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We know that the temperature in space (which has vacuum) is low. If I go to space will I feel sweaty and hot or chilly? I think I will feel sweaty and hot because the radiation (UV, IR, etc) of the sun is hitting me directly. This question is just theoretical so assume I go to vacuum without a space suit.

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    $\begingroup$ please anyone tell me why are the people giving me a downvote please :( Every question today i asked is getting a downvote within seconds. :( $\endgroup$
    – Bhavesh
    Apr 17 '15 at 14:55
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    $\begingroup$ possible duplicate of How can interstellar space have a temperature of 2-3K? $\endgroup$ Apr 17 '15 at 14:55
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    $\begingroup$ Not my downvote, but you are asking questions that could be answered with a bit of Googling or in this case searching this site. This site is intended for working physicists, and we expect members to have done some prior research before posting a question. $\endgroup$ Apr 17 '15 at 14:57
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    $\begingroup$ Well, you certainly won't feel sweaty because any sweat or surface liquids will evaporate almost immediately. You'd probably also die really quick, so you probably wouldn't have much time to notice feeling hot or cold. But space is hot. Satellites are usually faced with the problem of being too hot and needing to find a way to keep cool. $\endgroup$
    – Jim
    Apr 17 '15 at 15:16
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    $\begingroup$ @ACuriousJim Space isn't hot, or cold, or anything. Things that fall in vacuum long enough will reach an equilibrium with the radiation environment--they eventually find some average temperature where they radiate (black body) as much energy as they absorb. For a spacecraft orbiting the Earth, that temperature can be pretty high because of $1400 W/m^2$ of Solar radiation. The equilibrium temperature of some hypothetical rock hanging out where the only incoming radiation is the CMB and the faint light of distant galaxies would be somewhat cooler. $\endgroup$ Apr 17 '15 at 17:33
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If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only).

For this, we need the Stefan-Boltzmann expression for total emission at a given temperature:

$$E = \epsilon \sigma T^4$$

Where $\epsilon$ is the emissivity (which we will set to 1 - it doesn't affect the answer) and $\sigma$ is the Stefan-Boltzmann constant, $5.67\cdot 10^{-8} W/m^2/K^4$.

When we want to know how much heat goes from one body (the sun) to another (the sphere), we actually need to take into account both emissivities -the sun's (how much power is it emitting) and the sphere's (how much of that power is it absorbing). For the emission, we only need to know the emissivity of the sphere. If we set the emissivity of the sun = 1, then we can ignore the emissivity of the sphere completely (if we can assume it is constant with temperature - that is not the case and will change the real result).

Your sphere "sees" the sun (5778K) over a small solid angle $\Omega$, and the cold universe (2.3K) over the rest. We need to determine the temperature where equilibrium occurs. This will happen when

$$\epsilon_{sun}\epsilon_{sphere}\sigma T_{sun}^4 \Omega=\epsilon_{sphere}\sigma T_{sphere}^4 4\pi$$

This simplifies to

$$T_{sphere} = T_{sun}\sqrt[4]{\frac{\epsilon_{sun}\Omega}{4\pi}}$$

From earth, the sun's diameter is about half a degree across, so $\Omega$ is about $6.8\cdot10^{-5}$ steradians (see derivation here)

Plugging in the numbers, we obtain the temperature of the sphere as

$$T_{sphere}=279 K$$

As I said - pretty cold.

On Earth we are warmer because of the effect of the atmosphere and the heat from the Earth itself. The moon, lacking the benefits of the Earth, is a lot colder on average. Note that the moon's actual mean temperature is lower because it rotates slowly - and the side that is hotter will lose heat much more quickly. The above calculation is therefore an upper limit - and in reality you will be colder than that.

I ignored the fact that the cosmic background is 2.3 K and not 0 - it makes no difference to the answer.

Note also that if you used actual values for emissivity of human skin at different wavelengths, the answer will change. This was nicely explored by ACuriousJim in his comment, which I reproduce here because it is an important point (and comments can disappear over time):

I went ahead and looked up the absorptivity values of human skin in the visible spectrum and plugged those into calculations. For white people, you'd find an equilibrium of 235K and 259K for black people. Other races are in between that range. $α_W$=0.5, $α_B$=0.74, $ϵ_{all}$=0.99.

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    $\begingroup$ @ACuriousJim - If I add $\epsilon$ on both sides on my equation (heat in = heat out) it just cancels. Obviously if it's a function of wavelength it matters - that is the premise of the greenhouse effect which I am ignoring. Note that for "heat in" you really need the emissivity product of sun and sphere. As you say, we put sun = 1 $\endgroup$
    – Floris
    Apr 17 '15 at 15:21
  • $\begingroup$ @ACuriousJim - I don't doubt your expertise but that does not agree with what I thought I knew. Can you please expand - maybe edit my answer to make it more factually correct? Note that I do make the assumption that the surface temperature is constant - when it is not, I suspect the variability between hot and cold side will be greater, and so the actual temperature will be a function of emissivity. I just don't see how that is the case for the assumption of uniform temperature (relatively rapid motion). $\endgroup$
    – Floris
    Apr 17 '15 at 15:29
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    $\begingroup$ Ah, I didn't make the assumption of constant surface temperature. In that case, if absorptivity and emissivity are equal, then you are right. (also, it emits at the full $4\pi$ solid angle, not $4\pi-\Omega$). However, I think what is also forgotten is that the absorptivity is at visible wavelengths but emissivity is in IR. Those two are not equal for human analogs $\endgroup$
    – Jim
    Apr 17 '15 at 15:36
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    $\begingroup$ Depending on clothing and skin colour, absorptivity can range from low to high, but the emissivity of human skin in IR is always around 0.99 (it's high because we generate more heat than we absorb, so we have to radiate more away) $\endgroup$
    – Jim
    Apr 17 '15 at 15:39
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    $\begingroup$ Yeah, it reads better. I went ahead and looked up the absorptivity values of human skin in the visible spectrum and plugged those into calculations. For white people, you'd find an equilibrium of 235K and 259K for black people. Other races are in between that range. $\alpha_W=0.5$, $\alpha_B=0.74$, $\epsilon_{all}=0.99$. If you want to keep it assuming equal absorption and emission, that's fine. Just let people know that's what you're doing. I'll give my +1 on it though $\endgroup$
    – Jim
    Apr 17 '15 at 15:50

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