Wolfram Alpha:

```
Stefan-Boltzmann Law
--------------------
The Planck radiation summed over all wavelengths leads to Stefan's Law.  The energy
radiated by a blackbody radiator per second per unit area is proportional to the
fourth power of the absolute temperature and is given by

Q = AσT4   σ = Stefan's constant = 5.6703 x 10-8Wm2K4

Q is in Watts

For hot objects other than ideal radiators, the law is expressed in the form:

Q = eAσT4

where e is the emissivity of the object (e = 1 for ideal radiator).

If the object is radiating/absorbing energy to its surroundings at temperature
Tenvironment, the net radiation loss/gain rate takes the form:

Q = eAσ(Tobject4 - Tenvironment4)

If Tobject > Tenvironment the object is radiating.
If Tobject < Tenvironment the object is absorbing.

The Stefan-Boltzmann relationship is also related to the energy density in the
radiation in a given volume of space.

Ex.

A block of ice (with dimensions 14cm x 10cm x 6cm) at temperature T = 0 deg C is
sitting outside in the shade.  The temperature outside is T = 26 deg C.  Assuming
it is a perfect emitter and absorber of radiation (e = 1), how long (in hours)
will it take to completely melt?  Assume heating occurs via radiation only
and suppose that heat leaves and enters all six sides of the block.

A = 1 * (2 * 0.14 * 0.10) + (2 * 0.14 * 0.06) + (2 * 0.10 * 0.06) = 0.057m2

The rate at which heat enters the block is:

0.057 * 5.6703 x 10-8 * (2734 - 2994)

= 7.88J/s

Therefore, total Q = 7.88 * t

Now,

Q = mLf where Lf is the latent heat of fusion

= ρiceViceLf

= 931 * .14 * .10 * 0.06 * 334,000

= 2.6 x 105 J

Therefore,

t = 2.6 x 105/7.88 = 3.3 x 104s = 9.2 hours
```