What would be the rate of temperature loss for an average sized human in space without a suit? A human generates about 100 watt at rest. But how can we use that to calculate how fast the temperature will go down? Also how much heat would be absorbed by the sun? Assume the person is somewhere between the earth and the moon.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/26332/2451, physics.stackexchange.com/q/3076/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 9, 2013 at 15:54
  • $\begingroup$ Thanks but, no, not really. One comment said the energy loss was 1kW if the body was in complete darkness. That still doesn't mean anything to me. How much loss of body temperature is that? $\endgroup$
    – John
    Jun 9, 2013 at 16:17
  • $\begingroup$ Without a suit is a meaningless description. The equilibrium temperature and heat loss depend very much on: reflectivity & emissivity of the clothes/skin (have no data), the human being in the shadow of the Earth or not, distance to the Earth (since flux from the Earth matters), whether the human being is still alive... Heat loss is due to radiation only, so Stefan-Boltzmann law applies: $P=A\epsilon\sigma T^4$ (you have to take into account incident radiation in the visible and IR bands to get net loss/gain). $\endgroup$ Jun 9, 2013 at 18:03
  • $\begingroup$ As stated, this is an unsolvable question. What is the surface-area/volume ratio of the human? What is their % body fat, and where is that fat located (e.g., fat is an insulator). Is the person naked or clothed? If clothed, what type of clothing, how much layering, how much trapped air in the layering, etc.? Is there any incident radiation on this person? If so, how much and what wavelength? If there is incident radiation on the person, is the person rotating, or is only one side of the person exposed to the radiation? The list of questions could go on and on! $\endgroup$ Feb 6, 2018 at 1:27

2 Answers 2


Here's a quick and dirty approximation:

Approximate a human with a water-sphere of radius $r$. Assume that blood circulation keeps the body temperature homogenous and that the skin keeps the water from boiling off (this isn't realistic but I figured this was the kind of situation you were thinking of). If we assume the sphere is a perfect blackbody, the heat loss due to thermal radiation is given by the Stefan-Boltzmann law:

$$ P_r=\sigma T^4\cdot A=\sigma T^4\cdot 4\pi r^2 $$

where $T$ is the temperature of the water and $A$ is the surface area of the sphere. The rate of temperature change will be:

$$ \frac{dT}{dt}=-\frac{P_r}{mC}=-\frac{P_r}{\rho VC}=-\frac{3P_r}{4\pi r^3\rho C}=-\frac{3\cdot\sigma T^4\cdot 4\pi r^2}{4\pi r^3\rho C}=-\frac{3\sigma T^4}{r\rho C} $$

where $m$ is the mass of the sphere, $C$ is the specific heat capacity of water,$\rho$ is the density of water and $V$ is the volume of the sphere. Hence, we get a differntial equation of the form:

$$ \frac{dT}{dt}=-\frac{3\sigma T^4}{r\rho C} $$

If we make the ansatz: $$ T(t)=a(t+b)^n $$

and put it into the above equation, we get:

$$ an(t+b)^{n-1}=-\frac{3\sigma a^4(t+b)^{4n}}{r\rho C} $$

which gives:

$$ n-1=4n\Rightarrow n=-\frac{1}{3} $$


$$ an=-\frac{3\sigma a^4}{r\rho C}\Rightarrow a=\left(-\frac{nr\rho C}{3\sigma}\right)^\frac{1}{3}=\left(\frac{r\rho C}{9\sigma}\right)^\frac{1}{3} $$

If we assume $r$ is 0.5 m, we get:

$$ a=\left(\frac{0.5\cdot 1000 \cdot 4200}{9\cdot 5.67\cdot 10^{-8}}\right)^\frac{1}{3}\approx 1.6\cdot 10^4 $$

Hence T will change as:

$$ T(t)=1.6\cdot 10^4(t+b)^{-\frac{1}{3}} $$

If we assume that the temperature at $t=0$ is $37$ degrees C, which is approximately $310 K$, you get:

$$ 310=1.6\cdot 10^4b^{-\frac{1}{3}}\Rightarrow b=\left(\frac{1.6\cdot 10^4}{310}\right)^3\approx 1.38\cdot 10^5 $$

which gives:

$$ T(t)=1.6\cdot 10^4(t+1.38\cdot 10^5)^{-\frac{1}{3}} $$

Here I have plotted the time evolution of the temperature according to this formula:

As you see, it takes approximately 65000 s (almost 18 hours) for the sphere to freeze. This is obviously something of a worst case scenario. I have not included the effect of heat produced by the human body, incoming (solar) radiation, less than perfect emissitivty, possible insulation, etc. These could make it take much longer or even prevent cooling, depending on the assumptions.

  • $\begingroup$ Jkej,a nice analysis assuming no incoming radiation from Earth and Sun, which would increase your freezing time estimate. $\endgroup$ Jun 9, 2013 at 19:46
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    $\begingroup$ However the assumption that the skin withholds boiling will not hold in reality without a spacesuit. This will lower the temperature a lot in the beginning. And you will for sure die and therefore the bloodflow will stop, allowing the core to maintain a relative high temperature for quite a long time. Vsause also discussed this in one of his videos. $\endgroup$
    – fibonatic
    Jun 9, 2013 at 20:21
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    $\begingroup$ I like the spherical astronauts. ;-) $\endgroup$
    – Alexander
    Jun 9, 2013 at 21:58
  • $\begingroup$ It might be worth noting that humans die of hypothermia by the time they reach 300 K, no need to freeze solid. $\endgroup$
    – LeBleu
    Oct 5, 2016 at 20:04
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    $\begingroup$ "Approximate a human with a water-sphere of radius r" + in a vacuum. The perfect physics statement ! $\endgroup$ Feb 5, 2018 at 20:34

The answer by jkej is good. I will add onto the discussion by using numerical software. Here is syntax in Maple that solves the equation already discussed.

dsolve({diff(T(t),t)=-alpha*T(t)^4,T(0)=T0}) assuming alpha>0, T0>0;

The answer is:

$$ T(t) = \left( 3 \alpha t + T_0^3 \right)^{-\frac{1}{3}}$$

This is consistent with the other answer. For reference, $\alpha=1/(3 a^3)$. Since we have a value for $a$ we can get alpha trivially. Now let's try it with heat production! This is solving the following differential equation.

$$ \frac{d T}{dt} = - \alpha T(t)^4 + \beta $$

As the time limits to infinity, we expect the function to level out at $(\beta/\alpha)^{1/4}$. If you assume that the heat production is 100 Watts, then:

$$ \beta = \frac{P_p}{m C} = \frac{ 100 W}{( 1 \frac{g}{cm^3}) \frac{4}{3} \pi (0.5 m)^3 (4,186 \frac{J}{kg K} ) } = 4.56 \times 10^{-5} \frac{K}{s} $$

$$ T_{\infty} = \left( \frac{\beta}{\alpha} \right)^{1/4} = \left( 3 \beta a^{3} \right)^{\frac{1}{4}} = 153.8 \text{ Kelvin} $$

The syntax to solve the differential equation is:


Getting the answer to a coherent format takes some algebra wrestling. For simplification, I collected terms to insert the final temperature calculated before. The "solution" to the differential equation is then:

$$ 0 = -4 t \beta+ T_{\infty} \ln{ \left( \frac{T(t)+T_{\infty}}{T(t)-T_{\infty}} \frac{T_0-T_{\infty}}{T_0+T_{\infty}} \right) } + 2 T_{\infty} \left( \arctan{ \frac{T(t)}{T_{\infty}} } - \arctan{ \frac{T_0}{T_{\infty}} } \right) $$

This has to be solved at every time step. Here's what I produced:

cooling with power production

The time frame is still about the same. Nothing really interesting to speak of. It should be noted that the astronaut's mass comes out to like $520 kg$, which is fairly unrealistic. That's a major factor in the time frames being as long as they are.

In the above graph the astronaut freezes in about 11 hours. While there is some correction needed for the radiative area exposed to space, the linear dimension of the astronaut exceeds a meter in the vertical, so there are corrections in both directions. While the effective radiative area should still probably be revised downward, it won't compare to the downward revision to the mass. So in practice you will freeze sooner. Maybe you'll freeze in 4 hours or so. You'll be comatose at 31 degrees C, which linearly interpolated, only takes about 13% of the time, or 39 minutes. Hypothermia should only take 12-13 minutes. Loss of hope, probably sooner.

This made me think - if you're stranded floating in space, curl in the fetal position so you won't radiate heat as fast.

I realized the above graph is technically wrong. After it crosses the freezing point the heat production shuts down. Because you'd be dead.

  • $\begingroup$ Nice work! Yes, I should probably have made a reality check on the mass before choosing the radius (or better yet, chosen another geometry). $\endgroup$
    – jkej
    Jun 9, 2013 at 22:30

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