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?

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    $\begingroup$ This question appears to be off-topic because it is about problems faced in space exploration. Perhaps Space.SE might be better suited for this question? $\endgroup$
    – Kyle Kanos
    Commented Jan 15, 2015 at 15:29
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    $\begingroup$ Its a physics question. Are there not Physics in space? $\endgroup$
    – Jonathon
    Commented Jan 15, 2015 at 15:30
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    $\begingroup$ Yep, I just looked it up there is in fact physics in space, and physics.SE in face does not include any rule about space based physics being disallowed. $\endgroup$
    – Jonathon
    Commented Jan 15, 2015 at 15:32
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    $\begingroup$ Surely you could argue everything is physics, but we don't need to explicitly make "topic x" off-topic by rule, we can simply use knowledge-based decisions. Your question is about problems of containing/expelling heat while traveling in space; in my opinion, this is better suited at the Space Exploration site I linked than here. $\endgroup$
    – Kyle Kanos
    Commented Jan 15, 2015 at 15:51
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    $\begingroup$ In deep space, keeping warm is an issue. In Earth orbit or around a star, keeping cool is an issue. But in general, thermal control is not as big an issue as popular media makes it. It's usually a matter of how to paint the spacecraft $\endgroup$
    – Jim
    Commented Jan 15, 2015 at 16:14

2 Answers 2


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:

$$ \frac{r^2}{r_E^2} = \frac{1400}{1000} $$


$$ r = 1.18r_E $$

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    $\begingroup$ The edit to the question has slightly sabotaged my answer. The original version asks Or would body heat addition balance with the radiation at some survivable temperature?, which is why I've calculated the heat balance for a human body. The edit changes the question to be about heat balance for a spaceship. $\endgroup$ Commented Jan 15, 2015 at 17:26
  • $\begingroup$ My bad. That was the most extensive edit I've ever done on a question, because I wanted to strip away all of the Star Trek stuff and stick to the physics, otherwise the OP was likely going to continue to get downvotes. I think your answer is still great though, given the spirit of the question. +1 $\endgroup$
    – Sean
    Commented Jan 15, 2015 at 18:05
  • $\begingroup$ I'm a bit confused by this answer... maybe it is because the question was changed? The net power radiated away by an adult male is 120 W given an ambient temperature of 290 K (room temp). The net power radiated away by an adult male is 1 kW given an ambient temperature of 0 K. Where does the 880 K come from? Additionally, if you are equilibrating with space (3 degrees Kelvin?) your final temperature will be 3 K, not 180 K. $\endgroup$
    – pentane
    Commented Jan 15, 2015 at 19:40
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    $\begingroup$ @pentane The 880 K difference mean that your 290 K environment radiates back 880 W at you. - The equilibrum temperature of 180 K is based on the assumption that the bodily exothermal processes are still active and emitting 120 W. Of course, the fatality of cooling down even to 180 K will make these processes stop and then pave your way to 3 K. $\endgroup$ Commented Jan 15, 2015 at 22:25
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    $\begingroup$ A slight criticism is that 37 is the core body temperature, torso skin temperature varies between 33.5 and 36.9 depending on environmental conditions. Extremities are colder. Its not going to make a significant difference though. $\endgroup$
    – Dale M
    Commented Jan 15, 2015 at 22:55

If you are out of the sunlight, the main source of cooling will be radiation. The amount of net heat you radiate away depends on the temperature of your skin $T_{skin}$ and the ambient temperature of the surrounding environment $T_{ambient}$:

$\frac{Q}{t} = e \sigma A (T^{4}_{skin}-T^{4}_{ambient})$

$Q$ = heat loss in Joules
$t$ = time in seconds
$e$ = emissivity of skin ($\approx$ 0.98 for human body)
$\sigma$ = Stefan-Boltzmann constant
$A$ = surface area of human body

Plug $T_{skin}$ and $T_{ambient}$ into this calculator and it will calculate the heat loss in Watts using the equation above. For human body temperature ($T_{skin}$ = 34 °C = 307 K) and room temperatue ($T_{ambient}$ = 23 °C = 296 K), the heat loss is 133 Watts. For human body temperature and outer space ($T_{ambient}$ = -270 °C = 3 K), the heat loss is nearly 1,000 Watts.

Assume a human is 70 kg of water. Since water's heat capacity is 4.18 J/gK, if the person emits 1,000 Watts for ten minutes, that is only enough to decrease its temperature by 2 K. To balance the heat radiation away from your body you would need to consume food at 1,000 Watts = 14 food Calories per minute.


One of these 50 Calorie cookies every 2-3 minutes would do the trick.

In other words, you are right that space is not as "cold" as it is made out to be. Yes the temperature is low-- but there is not enough matter out there to cool down anything quickly. A Thermos is able to keep coffee hot for long periods of time by taking advantage of exactly this phenomenon.

  • $\begingroup$ Great, thanks. I have no idea how heat radiation work, but if it cares about ambient temperature should it not also care about the vacuum? Temperature is sort of irrelevant in a vacuum, right? There is no such think as temperature without mass. Space might not be a perfect vacuum, but it is pretty close. Meaning its temperature is undefined. Right? Or does it mean ambient radiation, or something? $\endgroup$
    – Jonathon
    Commented Jan 16, 2015 at 1:17
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    $\begingroup$ +1 Great answer, really gets the physics across. I'm pretty sure thought that in practice the human body cannot metabolise the cookies swiftly enough to make $1kW$, even if you were eating them swiftly enough. I think you'd just feel very sick and ultimately lethally cold after ten minutes or so. But your answer illustrates very well that things just don't "snap freeze" in space. $\endgroup$ Commented Jan 16, 2015 at 2:08
  • $\begingroup$ And does this mean that given a jacket, that reduced the emissivity, you would be perfectly fine? It also looks like people can produce over 1KW of heat, at least for a few hours at a time. So that would lead me to believe that space is sort of equivalent to something on the order of -10to-30C on earth. Pretty cold. You will die eventually without protection, but it is not some ridiculous coldness that a jacket would not cover. $\endgroup$
    – Jonathon
    Commented Jan 16, 2015 at 2:27
  • $\begingroup$ @WetSavannaAnimalakaRodVance But that is for a fat free cake. Assuming we using some sort of Eskimo food and were extremely fit, I think it might actually be within the realms of possibility (sleeping would likely be a problem). And this is without any insulating cloths. $\endgroup$
    – Jonathon
    Commented Jan 16, 2015 at 2:32
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    $\begingroup$ @JonathonWisnoski yes, the "ambient temperature" in that equation is also radiative. For "deep space" you could use the 3 Kelvin or so of the cosmic background radiation. If you're in the vicinity of any stars or other big radiators, you would take that into account and the ambient temperature would be higher. $\endgroup$
    – hobbs
    Commented Jan 16, 2015 at 5:12

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