Warming of an infinite slab by radiation
Assuming no heat loss
 Initial Temperature K   °C   °F
 Final Temperature K   °C   °F
Compute this when values change

 Radiation W/m2 <==> J/s/m2 Elapsed time sec days hr min sec Specific Heat J/gm K Density gm/cc Absorptivity Thickness mm
• Use a thickness of 1 millimeter (or less) to simulate a material with low thermal conductivity, such as a ceramic or insulator.
• When cooling, use 10 to 100 mm (1 cm to 10 cm) to simulate water, when heating water, most of the energy produces evaporation, not a temperature change
• This calculator assumes constant density and specific heat with a change in temperature - in reality, they aren't
• The model assumes no internal thermal gradients, only the volume matters
• When heating via electricity or conduction, set the Absorptivity to one (1)

Overview | Basics Validation | Water Specific heat | References

### Overview

How fast do objects warm (or cool) assuming a constant heat (or cooling) source?

This calculator is pretty straight forward

• It will compute one of the following
• The amount of time required for an object to warm from one temperature to another
• The expected temperature after absorbing radiation for a specified amount of time
• The amount of energy required per second to change the temperature in the specified amount of time
• The computed value depends on the
• Specific heat
• Density
• Absorptivity
• Thickness
of the selected substance
This application uses a simple linear model. Energy (heat) is measured in joules. A watt is the rate of energy transfer and is equal to one joule per second. Therefore, the amount of heat (number of joules) added (or removed) is simply the integral of watts over time. Since this model holds the source constant, it simply multiplies the input by the elapsed time. Note that cooling is modeled by supplying negative heat (the net flow is out of the system). I have kept the main power input as W/m2 because I intend to use this when analyzing climate issues where most values are normalized "per square meter".

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 example, 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 1.0*10-8 m3. Note - this value is just a swag (it gives reasonable results) and is not based on an actual calculation. Since light bulbs turn on in a fraction of a second, and because the calculator does not allow fractions of a second, increase the power to emulate shorter times - if the power is increased by 10, then one second displayed is actually 0.1 second elapsed.

• The Radiative Cooling Calculator assumes that heat is loss by radiation to a 0°K sink
• This calculator assumes a constant heat source or sink
• Emissivity has been renamed Absorptivity - but the values are still the same
• You can either set or compute the heating rate

### Basics

One watt per square meter, is the same as 1 joule per second per square meter.

 ```1 W/m2 <--> 1 J/s/m2 ```
Since there are about 86,400 seconds in a day, one watt per square meter is 86,400 joules per square meter per day.

These are typical properties for sand - the actual values depend on the type of sand and the data source.

 ```Density - 1500 kg/m3 = 1.5 gm/cc ref Specific heat - 830 J/kg/K = 0.830 J/gm/K ref ```
Assuming that only the top centimeter changes temperature in a "short" amount of time (reasonable because of low thermal conductivity)
 ```86,400 J/day / [830 J/kg/K * 1500 kg/m3 * (1*1*0.01 m3) ]= 6.9 K/day ```
To verify this calculation, in the calculator above, set
• Elapsed time to 1 day
• Specific Heat to 0.83 J/gm/K
• Density to 1.5 gm/cc
• Absorptivity to 1
• Thickness to 10 mm (1 cm)
Obviously, real materials don't have an absorptivity of 1.0, and they will lose heat via conduction, convection, and radiation. However, the purpose of this (extremely simple) model is just to provide a first order approximation to some of the claims made with respect to Greenhouse gases.

### Validation against an existing model

I never really trust my code and always try to validate it against someone else's data. This model was tested using the data and examples provided via Heat Storage in Materials. Select the Compute Radiation radio button and set the following
• Initial Temperature to 20°C
• Final Temperature to 40°C
• Elapsed time to 1000 seconds
• Specific Heat to 0.79 J/gm/K
• Density to 2.4 gm/cc
• Absorptivity to 1
• Thickness to 2000 mm (2 meters)
The computed radiation is 75,840 W/m2 which, when multiplied by 1,000 seconds yields
 ```75,840 W/m2 * 1,000 s = 75,840 J/s/m2 * 1,000 s = 75,840 kJ/m2 ```
which is the value provided in the example.

The second example requires a conversion (via google)

• 1 pound of water lb wt.= 15.34 US fluid ounces of water fl-oz ref
• 15.34 fl oz - fluid ounce US = 0.00 045 366 m3 - cubic meter ref
• 1 pound of water lb wt. = 453.59 grams of water g wt.
Using a thickness of 0.45366 mm, the calculator finds 0.9988 BTU - close. With a thickness of 0.4542 mm (15.358 fl oz), the result is exact. I assumed that this was because the volume of the water is based on the temperature - which was not specified in either the example or the conversion.

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. ref The imperial (avoirdupois, or international) pound is officially defined as 453.59237 grams. ref
The solution was to change the specific heat from 4.1813 J/gm/K to 4.187 J/gm/K - the value provided by the engineeringtoolbox. A detailed discussion of this is provided below.

### Water

Water is always a problem - every property depends on the temperature. They also depend on the amount of dissolved salts. To simplify the model, the calculator simply uses the values entered.

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.

### Specific heat

Specific heat indicates the amount of heat needed to change the temperature of one gram by one degree - in SI units, that is J/gm/K. In general, this value depends on the current temperature. Unfortunately, it also depends on the reference you get it from.

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.
On another page, the engineeringtoolbox provides
 ``` 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
The issue turned out to be the specific heat of water. The reference didn't bother to specify what value it was using ... and it was different from the value I was using.

### References

I love it - 5 references, 5 values.

Author: Robert Clemenzi