This calculator is pretty straight forward
This calculator ignores all methods of heat loss, including conduction, convection, and the heat loss expected via blackbody radiation due to the temperature of the object. As a result, the computed temperature will quickly build to an unreasonable value in a very short time - real world results will be quite different.
When absorbing radiation, Thickness is used to estimate the thermal conductivity and/or mechanical mixing of the substance. Basically, when heating an insulator (sand or concrete) the heat penetrates only a few millimeters per hour. On the other hand, water may be mechanically mixed (wind and waves) to a depth of several meters. Air is well mixed from one to ten kilometers. When heating via other means (such as electricity), Absorptivity should be set to one and Thickness is used to specify the volume of the substance.
Because the thickness is mainly a guess, the calculator does not give accurate results. Its intent is to provide ball park values and to help understand the effect of changing various properties. In addition, since this model assumes no internal thermal gradients, it simply gives a lower (or upper) bound (depending on what is being calculated).
When heating (or cooling) a finite object with electricity (such as a light bulb),
W/m2 no longer makes sense.
In these cases,
it is the total volume that matters, not the surface area.
Since the Thickness is in millimeters,
the volume in cubic meters will be the Thickness value times 10-3.
for a volume of 1 m3, set the thickness to 1000 mm.
Just adjust it as appropriate.
In the case of a light bulb, I suggest a Thickness of 0.00001 mm
which produces a total filament volume of
The calculator on this page is similar to (actually made from) that provided by my Radiative Cooling Calculator except that
One watt per square meter, is the same as 1 joule per second per square meter.
1 W/m2 <--> 1 J/s/m2
These are typical properties for sand - the actual values depend on the type of sand and the data source.
86,400 J/day / [830 J/kg/K * 1500 kg/m3 * (1*1*0.01 m3) ]= 6.9 K/day
Validation against an existing model
75,840 W/m2 * 1,000 s = 75,840 J/s/m2 * 1,000 s = 75,840 kJ/m2
The second example requires a conversion (via google)
However, since the density is set to 1gm/cc, a simple mass conversion to grams followed by a density conversion to volume should remove any temperature dependence. Thus, the temperature should not matter. Based on the conversions, 1.0 lb = 453.59 grams, but that still gave the wrong answer. Interesting problem! The references are clear.
One BTU is the energy required to heat 1 avoirdupois pound of liquid water by 1 degree Fahrenheit, at a pressure of one atmosphere.
The imperial (avoirdupois, or international) pound is officially defined as 453.59237 grams. ref
The current version of this calculator also ignores evaporation which also depends on the temperature as well as the temperature of the adjacent air, the relative humidity, the wind speed, and many other things.
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 cools. 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. For the sake of simplicity, this model completely ignores evaporation.
This calculator uses 4.1813 J/gm°C. I am no longer sure which reference this came from, but it is a common value.
According to the engineeringtoolbox the specific heat of water is 4.187 J/gm/K. Located just below that, on the same page, is a table showing how the specific heat varies with temperature, from a low of 4.178 at 30°C up to 4.219 at 100°C. and on up to 14.6 near the triple point (360°C). For climate studies, values above about 55°C (131°F) are not of interest.
The following is from Footnote 1 at the bottom of the table.
|The International Committee for Weights and Measures, Paris, 1950, accepted W. J. de Haas’s recommended value of 4.1855 J/gm°C for the specific heat of water at 15 °C.
Specific heat (kJ/(kg K)) Water, fresh 4.19 Water, sea 36°F 3.93
This wikipedia table shows 4.1813 @ both 25°C and 100°C for constant mass. However, the wikipedia values (in the same table) for constant volume are identical to the constant mass values given in the engineeringtoolbox table. Obviously one of those references is wrong.
So, what value should I use?
The value used in my calculator when the Water button is pressed is 4.1813 and (according to the table) is appropriate for water at about 23°C (73°F), 4.187 is appropriate for about 13°C (55°F).
I discovered this issue because I was having trouble validating the calculator against
|One BTU is the energy required to heat 1 avoirdupois pound of liquid water by 1 degree Fahrenheit, at a pressure of one atmosphere. ref