In a vacuum, the only methods of heat transfer are conduction and radiation. Since the crew cabin was fairly well isolated from the lunar surface, I think we can ignore conduction and focus solely on radiation.

[At this point I have to rely on external sources - I’m a code monkey, not an engineer or physicist]

This page shows how to compute the rate radiative heat loss using the Stefan-Boltzmann law. With some rearranging and integrating, you can compute the rough amount of time it takes for an object to cool down from a high temperature to a lower one.

Keeping things “simple”, a solid block of unpolished aluminum (emissivity 0.09) just 10 cm on a side will take around 8 *hours* to go from 300 K (about 80 deg F) to 255 K (just under 0 deg F). That’s...not that cold. To cool down to 100 K (roughly -280 deg F) would take around 13 days. And that’s assuming there’s nothing heating that aluminum block (internal electronics or the Sun).

IOW, this ain’t Hollywood. Things and people don’t immediately freeze upon exposure to space. It takes nontrivial amounts of time for objects to lose heat strictly through radiation.

Now, the crew cabin isn’t a solid block of commercial sheet aluminum 10 cm on a side - there are a bunch of different materials with different emissivities, densities, etc., and most of it’s built in thin sheets. It’s also full of electronics and heaters to maintain stable temperatures, and it’s standing in the Sun the whole time.

So evacuating the cabin for a few hours is simply not a big deal from a thermal management perspective.

I realized it would help to show the work (I typed the above from an iPad without a keyboard, which got ... frustrating when I tried to add the math):

We start with the Stefan-Boltzmann law as described

here:

P = dE/dt = emissivity * sigma * surface area * ( T

^{4}_{hot} - T

^{4}_{ambient} )

where

*P* is the total power emitted in Watts

*E* is the energy of the system in Joules (power is energy change over time)

*emissivity* is the effectiveness of emitting radiation (between 0 and 1, where 1 is an ideal emitter)

*sigma* is the Stefan-Boltzman constant 5.670374419 x 10

^{-8} W m

^{-2} K

^{-4} *surface area* is in m

^{2} *T* is the temperature of the object in Kelvin

With some rearranging and integrating (see the page linked above for details), we get

t = (Nk / (2 * emissivity * sigma * surface area) ) * ( T

^{-3}_{ambient} - T

^{-3}_{hot} )

where

*t* is cooling time in seconds

*N* is the number of particles in the system (atoms, molecules, etc.)

*k* is the Boltzmann constant 1.380649 x 10

^{-23} J/K

For my example, I computed the number of atoms in a 10 cm

^{3} block of aluminum by first computing it's mass (assuming density of 2.7 g/cm^3, that's 2700 g or 2.7 kg), multiplied that by Avogadro's number, then divided by the molar mass of aluminum (29.98 g/mol), giving me an N of 6.03 x 10

^{25}. I picked an emissivity of 0.09 for unpolished aluminum from

here. Plugging that into the above formula gave me

(6.03E25 * 1.380649E-23)/(2 * 0.09 * 5.670374419E-8 * 0.06)(255

^{-3} - 300

^{-3})

which gave me around 3.16 x 10

^{4} seconds, or 8.78 hours. Note that this is the minimum,

*ideal* cooling time - it will be longer in the real world.

This is consistent with the calculator on that page which uses a sphere, rather than a block, and the numbers are close

*enough* for me to be confident I'm doing it right.

So you can play around with this - pick different materials, different geometries, etc., and compute how long it would take for those items to radiate away all their heat.

And remember, a

*vacuum* isn't cold

*or* hot - the concept of temperature doesn't really apply to a vacuum. Things will get cold in space if they radiate away all their heat and there's nothing to warm them up again. But the mere absence of an atmosphere

*by itself* doesn't make things cold (otherwise Thermos bottles wouldn't work very well).