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How cold is space?
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07-01-2011 12:49pm#1I don't want to hear "three kelvin" as an answer or "2000K in direct sunlight", or anything like that. A simple temperature compares little with the human body's subjective idea of how warm or cold it is. I'm sure you can be perfectly comfortable in 20000C if the air is rarefied enough that it doesn't actually hold much energy per square meter to conduct to you as heat.
For example, an oven can be at 200C, but you can still open the door and stick your hand in it. Putting your hand on an iron bar at 200C is a completely different story.
So, you're floating in the void, such that you can receive no heating from conduction or convention from being in contact with a gas or other media. This means your only means of heat exchange with anything else in the universe is radiation. Furthermore, you're in "the shade", such that the only radiative heating your getting is from distant starlight sources, which we shall assume is negligible.
Specifically, what I want to know is: How cold do you feel? What temperature of still air at standard atmospheric pressure would you have to be immersed in on earth to incur the same rate of heat loss. Would it be like sitting in a room at 0C? 10C? 20C?
The equation that needs to be solved is simple I'm sure. You just need to equate a body's rate of radiative heat loss with the rate that air at standard pressure will conduct heat to a human body at various temperatures. But I don't know how to estimate these quantities. Does anybody have any ideas?1
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Found this:Rate of Heat Loss =A*e*s*Tb^4
A = surface area assume unity
e = emissivity assume unity
Tb = body temp = 310K (Space temp = 0K)
s = S-B const = about 6x10^-8
So P=about 550J/sec
Human body produces about 100J/sec (food metabolism)
Net heat loss = about 450J/sec (P decreases with Tb..but this is negligible for small DT)
Assume body mass = 70Kg
Energy loss 37 DegC to 32 DegC (normally coma and death)
E=4200*70*5=1470Kj (assuming tissue=water)
Time = 1470Kj/450 = just under 1 hour. I thought it would be a lot less..in water @ 10deg C it's about the same time, only you've got convection/conduction going on..
Not my own work or anything, just something I found on the interwebz!
Edit: I see that they've actually put the temperature of space as 0K which is clearly wrong. But not knowing much about physics, I can't tell if the formula is sound or not. I'll leave that to more learned people than myself.0 -
Not trying to be combative, but I really do not understand your "question."
You open up by telling us what you don't want. How about condensing the whole lot into a simple question.
Also, I am not trying to condescend, however, you appear to be confusing heat (energy) and temperature (a relative measure of hot hold or cold something is).
In the thermosphere, temperature increases with increasing altitude. Even at 80C (115km ASL) you would most definitely freeze to death. There are high speed particles, just not enough to warm you.
I have no doubt that if you were on Mercury in the shade you would freeze to death. In the sunlight you would burn to death in 10 seconds.
Be a bit more specific and we may help out.0 -
A crude estimation would be to use the Stefan-Boltzmann Law to find the amount of heat you will radiate. Then after taking away the energy from the light that is shining on you, you'll get the net power. This looks like what Improbable posted.0
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That's a good Last Word type question (the New Scientist forum for "everyday" science questions).
One answer, to a different though related question, suggests -150 deg C.0 -
How about condensing the whole lot into a simple question.
I do just that, I very explicitly state the question in no uncertain terms in my 4th paragraph.You open up by telling us what you don't want.
Experience with online forums and the types of answers people who dont really understand the question tend to give is what motivated my choice to do this.
I am genuinely shocked that you think I'm not being specific enough as I put a lot of thought into writing that question to ensure I was as clear as possible. Maybe you didn't read the question fully? I refer you back to paragraph four.What temperature of still air at standard atmospheric pressure would you have to be immersed in on earth to incur the same rate of heat loss. Would it be like sitting in a room at 0C? 10C? 20C?In the thermosphere, temperature increases with increasing altitude. Even at 80C (115km ASL) you would most definitely freeze to death. There are high speed particles, just not enough to warm you.
Sure, but if you neglect the net heat loss through radiation, then (provided your body can't manage to keep itself cool through sweating and evaporation) you will eventually warm to 80C (or more, since your body is producing its own heat too).
It's this net heat loss through radiation that I am trying to quantify in a meaningful way by expressing it in terms that we relate to in everday experience, such as an indoor air temperature.
Just giving a temperature doesn't tell the whole story as you yourself have attempted to illustrate with your 115km ASL example. To give another example, the outside air temperature may be 2C (above freezing), but if the sky is clear, the ground can still cool to below freezing and produce frost due to a net radiative heat loss into space.
Have you never noticed that if the sky is clear, it can FEEL like its colder than the reported air temperature.Be a bit more specific and we may help out.
I don't think I can be any more specific. But thanks for your take on it, I'm kind of disappointed that I wasn't able to present my question clear enough.
In any case, I believe Improbable (or more correctly, Improbable's source) did a very good job of addressing the question, although not with the strategy that I would have used.
He showed that by crudely estimating the surface area of a person to be 1m square and to radiate like a black body, the radiative heat losses far outweigh the rate at which a human body generates heat. He also puts a figure on the length of time it would take for such a rate of heat loss to reduce the temperature of a human body to a fatal level. The figure he gets is an hour.
By comparison with known average survival times of humans in cold water, he was able to estimate that it would FEEL like being in water at 10 degrees celcius.
To be honest I am quite surprised by this answer. It's a lot colder than I expected. I guess I have been underestimating the magnitude of radiative heat loss.0 -
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On a related topic. Here's a very interesting article I read a long time ago that dealt with the issue of net radiative heat loss. It's an article on how to keep your telescope lenses from fogging up on cold clear nights, by protecting them from exposure to what it called the "cosmic chill". I do like that term, cosmic chill.
http://www.skyandtelescope.com/howto/visualobserving/3304226.html?page=1&c=y0 -
Having read through my post again, I guess I neglect to explicitly state the postulate, that I assume, that a person's subjective measure of the hotness/coldness of an environment is an increasing function of the rate of heat gained/lost to that environment.
This proposal is interesting in itself and worthy of debate, but it is not relevant for answering the question "What temperature of still air at standard atmospheric pressure would you have to be immersed in on earth to incur the same rate of heat loss."0 -
I understand this thread has become a bit of a one way conversation now. But here's a little more I've dug up for anybody who has been following.
I discovered the links below, that back up some assumptions we have been making so far, and provide ways of estimating some of the things I'm currently unsure about.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/bodrad.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/bodcon.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/sweat.html
Firstly the links backup the black body assumption by stating that the emissivity of the human body is 0.97, making it a near ideal radiator in the infrared.
Secondly they say the typical human body area is 2m^2 instead of the previously estimate of 1m^2
Thirdly they use a more accurate estimate of skin temperature to be 34C = 307K instead of 37C
Since we have no conduction/convection into space, all heat lost is through radiation, and we get a total rate of heat loss in space to be
Qspace/t = e[sigma]A(T1^4 - T2^4) = (0.97)(5.67e-8)(2)(307^4 - 3^4)
= 977.098 Watts
So the total rate of heat loss in the void of space is 977 Watts.
The rate of heat loss in the room on earth will be a sum of quantities. Radiative heat loss, conductive/convective heat loss and cooling through perspiration. For simplicity, in the links provided above, they seem to assume zero convection (or simply put a question mark where convective heat losses should be), however I'm not sure how big the consequences are of this assumption.
so Qroom/t = Qrad/t + Qcond/t + Qperspir/t
We can use the same equation for radiative heat loss on earth, immersed in some kind of still air heat bath at an air Temperature of T.
Qrad/t = (0.97)(5.67e-8)(2)(307^4 - T^4)
For estimating the conductive heat losses, in the links given above, they make a crude approximation that there is a layer d=5cm out from your skin over which the temperature drops to the ambient air temperature and use the following heat transfer equation for conduction.
Qcond/t = k A (T1 - T2) / d
where A again is human surface area, and k is the thermal conductivity of air.
so plugging in our values
Qcond/t = (5.7e-5)(2)(307-T)/(0.05)
Finally we should consider heat lost through perspiration. Although, I suspect that the solution temperature T were solving for will be cold enough, that the body probably won't perspire, and that we can ignore this element from the full equation for Qroom/t.
This gives
Qroom/t = (0.97)(5.67e-8)(2)(307^4 - T^4) + (5.7e-5)(2)(307-T)/(0.05)
= 1.10e-7(8.88e+9 - T^4) + 2.28e-3(307-T)
= 977.798 - (1.1e-7)(T^4) - (2.28e-3)(T)
Equating Qroom/t = Qspace/t
Gives
0.7 - (1.1e-7)(T^4) - (2.28e-3)(T) = 0
The positive real solution to this is T = 47 K = -225C, which is very cold indeed.
The biggest source of error I suspect in the above analysis (apart from the estimate of human surface area) is the modelling of conductive/convective heat loss to the air. The idea of considering heat loss to the air as conduction through a 5cm layer of still air seems to me quite crude and it would not surprise me if it is grossly inaccurate.
Surely convection must play a bigger part. I suspect the equation for Qroom/t grossly underestimates heat loss to the air for this reason, and the true answer I seek is probably a little higher than -225C.
Edit:
Through further research, I have uncovered that the better way to model conductive/convective heat losses to still air may be to use an empirically measured heat transfer coefficient.
http://en.wikipedia.org/wiki/Convective_heat_transfer
Q/t = h A (T1-T2)
To quote wikipedia :
"The heat transfer coefficient depends upon physical properties of the fluid such as temperature and the physical situation in which convection occurs. Therefore, the heat transfer coefficient must be derived or found experimentally for every system analyzed. Formulas and correlations are available in many references to calculate heat transfer coefficients for typical configurations and fluids."
Not great news.
Thankfully I came across this paper.
http://www.springerlink.com/content/3mk538uvxmmek1ff/
They say in the abstract that previously published values for h in the literature are between 3.3 and 3.4 W/(m^2 K) for "natural" convection of heat from human skin, which I believe is just convection in still air.
Furthermore, they say that the radiative heat transfer coefficient is 4.7 W/(m^2 K). This is not quite consistent with my previous analysis since (4.7)(2)(307-3) = 2858 W which is a lot greater than 977 W. But perhaps the heat transfer coefficient approximation breaks down for such a large temperature differential. This makes sense because radiative heat loss really goes as T^4 and the linear approximation ought only hold when dt << T.
Using this method of estimation I get
977 W = [(4.7 + 3.35) W / (m^2 K)] * [2 m] * [307 K - T]
Solving for T I get
T = 246 K = -27 C
This is more in line with Improbable's answer. I would agree that you would probably only survive about an hour if you were sitting around naked in -27C air.
Still, due to the huge discrepancy between this answer and my previous, I'm not too confident about it's accuracy and have no notion of the error bounds on it. Anybody else have any ideas? have I made any mistakes?
Whatever the answer may be, I suppose one interesting conclusion is that there would be absolutely no point in trying to insulate yourself in space by traditional means, e.g. wearing insulating clothes. The best thing to do is to cover yourself in reflective material to trap your bodies radiation and keep your emissivity low. If you had one of those first aid/emergency foil blankets I'd say you'd be pretty comfortable in space.0 -
hi sev, i read your post with interest, i only did physics in first year in college, but i think i know what you are getting at. I suppose if you could reflect all of your own radiated heat you could remain in space for quite a while. As for how cold it "feels" i think at a certain temp the cold becomes hot or at least it feels hot or burns. I have invertantly tried this with liquid nitrogen many times0
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They say in the abstract that previously published values for h in the literature are between 3.3 and 3.4 W/(m^2 K) for "natural" convection of heat from human skin, which I believe is just convection in still air.
Furthermore, they say that the radiative heat transfer coefficient is 4.7 W/(m^2 K). This is not quite consistent with my previous analysis since (4.7)(2)(307-3) = 2858 W which is a lot greater than 977 W.
How can you have a factor for radiative heat loss in W/(m^2 K) when the radiative process is dependent on T^4, I mean that can't hold true for every range of temperatures cause it doesn't take into account your start and end temperature?0 -
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How can you have a factor for radiative heat loss in W/(m^2 K) when the radiative process is dependent on T^4, I mean that can't hold true for every range of temperatures cause it doesn't take into account your start and end temperature?
I dealt with that point somewhat already.But perhaps the heat transfer coefficient approximation breaks down for such a large temperature differential. This makes sense because radiative heat loss really goes as T^4 and the linear approximation ought only hold when dt << T.
If you taylor expand the function
dQ/dt = e A σ [(T+ΔT)^4 - T^4]
about the point T and ignore terms of higher order than linear in ΔT, then we get
dQ/dt = 4 e A σ T^3 ΔT
= h A ΔT
where h = 4 e σ T^3
This approximation (leaving out terms of higher order than linear) should be accurate for ΔT << T
Using the value of T = 293K = 20C (about room temperature), and an emissivity of 0.97, we get
h = (4)(0.97)(5.67e-08)(293)^3
= 5.54 Watts/m^2/K
This is very similar to the value of 4.7 they give in the paper. If I choose to let the basis ambient temperature be 5 degrees Celsius (not too far off the kind of temperatures were used to dealing with outdoors) then we get almost exactly 4.7 for the answer.
In the paper I referenced, I guess they measured the value of h experimentally for ambient air temperatures typical of human experience.0
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