so this article is about temperature regulation for a spacesuit on mars. we assume you want to walk around on the outside, doing light work like checking the terrain and maybe also heavy work like constructing houses and such.

there's 4 major ways to lose heat:

• radiation
• convection (through the wind)
• evaporation (water loss)
• conduction (through the ground)

the share of each way to heat loss on mars is as follows: assuming -60°C external temperature, almost all of the heat loss is through radiation; there's a very thin atmosphere (therefore almost no convection losses), we assume no evaporation losses (the suit is airtight), very small conduction losses (small feet area).

so sorry for posting chatgpt output here, i checked the text however and it's true, but it's neatly summed up and i'm too lazy to rephrase.

sources for the details:

• Stefan-Boltzmann-law
• Emissivity

emissivity of materials can vary from 0.03 to 0.99 so you can basically choose your emissivity. depending on the emissivity, the heat losses will be different. there's an almost linear connection there.

so for a typical white spacesuit, the losses will be around 600 W of heat on average; which can be reduced by an order of magnitude if you use tin foil as the surface material instead.

meanwhile the average human metabolic output is around 250 W throughout the day (assuming light activity like walking around), and it can peak at around 500 W during heavy work activity.

so you could choose a proper material to have around 400 W of heat loss on average, so you wouldn't overheat if you work hard but also not lose too much heat if you just walk around; in the latter case, you'd have to supplement the heat with a technical heat source, such as resistive heating. so the spacesuit would also have a battery attached to it as the power source; if you want the battery to last an hour before you have to change it for a new battery, it would have to have a capacity of around 100 Wh. if it's a Li-ion battery, it would have to weight around 400 g to provide that energy.