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
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 calculator on this page is similar to that provided by
except that I have
- 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
- 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
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.
- 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.
While their method provides good values for monatomic gases,
it produces the wrong results for other substances.(ref)
- 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
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
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).
- 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
In the hyperphysics calculator
In the calculator above
- 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
Set Density and Emissivity to the same values in both calculators.
(The exact values don't matter as long as they are the same.)
- 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
The Final temperature is
Close enough to validate the model.
- 240.54 K - hyperphysics
- 240.32 K - the calculator above
However, and this is important, the specific heat of water is 4.1813 J/gm K, but the hyperphysics calculator uses
Not even close!!!
12.5 J/mol K / (18 gm/mol) = 0.6944...