Space is a very low temperature environment, however it also has an extremely small number of particles per unit volume. This leads me to believe that, contrary to popular portrayals of heat loss in space, there would be very little heat loss due to conduction on a hypothetical spaceship in deep space. If you were put into space and provided with food, water, heat, and protected from the pressure would the spaceship cool due to head radiating away from the ship?
[Physics] Heat Transfer From a Spaceship in Deep Space
spacetemperaturethermal-radiation
Best Answer
Let's start by assuming you're in the shade, so you're not receiving any radiation (apart from the cosmic microwave background, which I think we can ignore). The amount of heat per unit area that you radiate is given by Stefan's law:
$$ J = \varepsilon \sigma T^4 \tag{1} $$
The emissivity of human skin is allegedly 0.98, and the area of skin of an adult male is around 2m$^2$, so feeding in $T = $37ºC gives us a total radiated power of about a kilowatt. The power produced by an adult male is about 120W, so at body temperature you're going to lose about 880W.
To work out what temperature you would cool to we just take equation (1), feed in $J = $60W/m$^2$ and we get $T = $180K. This would be fatal.
What's interesting is to see what happens when you're in direct sunlight. At the orbit of the earth the radiation from the Sun is around 1.4kW/m$^2$. Since only half your skin would be illuminated, you would be losing a kW and gaining 1.4kW for a net gain of 400W. To work out your equilibrium temperature we just feed $J = $1400W/m$^2$ into equation (1) and we get $T = $396K, which would again be fatal.
The distance from the Sun where the heat you radiate would exactly balance the heat you receive can be worked out using the inverse square law. If $r_E$ is the radius of Earth's orbit and $r$ is the thermal balance radius we get:
$$ \frac{r^2}{r_E^2} = \frac{1400}{1000} $$
or:
$$ r = 1.18r_E $$