Radiative Cooling Calculator

Cooling of an infinite slab to deep space (0 K) by radiation
Assuming no conduction, no convection, and no radiation from other sources

Initial Temperature K °C °F

Final Temperature K °C °F

Compute this when values change Elapsed time
Final temperature

Elapsed time	sec	days hr min sec
Specific Heat	J/gm K
Density	gm/cc
Emissivity
Thickness	mm

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 non-metallic 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 non-metallic 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 1m²
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

Hyperphysics provides derivations and calculators similar to the ones above: Specific Heat; Radiative Cooling Time - derives the formula for the amount of time to cool a substance
Emissivity: Concrete, Copper, Asphalt
Engineering Toolbox: This is one of several sources - as expected, they disagree
Densities of Solids Emissivity Coefficients Specific Heat

Author: Robert Clemenzi