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Heat transfer in space

  1. Jul 10, 2010 #1
    Hello. People always say that the space is 'cold', and I've always heared that on a satellite, the side facing the sun is really hot, while the dark side is very cold. What I dont understand is how can the dark side be cold if the heat it had couldnt escape in any way except radiation, which, to my understanding, is very small.
    Image this scenario: I'm orbiting earth in my new spacecraft, I grab a small oxygen tank that connects to my mask, and jump outside the craft without any special suitn just my breathing mask. Lets ignore all pressure effects and anything else not related to temperature and heat transfer. Will I feel a terrible cold outside? My first thought is that I wont feel cold at all, because as outside is vacuum, the heat from my body cant go anywhere (except the small amount that irradiates).

    Thanks
     
  2. jcsd
  3. Jul 10, 2010 #2
    I believe that there are two factors cooling you in space, as within an atmosphere or other relatively dense medium: 1) convection/conduction and 2) radiation. I lump convection and conduction together as both being transfers of heat as kinetic energy between material substances. Since space is a relative vacuum, I think you may be right that the vacuum may actually insulate you from heat-loss, similar to the way a vacuum thermos prevents thermal transfer through the vacuum wall of the container. The heat lost by radiation in such a thermos is reflected back into you coffee or soup or whatever by the shiny chromed glass inner lining that always seems to shatter for some reason that someone else will have to explain for me (I'm sure it has something to do with pressure-differential).

    Anyway, I think you're right that for thermal equilibrium to be achieved between two substances, one of the substances can't be relatively empty space. So maybe Khan was wrong in Star Trek II when he said that "revenge is a dish best served cold and it is very cold in space." Maybe he should have said that the coldness of his revenge was well-preserved in the vacuum thermos of space, but that's not as dramatic - although it may have been more apt for his "genetically superior intellect" lol.
     
  4. Jul 10, 2010 #3

    russ_watters

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    Radiation is actually a lot more significant than you realize, even at the temp of the body. It isn't that hard to use the Stefan Boltzman equation - give it a try.
     
  5. Jul 10, 2010 #4
    Are you talking about our "auras?" How is that significant in space, though, provided your space suit is lined with some kind of reflective material?
     
  6. Jul 10, 2010 #5

    russ_watters

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    Huh? Auras? Anyway, the OP asked how you'd do without a space suit and also how significant is it to the space ship if it isn't wearing a space suit either.

    Here is the equation: http://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/stefan.html

    Give it a shot: plug and chug!
     
    Last edited: Jul 10, 2010
  7. Jul 10, 2010 #6
    After re-reading the OP, you're right. I was thinking the bit about not having a space-suit just referred to air-pressure problems.

    Haven't you ever seen those pictures of people's "auras" with special cameras? I assume that these are made by using infrared photography and then superimposing the person's visible-light image over their thermal signature. I guess I should have done more research before assuming, but it just struck me as amusing that you were talking about EM radiation emissions from organic bodies and that these could be photographed and new agists would attribute metaphysical meaning to them.

    edit: I just googled it and "Kirlian photography" seems to involve registering the electric or magnetic field of an object through direct contact with a metal plate or something like that. So apparently it is not thermal photography or otherwise registering EM emissions, unless these correspond with the EM fields of the body, which may well be the case, no? Sorry, this has turned into a thread-hijack; although not my intent since I really thought that "aura-photography" was just thermal photography propped up as something mystical.
     
    Last edited: Jul 10, 2010
  8. Jul 10, 2010 #7
    Since a vacuum contains no particles, can it contain any heat? If not, then the heat lost by any object to a vacuum by radiation would be extreme right?

    For example, even a 1 meter cube of ice would lose significant heat:
    6 * 0.000000056703 * 1 * (273.15^4) = ~1900 watts
     
  9. Jul 10, 2010 #8
    So the same object at the same temperature radiates more energy in a vacuum than in pressurized surroundings?
     
  10. Jul 10, 2010 #9

    russ_watters

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    Yep.

    Now in reality, you have to be away from/shielded from the sun and nearby planets to radiate that much, but otherwise, yeah, it really is that much.
     
  11. Jul 10, 2010 #10

    russ_watters

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    It's not the pressure, it is the absence of other objects to radiate heat back at it.
     
  12. Jul 10, 2010 #11
    I will try to derive how much time it would take to an average men to radiate its thermal energy from 37°C to 0°C

    Average human body surface area = 1.9m2
    average human body specific heat capacity = 3470 J/Kg°C
    Average human body emmisivity = .97

    Q = CmT
    Q = 3470*90kG*37
    Q=1.16E7 J

    P = A[tex]\epsilon[/tex][tex]\sigma[/tex]T4
    P = (1.05E-7)(T4)

    Q= [tex]\int[/tex](from 0 to t) (1.05E-7)(T4) dt = 1.16E7 J

    Q = 1.04x10^-7(t) (273^4) - 1.04x10^-7(0s) (310^4) = 1.16E7

    :yuck: Here I think I fail because the time t appears to be independent of the initial temp T (310K)

    I'm not quite sure if I know how to integrate something using two variables (time and temperature), that might be why I failed.
    Also sorry because of my real messy Latex, its my first time using it.
     
    Last edited: Jul 11, 2010
  13. Jul 11, 2010 #12
    I'm not that good with math, but would it help to calculate an amount of watts like with the ice-block example above? I checked my conversion program and watts converts to btu's/minute. So if you have the number of btu's in a human body at 37C, couldn't you just divide that by the btu's/minute rate to get the time to 0C? And, btw, what's the relevance of 0C? Doesn't water freeze and boil at different temperatures depending on air pressure, rendering 0C and 100C arbitary numbers in space?
     
  14. Jul 11, 2010 #13
    Watts = joules/second
    I think we cant do it the simple way because the power emitted changes with a change in temperature, which changes with time (decreasing by the irradiated heat). I'm pretty sure we need some integral here
     
  15. Jul 11, 2010 #14
    Your "human body" is like a big stone :biggrin:
    [tex]\epsilon \sigma AT^4dt = -mCdT[/tex]

    [tex]t = -\frac{mC}{\epsilon \sigma A}\int _{37+273}^{273}\frac{dT}{T^4}[/tex]

    t is around 4.3 hours!

    However human body isn't like stones. We eat, and we store energy. A recommended daily energy intake value for men is about 10^7 J, which is about the same amount of heat you lose if your temperature decreases from 37C to 0C. So if you have a big meal before jumping into the space without suffering from the pressure effect, I think you hardly feel any change, especially when a 90-kg man most likely stores a substantial amount of fat.

    Just my 2 cents :smile:
     
    Last edited: Jul 11, 2010
  16. Jul 11, 2010 #15
    Thanks!
    So space is not cold and I wont turn into a block of ice if i go there
     
  17. Jul 11, 2010 #16
    What would be really useful would be to compare the rate of heat-dissipation to some comparable terrestrial temperature situation, e.g. -40C at sea level.

    edit: here is an emissivity calculator online. I can't figure out how to use it. Can it solve this problem?
    http://www.pyrometer.com/cgi/pyrometer.cgi
     
    Last edited: Jul 11, 2010
  18. Jul 11, 2010 #17

    russ_watters

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    Our metabolism doesn't increase substantially after we eat. Humans generate an average of about 150w, so losing 1900 means getting cold very fast.
     
  19. Jul 11, 2010 #18

    russ_watters

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    Yeah, space is cold and you would turn into a block of ice if you go there. That's the whole point of what everyone has been saying (except that one post)! It just takes a while to freeze solid, but the cold could kill you long before you freeze solid. You'd be shivering in seconds and have hypothermia in minutes.
     
  20. Jul 11, 2010 #19
    Things are different in 2 cases: one in vacuum where the body only emits radiation (and self-generate heat), and the other on the Earth where there are not only radiation from the body, but also conduction, convection (wind speed is a factor), and radiation from the surrounding (and even the sunlight) that the body absorbs.

    It calculates emissivity, not the heat loss due to radiation. And I don't think it can calculate e for human.

    Thanks!
    So if it's like what you said, i.e. human heat generation rate is about 150W, time taken for our body to decrease from 37C to 35C, at which hypothermia takes place, is about 13min.
     
    Last edited: Jul 11, 2010
  21. Jul 11, 2010 #20
    The amazing thing, though, is that you don't need any insulation in your space-suit to retain warmth; only reflective material. If there is no cold air surrounding you, the kinetic energy heat inside your suit can't conduct or convey itself outside the suit, right? You could be wearing reflective mylar or some other incredibly light fabric, as long as it could contain the pressure and resist punctures and unfiltered radiation damage, correct?

    So, ignoring pressure and radiation exposure, you could run outside to get the mail in your pajamas and make it back inside the spacecraft without endangering your health?
     
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