# Temperature of thermally isolated space region

If we thermally isolate a region in space, say using a hypothetical material of $0$ conductivity, and measure the region's temperature, will it be 2.7K?

• You need to get a better definition: the temperature of "deep space" is generally related to the wavelengths being emitted by the material within that space. If the material is at X Kelvins, it'll stay there if your "0 conductivity" box is also 100% reflective over the entire EM spectrum. – Carl Witthoft Dec 24 '13 at 16:16

If you isolate the region by surrounding it with a $\kappa=0$ material (using $\kappa$ for thermal conductivity), then no heat can enter or escape that region and so it would remain at whatever temperature it happened to be when you enclosed it.

• So the temperature would be ? – karthikeyan Dec 25 '13 at 7:20
• It would be whatever temperature it started at. If you took 400 K oven and enclosed it, it would remain 400 K for as long as it was enclosed. – Kyle Kanos Dec 25 '13 at 15:15

I think it will be 0K

Because if said hypothetical material conductivity of 0, it cannot conduct any heat, therefore the temperature must remain at 0K

• Can anyone give me a reason as to why this has been downvoted??? – Joe Hilton Dec 24 '13 at 16:21
• Because it's wrong? No heat can enter or escape the region, so whatever $T$ it started at is what it will be later on. – Kyle Kanos Dec 24 '13 at 16:49
• Ahh right okay. Clearly you know more of the subject than I but thanks for clearing it up :-) – Joe Hilton Dec 24 '13 at 17:37
• @kyle kanos-but how come space have temperature? To my understanding, the temperature of space is attributed to cosmic background radiation. Once it is cut off, then there is no meaning of temperature of space..am i right? – karthikeyan Dec 25 '13 at 7:23
• What do you mean 'once it is cut off '? – Kyle Kanos Dec 25 '13 at 15:17

It would be the temperature corresponding to the current mass-energy density of the universe, which is $1.003^{+0.013}_{-0.017}ρ_c\approxρ_c = \frac{3H_0^2}{8πG_N} = 1.053 69(16) × 10^{−5} h^2 (\text{GeV/c}^2 ) \text{cm}^{−3}$, where $\rho_c$ is the universe's critical mass-energy density (the amount of density required to make stop the universe from expanding indefinitely), $H_0=100 h \text{ km s}^{−1}\text{ Mpc}^{−1}$ is the present Hubble expansion rate, where $h=0.73^{+0.04}_{-0.03}$, and $G_N$ is Newton's gravitational constant. (source: PDG's astronomical parameters table)

• I'm not the downvoter, but i can suggest that the "global" parameters of the universe are not the right things to apply to a local volume. – Carl Witthoft Dec 24 '13 at 17:35