }= 81, etc. The Stefan-Boltzmann
Law is valid only for perfect radiators (called "blackbodies"). Actual radiating
surfaces are not perfect radiators, and will always radiate ** less**
than the luminosity given by the S-B Law -- typically some 10 - 80%. But the
radiating surfaces (called "photospheres") of stars are fairly good approximations
to black bodies, and typically radiate a luminosity of more than 90% of the
value given by the S-B Law.

The value of the Stefan-Boltzmann constant, s
= 5.67 x 10^{-5} ergs cm^{-2}
s^{-1} K^{-4}, is determined by experiments
in Earth labs. We assume that the S-B Law, with the same value of s,
is valid everywhere in the universe. That assumption is one example of the Principle
of Universality of Physical Laws.

1. The Earth:
The average temperature of the Earth's surface must be about 10 C. We must first
convert this to the Kelvin scale: T(K) = T(C) + 273 = 283 K. The Earth's surface
area is about A = 4p R^{2} = 4p
(6.4 x 10^{8} cm)^{2} = 5.15 x 10^{18} cm^{2}.
Now we can calculate the radiated luminosity of the Earth: L = 5.15 x 10^{18}
x 5.67 x 10^{-5} x (283)^{4} = 1.87 x 10^{24} ergs s^{-1}.
To change ergs s^{-1} to Watts, multiply by 10^{-7}. So, L =
1.87 x 10^{17} Watts. (Actually, the Earth radiates only about half
this much luminosity into outer space because it is not a black body.)

2. The Sun:
The temperature of the Sun's photosphere is 5800 K and its radius is 7 x 10^{10}
cm. So the Stefan-Boltzmann Law gives a luminosity L = 4 x 10^{33} erg
s^{-1} = 4 x 10^{26} Watts.

3. Your body: (This calculation will be a bit more subtle because you must correct for the temperature of the environment.)

Your body temperature is about T(F) = 98.6 F. To use the S-B
Law, we must convert this to the Kelvin scale. To do this, first convert your
body temperature to Centigrade: T(C) = (5/9)[T(F) -32] = 37 C. Then, to convert
this to Kelvin, add 273: T(K) = T(C) + 273 = 310 K. Now, let's estimate your
body surface area. Approximate your body shape roughly by a rectangular block,
of height 180 cm, width 30 cm, and depth 15 cm, giving a surface area A = 2[180
x 30 + 180 x 15 + 30 x 15] = 17,100 cm^{2}. Now we can calculate the
luminosity of your body: L = 17,100 x 5.67 x 10^{-5} x (310)^{4}
= 8.95 x 10^{9} ergs s-1. So, according to the S-B Law, your body could
radiate a luminosity L = 895 Watts. That's a lot! But this would only be true
if your body was in outer space where the temperature of the sky is very cold
-- about 5 K -- and if you were naked. If you wore an insulating space suit,
you would radiate much less than this because the surface temperature of the
suit would be much colder than your body. On Earth, you won't lose nearly as
much heat as this, because your body is absorbing heat from its surroundings
at the same time it is radiating. To correct for this absorption, you should
use a modified Stefan-Boltzmann Law:

L = As
(T^{4 }- T_{env}^{4})

Where T_{env}
is the temperature of the environment. Let's say that T_{env} = 290
K (about 63 F). Using this formula, the calculation gives L = 210 Watts. But,
if you put your clothes on, you can cut this radiation down by a big factor
-- say, a factor 4 with light clothes and a sweater. Then you are losing about
50 Watts, the equivalent of about 1 light bulb.