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/sci/ - Science & Math

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>> No.9586328 [DELETED]  [View]
File: 49 KB, 790x433, black-body-radiation-curves.png [View same] [iqdb] [saucenao] [google]
9586328

At least pretend you know why you're sad.

>> No.9404710 [View]
File: 49 KB, 790x433, black-body-radiation-curves.png [View same] [iqdb] [saucenao] [google]
9404710

>>9404675

Partly. The light of the sun is black body radiation, meaning it simply glows because it's hot. It's the same kind of glow that you get when you heat something in a furnace to glowing (only hotter, surface of the Sun is 6000K). This is a continuous band of radiation across many wavelengths (see figure).

A flame also gives off black body radiation, because the air is heated to glowing. But there is light also from the chemical reactions happening (oxidation of the fuel), which has just a few very well-defined wavelengths in which the color peaks. This is why manganese burns bright white while sodium burns yellow, it all depends on the relative energy levels of the oxidation reaction.

It's a bit of a complex mess since there's burning soot flying around and gas-phase reactions etc.

But short answer: a flame is not as hot as the sun, so the black body radiation will be less intense, and you have chemical reaction induced radiation as well.

>> No.8083246 [View]
File: 49 KB, 790x433, black-body-radiation-curves.png [View same] [iqdb] [saucenao] [google]
8083246

>>8082948
You must be new here

>> No.6973249 [View]
File: 49 KB, 790x433, black-body-radiation-curves[1].png [View same] [iqdb] [saucenao] [google]
6973249

objects at room temperature (~300 K) emit quite a lot of radiation.

>> No.6654785 [View]
File: 49 KB, 790x433, bull.png [View same] [iqdb] [saucenao] [google]
6654785

When graphing the spectral distribution for a blackbody you get something like pic related, they are using arbitrary units for intensity, how are you supposed to interpret and measure arbitrary units?

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