# Computing delta temperature based on radiation and heat loss

Is there a formula that gives me the instantaneous change in temperature under ideal circumstances? Details:

On a cloudless day, temperature is affected by two major things(?):

• While the Sun is up, incoming solar radiation, which increases the temperature at a known rate(?) depending on the sun's angle.

• Heat loss to "outer space". "Outer space" is colder than the ground, so heat is always flowing away from the ground into outer space. The hotter it is on the ground, the faster the heat flows away.

Is there a semi-accurate mathematical function that models these temperature changes for a given location on a given cloudless day?

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Your assumptions seem to be problematic. The thickness of the should have significant influence and you should consider the wind... –  hwlau Dec 16 '10 at 0:16
OK, let's make it a windless day as well. The thickness of what? –  barrycarter Dec 16 '10 at 0:19
I mean "of the cloud". I just want to point out that there are many factors affecting the temperature of a small location. –  hwlau Dec 16 '10 at 1:55
I'm considering a cloudless day, however. Also at sea level just to avoid altitude issues. –  barrycarter Dec 16 '10 at 2:04
hwlau's right, there are plenty of things that affect temperature in a small location, most notably wind (a.k.a. convection), but also heat conduction to/from the ground or other surroundings (cities are notorious heat sources). If you ignore all that stuff, you can make a simple model based only on thermal radiation, but it's pretty inaccurate. (For an example of how badly this fails, see my latest blog post - coincidence I swear :-P) It does work better for the Earth as a whole, though. –  David Z Dec 16 '10 at 2:14

The radiative portion is actually pretty straightforward. For the absorption of sunlight you multiply the strength of the sun (usually taken as 1000 watt/meter**2 at sea level by 1 minus the albedo, and multiply by the cosine of the angle. For IR, you need to know the emissivity (but for most surfaces .95 to 1 can be assumed), and use the Stefan Bolzmann law. You also have downgoing IR radiation from the sky, this flux will depend upon the temperature structure of the atmosphere and the atmospheric humidity. I have an IR thermometer, and typical clear sky temperatures are rougly -20-30F during the summer, and -60F during the winter. This is typically 130watts (at -60f) and 200watts at -20f, so it is significant.

Heat transport to the atmosphere is more difficult, as transport through a thin boundary layer will dominate. Although typically well more than half of any excess flux will be accounted for by the net IR flux.

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