Radiative Cooling Calculator
 Use a thickness of 1 millimeter (or less) to simulate a material with low thermal conductivity,
such as a ceramic or insulator.
 Use 10 to 100 mm to simulate water
 The radiating area does not matter
 Radiation is from one side of the slab only, cut the thickness in half to simulate
radiation from both sides of the slab
 This calculator assumes constant density and specific heat
with a change in temperature  in reality, they aren't
Overview
Using just radiation  How fast do objects cool?
This calculator simulates cooling to space (at 0K) by radiation alone 
it ignores conduction and convection,
assumes no internal heat source, and assumes that no heat is added
from any outside source.
Using a simple formula (which has a slightly confusing derivation),
it is possible to compute either
 The amount of time required for an object to cool by radiation alone, or
 The expected temperature after cooling for a specified amount of time
The calculator on this page is similar to that provided by
Hyperphysics
except that I have
 Provided a more dynamic user interface
 Used specific heat instead of molar mass
 Added a thickness value to help simulate different thermal conductivities
 Standardized the results for any flat radiating surface (slab) instead of a sphere
Water
Standing water cools much slower than a solid nonmetallic surface.
This is because the effective thermal conductivity is significantly different.
With low thermal conductivity, there is a large internal thermal gradient
because the surface cools much faster than the bulk of the material.
To simulate this for materials like sand or concrete, set the thickness to 1mm or less.
Smaller values simulate better insulators.
The thermal conductivity of water is not a lot greater than that of nonmetallic solids.
However, above about 4°C, as water cools, its density increases and the cooler surface layer will sink.
As a result, the surface layer is mixed to some depth as it cools.
This means that a larger volume of material must cool during the same amount of time
that a very small volume of a solid material.
This is what I mean by effective thermal conductivity.
For calm water in a lake or ocean, I suggest setting the thickness to a value between 1 and 10 cm
(10 and 100 mm).
For streams, set it to the depth of the stream.
When adding heat to water via IR radiation from above, the opposite is true 
Only the top millimeter or so warms and a significant amount of the absorbed heat
produces evaporation and, therefore, does not actually increase the temperature.
Future Improvements
 Use thermal conductivity to model thicker materials
 Add evaporation to water to see (demonstrate) which removes more heat
 Add constant radiation from the sun to heat the body to equilibrium
 Cycle the power from the Sun to determine the min, max, and average temperature
Validation against an existing model
When I create a new model, I like to validate the results against existing models.
For this calculator, I tried to validate it against
the one provided by hyperphysics
which uses a sphere.
Initially, I got radically different results  but I eventually found out why.
 I use measured (approximate) specific heats from tables
 However, hyperphysics uses the equipartition of energy formula
to predict the specific heat from the molecular weight
While their method provides good values for monatomic gases,
it produces the wrong results for other substances.(ref)
As a result, I decided to check the results with argon  40 gm/mol.
For an object with no internal thermal gradients, only the surface area and mass (volume) affect the rate
of cooling via IR radiation.
As a result, it is possible to emulate a sphere with a surface area of one square meter by simply
evaluating a slab of the appropriate thickness.
For a Sphere
 Area = 4pi r^2
 Volume = 4/3 pi r^3
 Volume = Area * r/3
 For an area of 1 m^2  r = sqrt(1/(4pi))
 Volume = 4/3 pi 1/(4pi) sqrt(1/(4pi)) = sqrt(pi)/6pi = 0.09403 m^3
 Which yields a 1.0 m^2 slab thickness of 94.03 mm for the same volume
Remember, this value assumes that the temperature of the sphere is uniform (no thermal gradients).
Otherwise, the model simply gives a lower (or upper) bound (depending on what is being calculated).
In the hyperphysics calculator
 Set Area to 1 m^2
 Set molar mass (M) to 40 gm/mol
 Set High temperature to 300 K
 Set Cooling time to 6,000 sec
In the calculator above
 Set Specific Heat to 0.31 J/gm K  (12.5 J/mol K / (40 gm/mol))
 Set Thickness to 94.03 mm to simulate (have the same volume as) a sphere with an area of 1m^{2}
 Set High temperature and Cooling time to the same as above
Set Density and Emissivity to the same values in both calculators.
(The exact values don't matter as long as they are the same.)
The Final temperature is
 240.54 K  hyperphysics
 240.32 K  the calculator above
Close enough to validate the model.
However, and this is important, the specific heat of water is 4.1813 J/gm K, but the hyperphysics calculator uses
12.5 J/mol K / (18 gm/mol) = 0.6944...

Not even close!!!
References
Author:
Robert Clemenzi