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.