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I have an objected suspended in a vacuum by a very thin point. Thus, it can be heated only at the point of contact? Am I correct in assuming that the object cannot dissipate heat into the vacuum- therefore the vacuum doesn't necessarily "cool" the object? A vacuum does not have a temperature because there are no air molecules to be moving, am I thinking about this situation in the correct way?

If I put a thermometer in the vacuum, what would it read (not talking about CMB)?

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    $\begingroup$ Would you say the sun is able to transfer heat to the earth through a vacuum? $\endgroup$ – BowlOfRed Jun 20 '19 at 23:17
  • $\begingroup$ ah good point, thanks. I didn't consider EM radiation $\endgroup$ – user203234 Jun 20 '19 at 23:20
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    $\begingroup$ Conduction, convection, and radiation... $\endgroup$ – Jon Custer Jun 20 '19 at 23:34
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With regard to thermodynamic heat transfer (i.e. $Q$), energy can be transmitted through a vacuum, but in only one way: radiation - the creation of particles which then get sent out from the object in question, which under all ordinary conditions means effectively just photons, i.e. electromagnetic radiation. The other two mechanisms for heat transfer - conduction and convection - require contact with some sort of preexisting outside matter and effectively are what you're talking about in terms of setting molecules into motion (namely, the molecules of the hot object collide with and transfer momentum and energy to that medium). In that case, yes, by those two mechanisms, heat will only transfer out that tiny point of contact.

However, radiation always occurs, and is omnidirectional, though not all materials are equally good at radiating. The amount of power sent out by radiation depends on both the temperature and the material, in particular, the equation is

$$I(\epsilon, T) := \epsilon \cdot \sigma T^4$$

where $\epsilon$ is the overall emissivity of the material, a number from $0$ to $1$, indicating how good it is at emitting ($1$ and $0$, respectively a perfect emitter and a material that does not radiate at all, are ideal values that do not exist at all in the real world). $T$ is the temperature, which must be on an absolute scale, i.e. zero is zero, e.g. Kelvin. $\sigma$ is what is called as the Stefan-Boltzmann constant, about $5.67 \times 10^{-8} \mathrm{\frac{W/m^2}{K^4}}$. Due to the fourth power, thus, radiation rapidly increases with temperature - in fact, even when other mechanisms are present, at high enough temperatures radiation will eclipse them all. A doubling of temperature multiplies radiation by 16x, and a 10-fold increase in temperature is a 10,000-fold increase in radiation power.

Likewise, just as energy goes out by radiation, energy can also be absorbed by an object through radiation, and in a vacuum environment, the final temperature of an object will be determined by the balance between outgoing and incoming radiation: when the two are equal, that is the temperature it will remain at if nothing changes. In the case of outer space, such incoming radiation may be provided by a star, like the Sun, which is how the Earth can be warm at all despite being surrounded by the vacuum of space. In the case of a vacuum chamber as you are suggesting, the chief source of incoming radiation will be the chamber walls.

This also is the answer to your question about the thermometer. The thing to remember about thermometers is they don't measure the temperature of other objects (unless it's, say, an infrared eye thermometer, but given the question you're asking I suspect this is not the kind you have in mind) directly, but instead measure their own temperature. They can only measure the temperature of another object by absorbing energy from it to equal its temperature: at that point, the temperature of the thermometer equals the temperature of the object. Of course, there is also an observer effect here in that the thermometer has to suck some heat out of the object we are taking the temperature of (or deposit some heat if the object is cold), so the imposition of the instrument changes what is being measured, hence the reading it gives must not be the original temperature. But in principle, we can reduce this by making the thermometer have as little thermal mass as possible and most thermometers of this type are not accurate enough to notice.

Hence, when a thermometer, which is a material object just same, is put into a vacuum chamber, what it will read after a suitably long time is then its equilibrium temperature, set by the balance of incoming and outgoing radiation (and perhaps also any contribution from a point of contact with the chamber walls or some other object, which can be minimized through the use of suitable insulating material). If the incoming radiation is all coming from the chamber walls, effectively the thermometer will come to tell you the temperature of those chamber walls. In outer space, it will be chiefly determined by whether or not there is a star or other large thermal source present: if the thermometer is orbiting the Sun at the distance of Earth (150 Gm), it will actually register quite high (about 400 K - the reason Earth doesn't get this hot is its rotation: at night, it dumps heat back into space). If the thermometer is in inter-galactic space, then it will, indeed, ultimately register the CMB temperature, about 2.7 K.

(Note though that this assumes ideal emitters and absorbers: i.e. $\epsilon = 1$ for both thermometer and walls. If that is not the case, then the reading can/will be different, but the same principle will apply, that it is the walls which set the temperature for any given instrumental setup.)

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