How Can Space Be Cold in a Vacuum Despite Molecular Vibration?

[bassplayer142]

Heat is the the vibration of molecules in a medium. If this is so, then how is space cold if it is a vacuum. I'm aware that space has very few particles but I don't understand how heat is emitted. I'm guessing that it might have to do with radiation.

 

[G01]

Yes heat can travel through a vacuum as radiation. The thermal energy of atoms and molecules (made of charged particles) causes then to vibrate, which causes them to give off energy as electromagnetic radiation. The electromagnetic radiation can travel through a vacuum, taking the energy with it. This is how thermal energy is transmitted through a vacuum.

 

[Dale]

There are three mechanisms of heat transfer: conduction, convection, and radiation. Conduction and convection do not occur in space, but radiation does. The reason space is said to be cold is because it radiates like a black body at about 4 Kelvin.

 

[bassplayer142]

Okay, so your saying that on Earth as you radiate heat other things are radiating heat onto you. While in space you alone are radiating heat and nothing but distant stars radiate heat on you?

 

[DaveC426913]

Okay, so your saying that on Earth as you radiate heat other things are radiating heat onto you. While in space you alone are radiating heat and nothing but distant stars radiate heat on you?

Correct, with one added detail: You are bombarded by microwave radiation at 4 degrees K.

 

[Dale]

Yes, and as long as your temperature is >4 K you will continually be radiating away more heat than space is radiating back on you.

 

[Repetit]

What it the timescale of this radiation? I mean, if I jump into water at ~0 Celcius it will immediately feel extremely cold because heat is being rapidly removed from my body and transferred to the water. How will this feel in space? I am guessing that heat transfer by radiation is quite slow at temperatures of 37 C so that it won't feel very cold.

 

[Dale]

There is a nice explanation and calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

Using a human body-surface area of about 1.5 m^2 and an emissivity of about .7 you get something like 500 watts of heat loss. To put that in perspective, that is about 10000 food Calories per day, so you would have to eat at least 3 times the normal American diet just to get enough energy to compensate for heat loss, let alone the extra energy needed for other important things like breathing etc.

 

[russ_watters]

Correct, with one added detail: You are bombarded by microwave radiation at 4 degrees K.

Why don't we expand on that a little - it seems to contradict what bassplayer said if not fully explained...

The microwave backtround radiation is left-over energy from the big-ban that is still flowing through the universe. It has been around so long and dissipated so much due to the expansion of the universe that it is now very "cold".

 

[russ_watters]

There is a nice explanation and calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

Using a human body-surface area of about 1.5 m^2 and an emissivity of about .7 you get something like 500 watts of heat loss. To put that in perspective, that is about 10000 food Calories per day, so you would have to eat at least 3 times the normal American diet just to get enough energy to compensate for heat loss, let alone the extra energy needed for other important things like breathing etc.

Very cool and good to know. At rest, a person dissipates somewhere around 70 watts. 500 is probably near the maximum that an in-shape athlete could sustain for much time.

So you would start to feel cold relatively quickly - just a couple of minutes.

 

[DaveC426913]

So you would start to feel cold relatively quickly - just a couple of minutes.

I'm fairly certain that vacuum would be barkin' cold. Like frostbite cold instantly.

 

[DaveC426913]

The microwave backtround radiation is left-over energy from the big-ban that is still flowing through the universe.

Actually, that phenomenon was fairly recent in time and limited in scope to the PF forum...

 

[russ_watters]

I'm fairly certain that vacuum would be barkin' cold. Like frostbite cold instantly.

I don't see how...

 

[Loren Booda]

While wearing an unheated, oxygen supplied, airtight suit of moderate albedo in 2.735^o K space, one would lose less heat than if one were to fall into ice-cold salt water?

 

[Dale]

Even an unheated suit dramatically changes the thermal situation. Most importantly the body is thermally interacting with the suit rather than space, the 500 W radiated by the body could conceivably rapidly heat the suit to the point that it was radiating a significant amount of energy back to the body. Also, when wearing a suit conduction becomes a major mechanism of heat transfer.

I think the comparison that you really want here is simply to wave hands and dismiss the other lethal effects of vacuum and concentrate purely on the heat transfer. So the question is if ice-cold salt water removes heat faster than 500 W. I don't know the answer to that, but the mechanisms are very different. In space the heat transfer mechanism is limited to radiation, in the water you have less radiation, but you add both conduction and convection. Also, water has a very high thermal conductivity, much higher than air. If the water is flowing then the convection is even more problematic since the body will not be able to warm the nearby water. It is quite possible that ice water would remove more than 500W of heat.

 

[Loren Booda]

Thank you, DaleSpam. An accurate synopsis.

 

[Mephisto]

I'm fairly certain that vacuum would be barkin' cold. Like frostbite cold instantly.

That was in one movie that I saw, but i can't remember which one now... Appolo13 or Armageddon or something? Someone's shield broke and he froze instantly in space. Then later on I read that that's actually wrong, and then I read a whole article from NASA web-page saying that if you were exposed to pure space vacuum, the thing that would kill you soonest would be the fact that you would suffocate from lack of oxygen... so I guess that implies cold is not the biggest issue at least for a couple of minutes.

 

[Huckleberry]

How much radiative heating would a body in space receive from the sun if it were in orbit around Earth?

 

[TVP45]

What it the timescale of this radiation? I mean, if I jump into water at ~0 Celcius it will immediately feel extremely cold because heat is being rapidly removed from my body and transferred to the water. How will this feel in space? I am guessing that heat transfer by radiation is quite slow at temperatures of 37 C so that it won't feel very cold.

You're generally correct in thinking heat transfer by liquid is quite efficient. But, there are two examples I can readily think of to show the sensation of cold by radiation: (1) go into the desert on a clear night and the sky feels cold; or (2) there is a classic demonstration in which you take 2 parabolic mirrors and place your eyeball at one focus and an icecube at the other and pretty quickly your eye feels cold.

 

[russ_watters]

How much radiative heating would a body in space receive from the sun if it were in orbit around Earth?

1200w/m^2. There isn't a significant amount absorbed by our atmosphere.

 

[Dale]

How much radiative heating would a body in space receive from the sun if it were in orbit around Earth?

The solar irradiance at 1 AU is usually a little under 1.4 kW/m^2. Of course, in this case it is not the total surface area of about 1.5 m^2 that is important, but only the cross-sectional area. That obviously depends on the orientation (i.e. much less cross-section in the feet-down position than in the face-down position), but it will always be less than half. Let's estimate it at one third in the face-down orientation. In that case the amount of solar irradiance at 1 AU is going to about the same as the amount of heat radiated away.

Of course, in these extreme environments (subject to 5800K radiation from one direction and 4K radiation from everywhere else) small deviations from "about the same" can lead to pretty large differences in terms of survivability. However, the bottom line is that 500 W plus or minus will not kill you before the vacuum will. All it will do is determine if your corpse will be freezer-burned or well-done.

 

[Dale]

Thank you, DaleSpam. An accurate synopsis.

Very welcome!

 

[DaveC426913]

That was in one movie that I saw, but i can't remember which one now... Appolo13 or Armageddon or something? Someone's shield broke and he froze instantly in space. Then later on I read that that's actually wrong, and then I read a whole article from NASA web-page saying that if you were exposed to pure space vacuum, the thing that would kill you soonest would be the fact that you would suffocate from lack of oxygen... so I guess that implies cold is not the biggest issue at least for a couple of minutes.

Yes, I would agree on all counts.

A broken mask would not instantly freeze you - I wasn't suggesting that. Nor would cold be a big concern at that point - suffocation is your only real concern.

But that's sort of complicating the original question, which was simply about how cold is it.

Imagine a glove came off. I suspect it would be somewhat like Antarctica in a winter storm.

 

[phlegmy]

lol i had an argument with a thermodynamics lecturer in the middle of a lecture over this.
he posed the question

"what is the temprature of space"
nobody answered so i offered
"that sir, is a nonsense question"
"what do you mean its a nonsense question"
"you may as well ask what is the voltage between a square and a circle"
"you're wrong, the temprature of space is aprox 4 kelvin"

i offered the argument that space is defined by an absence of matter and temprature was a property of matter. i suggested that perhaps an object in space would reach 4 kelvin but space itself could not have a temprature. he refused to conceed, as did i. but i still maintain he is wrong, in the wording of the question at least

 

[sneez]

Imagine a glove came off. I suspect it would be somewhat like Antarctica in a winter storm.

How can that be? In Antarctica the ambient air has temperature and hence radiates back to you. Space has no temperature, hence it should feel a lot worse ...? Assuming -80 C in storm conditions, you get back around 70 W m-2 from the air in radiation.

 

[Proggle]

How can that be? In Antarctica the ambient air has temperature and hence radiates back to you. Space has no temperature, hence it should feel a lot worse ...? Assuming -80 C in storm conditions, you get back around 70 W m-2 from the air in radiation.

You're only counting radiation. Even though the air is radiating back to you, it is also taking heat through conduction and convection at a much higher rate.

In other words, your losses through radiation on Earth are smaller than they are in space, but two other mechanisms exist that do not happen in space. In circumstances of extreme cold, those two vastly overwhelm your relative gains from radiation.

 

[DaveC426913]

How can that be? In Antarctica the ambient air has temperature and hence radiates back to you. Space has no temperature, hence it should feel a lot worse ...? Assuming -80 C in storm conditions, you get back around 70 W m-2 from the air in radiation.

I'm not sure if it was clear from Proggle's or not, so I'll paraphrase:

[B]Loss through    Antarctica   Space[/B]
Radiation         low        high
Conduction        medium     zero
Convection        high       zero*

*actually conductive loss would not be zero - you would start to have sublimation since liquids would have a tough time staying liquid in vacuum. This would result in a slow evaporative** loss, which is a kind of conductive loss.

**evaporative? Would it be more generally ablative, since the material is actually leaving with the heat? Hmm.

 

[D H]

I disagree with the low/high designation. Applying the Stefan-Boltzmann Law, the ratio of radiative loss in space to the radiative loss in an Antarctic blizzard is

\frac{P_{\text{space}}} {P_{\text{blizzard}}} =<br /> \frac{T^{\;4}_{\text{body}}-T^{\;4}_{\text{space}}}{T^{\;4}_{\text{body}}-T^{\;4}_{\text{blizzard}}}

Using T_{\text{body}} = 310K, T_{\text{space}} = 4K, and T_{\text{blizzard}} = 193K yields a ratio of 1.18. I would not classify an 18% increase in cooling as the difference between low and high.

Moreover, the radiative cooling is a small portion of the total heat loss in a blizzard. The most important factor in surviving a blizzard is getting out of the wind.

 

[Gokul43201]

Y'all've way underestimated the heat loss to forced convection.

It was 50F (10C) outside a little while ago, when I went out to grab dinner. There was a steady breeze of about 15mph (~8m/s). If I'd stood outside and taken my clothes off, I would have lost heat at a rate of at least about 500W (that's very rough - it could be anywhere from 300W to 1000W).

Now, I'd certainly be feeling pretty cold (especially since it's been in the 80s and 90s till just a couple days ago), but I don't think it would be instant frostbite. So, anyway...that's how cold space feels - a 50 degree day with a moderate breeze.

 

[DaveC426913]

I would not classify an 18% increase in cooling as the difference between low and high.

That depends on the endpoints of your scale doesn't it? Radiative cooling in a vacuum would more accurately be called 'maximum', since you're not going to find anything better. And radiative cooling at body temp would be 'zero'.

Kelvin           :  310 300              200 193          100          4  0
relative cooling :  zero                     low                    high   max

Compared to maximum radiative cooling in a vacuum, the loss in Antarctica is low(er).

Moreover, the radiative cooling is a small portion of the total heat loss in a blizzard.

I don't think anyone said otherwise.

 

[Gokul43201]

Compared to maximum radiative cooling in a vacuum, the loss in Antarctica is low(er).

It is lower, but certainly not worthy of the "low" label it was given. Even "medium" would have been a poor characterization.

PS: When we can actually get quantitative comparisons for cheap, why resort to vague, qualitative descriptions?

 

[DaveC426913]

PS: When we can actually get quantitative comparisons for cheap, why resort to vague, qualitative descriptions?

True, but considering the OP's knowledge of the subject, I think numbers don't tell the whole story and need some framing.

To some extent, 'low' could be treated as "a small component of the total cooling (i.e. in Antarctica)", whereas 'high' could considered "a large component of the total cooling (i.e. in space)."

 

[sneez]

Can we pose it as such?
Im standing in the storm at T=220K (naked :D) and there is a wind blowing at 20 m/s. The ambient pressure is 1atm and assume Earth atmosphere with air weight 0.029kg/mol .

Assume crossectional area of my body is 1m2, Cp (air capacity at 200K) = 1650 J/kg.K, Cp_body= 3500 J/kg.K . My skin has emissivity 0.7 (?) My initial body temperature is 310K . [we are concerned with conduction which should be negligible, convection which should be somewhat important and radiation] ? Is there anything which I did not consider before I embark?

 

[sneez]

Now that I am thinking about it, is the wind actually relevant? The wind acts only to slow down the process of equilibriation between body and the air, by keeping theT gradient between the body and air by blowing off the near skin molecules of air that my body heated. The body, even in hight wind, cannot cool bellow the ambient air temperature, right?

 

[Integral]

Now that I am thinking about it, is the wind actually relevant? The wind acts only to slow down the process of equilibriation between body and the air, by keeping theT gradient between the body and air by blowing off the near skin molecules of air that my body heated. The body, even in hight wind, cannot cool bellow the ambient air temperature, right?

You may want to spend a few minutes searching on "wind chill"

 

[sneez]

I sure did, is it than reasonalbe to assume the windchill temperature rather than the air temperature in this example?

 

[sneez]

Well, quick back of envelope results given convection heat loss is 1/2 of radiative loss in wind from http://www.drphysics.com/convection/convection.html,where 1/3 of heat loss is due to convection without wind condition.

I got something order of 1000 W , due to conduction, convectin, radiation. No condensation, evaporation, prespiration assumed.

It seem that somebody was very interested in the body heat loss in the year 1937 :)
JSTOR: http://www.jstor.org/view/00278424/ap000883/00a00090/0?frame=noframe&userID=89c65748@hamptonu.edu/01cce4405b00501cbca2a&dpi=3&config=jstor

 

[russ_watters]

Now that I am thinking about it, is the wind actually relevant? The wind acts only to slow down the process of equilibriation between body and the air, by keeping theT gradient between the body and air by blowing off the near skin molecules of air that my body heated.

Stop and think about that for a sec - you've felt wind before, right? You've worn heavy clothes, right? The purpose of heavy clothes (insulation) is to create a shallow temperature gradient. With no clothes and a heavy wind, you have the steepest possible temperature gradient and since heat loss is proportional to that gradient, you have the maximum possible heat loss.

You have it exactly backwards.

The body, even in hight wind, cannot cool bellow the ambient air temperature, right?

Of course. But that's the steady-state condition where you are frozen solid and in a normal winter, that takes hours. We're talking about the heat loss in the beginning, when your body temperature is near normal and you are not dead.

I sure did, is it than reasonalbe to assume the windchill temperature rather than the air temperature in this example?

I don't think that's Integral's point. The point is that "wind chill" is a measure of convective heat loss. It is basically how cold still air would have to be to provide the same convective heat loss as windy air at a higher temperature.

 

[russ_watters]

Well, quick back of envelope results given convection heat loss is 1/2 of radiative loss in wind from http://www.drphysics.com/convection/convection.html,where 1/3 of heat loss is due to convection without wind condition.

I'm not sure where you are getting the 1/2 number, and in the "other work" section, others who have calculated the ratio got .7 instead of 1/3 (2nd and fourth paragraphs). With wind, the first saw a 40% increase in convective rate with a 1mph wind (very, very low wind) and the fourth saw a 178% increase (10W/m^2) in a 3 m/s (2 mph) wind. So as that shows, the effect of wind is huge. With a very light wind (ie, just by walking), the convective heat loss rate swamps the radiative heat loss rate.

Note that these calculations are for room temperature air...

 

[sneez]

thanx for the input russ

Stop and think about that for a sec - you've felt wind before, right? You've worn heavy clothes, right? The purpose of heavy clothes (insulation) is to create a shallow temperature gradient. With no clothes and a heavy wind, you have the steepest possible temperature gradient and since heat loss is proportional to that gradient, you have the maximum possible heat loss.

Thats exactly what I thought I said.

Of course. But that's the steady-state condition where you are frozen solid and in a normal winter, that takes hours. We're talking about the heat loss in the beginning, when your body temperature is near normal and you are not dead.

1st point:Are you implying that it is possible for anybody temperature to cool below the T of the air due to wind effects at certain conditions?

2nd: explain how that is not applicable in the beginning

I don't think that's Integral's point. The point is that "wind chill" is a measure of convective heat loss. It is basically how cold still air would have to be to provide the same convective heat loss as windy air at a higher temperature.

Thats exactly why I posed it as a question , do you have an answer? And due to its definition I think its reasonable to use the concept in the wind condition. Basically, I use wind chill temperature in calculation as a temperature surrounding air would "have" given no wind.

I'm not sure where you are getting the 1/2 number, and in the "other work" section, others who have calculated the ratio got .7 instead of 1/3 (2nd and fourth paragraphs). With wind, the first saw a 40% increase in convective rate with a 1mph wind (very, very low wind) and the fourth saw a 178% increase (10W/m^2) in a 3 m/s (2 mph) wind. So as that shows, the effect of wind is huge. With a very light wind (ie, just by walking), the convective heat loss rate swamps the radiative heat loss rate.

I sort of arbitrarily assumed it. The article claims 1/3 and other sources also cite that, but that's for no wind condition, I thought with wind convection gets more and more important as you say, so I gave it more importance .

 

[russ_watters]

Thats exactly what I thought I said.

You said the wind slows the body reaching equilibrium with the air - it speeds it up.

1st point:Are you implying that it is possible for anybody temperature to cool below the T of the air due to wind effects at certain conditions?

Certainly not.

2nd: explain how that is not applicable in the beginning

The body isn't frozen solid, so talking about how heat would be removed when it is frozen solid is irrelevant.

Thats exactly why I posed it as a question , do you have an answer?

?? An answer to what? If we should assume wind chill temp? I guess if your equation doesn't have a term for wind speed, you can use the wind chill temp instead.

 

[Dale]

1st point:Are you implying that it is possible for anybody temperature to cool below the T of the air due to wind effects at certain conditions?

Yes! That is why we sweat. Otherwise everyone in Phoenix would die.

However, I think sweat would be a pretty small factor in Antartica :-)

 

[sneez]

Right, speed up by faster heat removal, i stand corrected.

DaeSpam, sweat is cooling by evaporation, I was trying to see if object will cool below the air temperature just due to wind alone.

(I don't think I am understanding the russ's comment on the steady state of frozen solid)

 

[Dale]

DaeSpam, sweat is cooling by evaporation, I was trying to see if object will cool below the air temperature just due to wind alone.

So neglecting evaporation just considering convection? No, I don't think that convection can cool things below the air temperature. I think the only way that would even be theoretically possible is for the the flow to cause some unusually low pressure region which would experience an adiabatic drop in temperature. But surely such effects would be negligably small and very difficult to model for a human. The wind-tunnel tests might be fun though!

 

[keinve]

that depends on how you define heat. if you mean hot or warm for a human, then there is very little heat in the universe. if you mean above absolute zero, then most of the universe has heat.