If we go to space why isn't the temperature high?

Question

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.

Answer 1

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.