How does heat behave in no atmosphere?

Huh. Cool. (No pun intended.)

So when Ali G. asked Buzz Aldrin if he could walk on the sun (“in winter,” if necessary) it’s doable, at least before you sink in the boiling granules:

The Sun’s photosphere has a temperature between 4,500 and 6,000 K (4,230 and 5,730 °C)[3] (with an effective temperature of 5,777 K (5,504 °C))[4] and a density of about 2×10−4 kg/m3;[5] other stars may have hotter or cooler photospheres. The Sun’s photosphere is composed of convection cells called granules—cells of plasma each approximately 1000 kilometers in diameter[6] with hot rising plasma in the center and cooler plasma falling in the narrow spaces between them. Each granule has a lifespan of only about eight minutes, resulting in a continually shifting “boiling” pattern. Grouping the typical granules are super granules up to 30,000 kilometers in diameter with lifespans of up to 24 hours…

Boiling granules: band name.

Two language queries; I’m not being a nitpick nudnik here, nor of course questioning your knowledge:

“absorbtivity”–is this the correct word?

“rejected”–ditto

Conventional oxy-fuel welding torches, like oxygen-acetylene torches, are close to the Sun’s temperature, but not quite. But plasma torches are much hotter than the Sun’s photosphere.

Usually spelled “absorptivity” but yes, that is the correct term. Here’s a table of absorptivity of various surfaces.

I think this is a colloquial term, not a specific technical term.

Sort of weird and cool. The ~ule suffix is meant to be a diminutive - so it’s odd to hear the term being used to describe things that are orders of magnitude bigger than Rhode Island. Also, ‘super granules’ is like ‘giant baby sweetcorn’.

I’m not sure where one ought to draw the line between colloquial and technical, but heat rejection is pretty common terminology, referring to waste heat discarded by thermodynamic systems. Example, the temperature of exhaust gas is a manifestation of heat that’s been rejected by your car’s engine.

<dogpile> … er … radiation rate … </dogpile>

Sorry, Stranger, but otherwise a nice post, thanx for the clarifications …

ETA: We’ve all seen things heated up to red-hot … heat it a little more and it becomes yellow-hot, like the Sun … heat it even more and it’s white-hot … even more and the white will take on a blue hue … and that color is actually a very good indicator of the temperature … and here’s the coolest part, it doesn’t matter what the substance is; hydrogen, iron, carbon, plutonium … at the same temperature, every substance will radiate the at the same frequencies …

Thus, for instance, a red-hot star (like Betelgeuse, say) is at about the same temperature as a red-hot stove burner.

This is an over-simplification. Most real-world objects are not blackbodies. If they were, everything would have an emissivity of 1, and absorptivity of 1. Most real-world objects have characteristic emission lines and/or absorption lines on top of the blackbody radiation. In many cases these line emissions dominate - e.g. neon signs, fluorescent lights, and fireworks. Even the Sun is pretty far from a blackbody - there are numerous absorption lines in visible light, and excess UV and X-ray emitted by the chromosphere and corona.

I’ll go check the recent flame thread, but doesn’t a lit match have the colors from the base of the flame blue-(clear/no color]-white, and the blue–ie, not the white–is hottest?

Most things aren’t blackbodies, but for most things blackbody is a pretty good approximation.

This is part of a chemical reaction … where we have molecules deionizing … electrons are cascading down the shells releasing energy at frequencies specific to the transition … that’s a different process than just a lump of 100 K iron sitting in 3 K space …

The important part here is that objects at a higher temperature than it’s environment will transfer energy to it’s environment … and without a conductive or convective medium, then this transfer will through electromagnetic radiation … an ideal substance will radiate exactly like a blackbody … but most non-ideal substances will also radiate something close to a blackbody … our 100 K lump of iron will eventually radiate enough energy to be at 3 K even in the remotest corner of the universe …

The blue part of the flame is very far from a blackbody. It’s blue because of emission lines. You can see a spectrum here.

The white/orange parts are very close to blackbodies - it’s particles of solid carbon (soot) which are getting hot and glowing.

This is why a properly adjusted gas flame is mostly blue/clear. With proper gas mixture, all the gas burns without turning into soot.

This. A theoretically ideal “blackbody” emits electromagnetic waves/photons over distribution of wavelengths and intensities that vary as a function of its temperature. See this page, and take a look at the first plot you see. At low temperatures, a blackbody emits little total power (per unit area), and the peak of its emission spectrum is in the infrared range; the net result is that it doesn’t emit enough light in the visible range for you to see it. As a blackbody warms up, the total emitted power per unit area increases, and the peak of its emission spectrum moves to shorter and shorter wavelengths; at some point, you see a dim red glow, and if you warm it up further, you see it emitting a bright bluish-white light. Note that if a blackbody is hot enough to appear white or bluish-white, it’s also emitting a good deal of ultraviolet radiation, which means it’s harmful for your skin and eyes.

Note from that plot that for a blackbody at 5000K, the peak of the emission spectrum is pretty much in the center of the visible spectrum. This is nearly the temperature of the surface of the sun, and a 5500K blackbody has a spectrum that pretty closely matches the solar spectrum (before the atmosphere filters it). This brings up color temperature, which is an indication of the general color of a light source; A light bulb/LED with a color temperature of X appears (to your eye) to have a color similar to a blackbody at that temperature (though unlike a true blackbody, commercial available light bulbs don’t emit significant amounts of UV radiation). A light source with a color temperature of 5500K would be expected to appear the same color as the sun.

Note that you can measure the temperature of surfaces remotely using a pyrometer, which makes use of blackbody behavior. Cheap pyrometers simply measure the intensity of infrared light coming off of a surface at a particular wavelength; they have limited accuracy due to their built-in assumption of a particular degree of blackbody emissive behavior on the part of the target (as noted upthread, many materials and surface finishes deviate significantly from the ideal blackbody), but they’re awfully handy for measuring really hot things, things you can’t reach, and things you don’t want to touch, and they give you an instant result. If you’ve got a little more money, you can get a two-color pyrometer, which measures the relative intensities of IR light at two different emitted wavelengths, and calculates a temperature based on the ratio of those intensities.

For general heat transfer calculations (in a bulk power sense), the radiative flux of a hot surface scales with the fourth power of absolute temperature: an object emits X watts per square meter at room temperature (293K), and if you warm it up to 586K, it will emit 16X watts per square meter.

Steady-state temperatures and net heat fluxes for two interacting radiative surfaces can be calculated via the use of view factors, which help describe what fraction of an emitting surface’s power output is absorbed by a nearby absorbing surface; this varies depending on the shape, size, and distance of the two objects. If you flip to the back of any heat transfer textbook, you usually find an appendix containing instructions for how to calculate view factors for various configurations of interacting surfaces.

Well, I’m not quite sure about that.
Let’s try to look at some rough numbers. An average human body generates about 60Watts of heat at rest (i.e. an old-fashioned light bulb), call it a round 100W for a spacewalk. That’s 100 Joules/sec or 24 calories (small-c calories)/sec. One calorie is the heat necessary to raise 1 gram of water by 1 degree C. Let’s assume our astronaut is a svelte 50 kg. That means the 24 calories/sec of heat generated, if absolutely no heat was lost, would raise the astronaut’s body temperature by [24/50,000=] 0.0005 degrees (C) per second. It would take 2,000 seconds to raise their temperature by 1 degree C (1 degree C rise in body heat is mildly stressful, but tolerable for a while).
I’m pretty sure I’d suffocate in half an hour without air, well before heat build up would be permanently or even temporarily damaging.

Hamburger, fries and a bowl of chili for example. How long might this stay above 100 degrees if left outside the ship?

The problem is that all the volatile compounds in the food are going to boil out. These volatiles will have a higher average kinetic energy than the non-volatiles, and so the food gets freeze-dried very quickly.

So the meal is not going to stay warm for more than a few seconds. It would stay warm a lot longer if it was sealed in a container strong enough to keep in the volatiles.

So in this case you have a form of convection, since warmer particles are outgassed from the food leaving cold solids behind. And, as the cloud of gas created by the food expands, it cools. This is known as adiabatic cooling, and it’s the same reason a compressed air spray duster gets colder as you use it.

It would be a different story for a 100 F rock. That would stay warm for quite a while longer. But the rock would only cool if it were in the shade. If your rock is floating in space at the same distance from the sun as the Earth, it’s going to get hot, just like rocks baking in the sun in the desert get hot. If it’s in the shade, like in the shadow of the ship, it will get very cold as it radiates heat into deep space.

So think of it this way. The rock is absorbing energy from some things, and radiating energy to other things. If the radiant energy source and energy sink remain constant, eventually the body will reach equilibrium. That could be very hot for an object in direct sunlight, or very cold for an object constantly in shade.

I don’t think **Nickerbot **was even thinking about heat buildup due to vacuum insulation, but instead was labouring under the false impression that when things boil, it means they are hot.
That’s incorrect - water will boil at room temperature once the pressure is sufficiently low - in fact, it’s doubly wrong, because once it starts boiling, it will also get colder due to latent heat of evaporation.

When you say “compressed-air spray duster,” if you’re truly talking about compressed air, then yes - expansion of compressed air results in adiabatic cooling. This is the same reason tire valve stems get cold when letting air out of your tires.

OTOH, if you’re talking about “canned air” that you buy at office supply stores for blowing potato chip crumbs out of your keyboard, then these cool via evaporative cooling: when you let gas out, you reduce pressure in the can, and some of the liquid boils, restoring gas pressure in the can and lowering the temperature of the liquid.

You’re right–if the pressurized whatever in the can is a liquid, you’ve also got the heat of vaporization. It’s only adiabatic cooling if the fluid is a gas through the whole process.