5 ms·

> But... there is no temperature in space ... That's misleading (and the linked article offers no support for your claim). A vacuum has no temperature, but obj

by lutusp 13y ago

> But... there is no temperature in space ... That's misleading (and the linked article offers no support for your claim). A vacuum has no temperature, but objects in space certainly do have a temperature and all the normal thermodynamic rules apply. If what you said were true, you could safely approach the sun as long as you didn't come into contact with it. But that's not true. In the neighborhood of the non-planet Pluto, the median temperature is just 33 Kelvins. In the extremes of space, between the stars, the temperature of an object falls to the CMB (http://en.wikipedia.org/wiki/Cosmic_microwave_background http://en.wikipedia.org/wiki/Cosmic_microwave_background), about 2.7 Kelvins. The linked article makes the same points.

IgorPartola 13y ago
Of course. The way I think about it is this: when you go for a space walk, first you are inside the airlock where the temperature is about our usual 293 K or so. Now, you suck all the air out and are in a vacuum. You are insulated, except if you are touching another object, and the heat you are losing through radiating it. So the question is: given a GoPro camera strapped to your wrist, and say, a 30 minute space walk, what temperature will it be once you are back in the airlock and before you pump all the air back in? My hunch is that it will be pretty darn high. After all, it is connected to your wrist, and the inside of your space suite must be at least above 273 K for you to be able to do anything.
- lutusp 13y ago
  > So the question is: given a GoPro camera strapped to your wrist, and say, a 30 minute space walk, what temperature will it be once you are back in the airlock and before you pump all the air back in? My hunch is that it will be pretty darn high. Think about this. One of a spacesuit's key features is that it uses insulation to protect the astronaut from the temperature of space (meaning those objects, however distant, that exchange heat energy with the spacesuit via radiation). This means the spacesuit's exterior will quickly rise or fall to the temperature of its environment. That, in turn, means the GoPro camera will also rise or fall to the temperature extremes of its environment. So, if the astronaut is located in a shadowed environment -- let's say on the moon -- the spacesuit exterior and the GoPro camera will fall close to the temperature of space, or the temperature of the shadowed areas of the moon: http://www.nature.com/news/2009/091217/full/news.2009.1149.html http://www.nature.com/news/2009/091217/full/news.2009.1149.h... Quote: "Previous findings had identified the Moon as the coldest place in the Solar System, but the latest results push the temperature even lower, all the way to 26 kelvin ..." The above refers to the shadowed areas of the moon. Pretty damn cold. :) But -- if the astronaut walks around in the sunlit parts of the moon, the temperature will be very high. My point is the astronaut's temperature inside the suit isn't the issue, and the more efficient the suit, the less its external temperature reflects that of its occupant. So the GoPro camera would have to tolerate a very wide temperature range. > After all, it is connected to your wrist, and the inside of your space suite must be at least above 273 K for you to be able to do anything. A spacesuit that externally radiated any significant energy at room temperature would be a very poor design indeed. And the GoPro camera isn't attached to the astronaut's wrist, it's attached to the spacesuit wrist area.
  - IgorPartola 13y ago
    I understand what you are saying about things like the Moon, Pluto, etc. If a bury a thermometer on the moon, and dig it up 5 minutes later, it will be very cold. But we are talking about a space walk here. Let's phrase the question this way: if I have a 1 kg mass of plastic and metal in some proportion and I throw it out of the airlock with the initial temperature of 293 K, what will its temperature be 30 minutes later if it was (a) in the shadow and (b) in the sunlight? This is a different case than being in contact with a hot or cold satellite or planet. It's a rate problem. If we assume a near-perfect vacuum, the only heat going in and out is via radiation. So the question is, how fast does a hunk of metal/plastic radiate heat at 293 K. Now, the radiation heat transfer out of a black body will be: q = σ T^4 A, where q is the Watts of heat transferred, σ is the Boltzman constant, T is the temperature in Kelvin, and A is the surface area. Roughly: 5.6703e-8 x 293 x 0.1 for a GoPro camera. That's 1.66e-6 Watts, or 1.66 microwatts. That's a pretty damn low transfer rate: we are talking about roughly 3 mJ of energy lost over 30 minutes. Now, the delta T here is equal to Q/(mC), where Q is our 0.003 Joules, m is the mass, 0.135 kilograms, and C is the specific heat. The specific heat is tricky but let's assume this thing is made out of copper: 390 J/(K kg). That gives us a delta T of, drum roll please, 5.7e-5 degrees. So assuming we are in the shade and are not touching cold objects, the GoPro will come back at the exact temperature it left after a 30 minute space walk. We can of course figure out what it will be in the sun, where I expect it'll heat up quite a bit, which is probably the more dangerous thing for a piece of equipment. The equations are nearly the same.
    - lutusp 13y ago
      > Let's phrase the question this way: if I have a 1 kg mass of plastic and metal in some proportion and I throw it out of the airlock with the initial temperature of 293 K, what will its temperature be 30 minutes later if it was (a) in the shadow and (b) in the sunlight? To answer, I have to say first that radiation is a much more efficient way to transfer heat energy than convection, and in some contexts it rivals conduction. If it's in shadow, exposed to space and not the sun, the object in your thought experiment will radiate most of its heat energy rather quickly, and will ultimately fall to a temperature near the CMB, i.e. 2.7 Kelvins. > Now, the radiation heat transfer out of a black body will be: q = σ T^4 A, where q is the Watts of heat transferred, σ is the Boltzman constant, T is the temperature in Kelvin, and A is the surface area. Roughly: 5.6703e-8 x 293 x 0.1 for a GoPro camera. That's 1.66e-6 Watts. In your calculation, you failed to take the fourth power of the temperature. http://en.wikipedia.org/wiki/Black-body_radiation#Stefan.E2.80.93Boltzmann_law http://en.wikipedia.org/wiki/Black-body_radiation#Stefan.E2.... j = σ T^4 (j = radiated power watts) = 5.67 * 10^−8 * 295^4 Watts from a "unit square", presumably a square meter, or about 430 watts. Remember that three-dimensional objects lose their heat energy more quickly as they become smaller in size (because their dimensions decrease proportional to the square of their dimensions, but their volume declines as the cube). This means the GoPro camera, initially at room temperature, exposed to space would radiate away its heat energy very quickly. The linked article suggests that a person (at body temperature and in a normal earthly environment) radiates away a net power of about 100 watts. Remember about this figure that is is a net (radiation minus absorption) for a person of about 2 square meters surface area at room temperature. If we calculate the example of a person exposed to space (in shadow) directly without any heat inflow, the radiated power would be about 860 watts. If a person were shaped like a GoPro camera (with the same density) and we scaled it down proportionally, the rate of heat loss would increase (even though the power would decrease) for reasons given above. That consumer camera would not be long for this world -- in fact, it would probably expire faster than the hapless human in the above example. > So assuming we are in the shade and are not touching cold objects, the GoPro will come back at the exact temperature it left after a 30 minute space walk. Do take the fourth power of temperature. See how that turns out. :)

by lutusp 13y ago

IgorPartola 13y ago

Of course. The way I think about it is this: when you go for a space walk, first you are inside the airlock where the temperature is about our usual 293 K or so. Now, you suck all the air out and are in a vacuum. You are insulated, except if you are touching another object, and the heat you are losing through radiating it. So the question is: given a GoPro camera strapped to your wrist, and say, a 30 minute space walk, what temperature will it be once you are back in the airlock and before you pump all the air back in? My hunch is that it will be pretty darn high. After all, it is connected to your wrist, and the inside of your space suite must be at least above 273 K for you to be able to do anything.

lutusp 13y ago
> So the question is: given a GoPro camera strapped to your wrist, and say, a 30 minute space walk, what temperature will it be once you are back in the airlock and before you pump all the air back in? My hunch is that it will be pretty darn high. Think about this. One of a spacesuit's key features is that it uses insulation to protect the astronaut from the temperature of space (meaning those objects, however distant, that exchange heat energy with the spacesuit via radiation). This means the spacesuit's exterior will quickly rise or fall to the temperature of its environment. That, in turn, means the GoPro camera will also rise or fall to the temperature extremes of its environment. So, if the astronaut is located in a shadowed environment -- let's say on the moon -- the spacesuit exterior and the GoPro camera will fall close to the temperature of space, or the temperature of the shadowed areas of the moon: http://www.nature.com/news/2009/091217/full/news.2009.1149.html http://www.nature.com/news/2009/091217/full/news.2009.1149.h... Quote: "Previous findings had identified the Moon as the coldest place in the Solar System, but the latest results push the temperature even lower, all the way to 26 kelvin ..." The above refers to the shadowed areas of the moon. Pretty damn cold. :) But -- if the astronaut walks around in the sunlit parts of the moon, the temperature will be very high. My point is the astronaut's temperature inside the suit isn't the issue, and the more efficient the suit, the less its external temperature reflects that of its occupant. So the GoPro camera would have to tolerate a very wide temperature range. > After all, it is connected to your wrist, and the inside of your space suite must be at least above 273 K for you to be able to do anything. A spacesuit that externally radiated any significant energy at room temperature would be a very poor design indeed. And the GoPro camera isn't attached to the astronaut's wrist, it's attached to the spacesuit wrist area.
- IgorPartola 13y ago
  I understand what you are saying about things like the Moon, Pluto, etc. If a bury a thermometer on the moon, and dig it up 5 minutes later, it will be very cold. But we are talking about a space walk here. Let's phrase the question this way: if I have a 1 kg mass of plastic and metal in some proportion and I throw it out of the airlock with the initial temperature of 293 K, what will its temperature be 30 minutes later if it was (a) in the shadow and (b) in the sunlight? This is a different case than being in contact with a hot or cold satellite or planet. It's a rate problem. If we assume a near-perfect vacuum, the only heat going in and out is via radiation. So the question is, how fast does a hunk of metal/plastic radiate heat at 293 K. Now, the radiation heat transfer out of a black body will be: q = σ T^4 A, where q is the Watts of heat transferred, σ is the Boltzman constant, T is the temperature in Kelvin, and A is the surface area. Roughly: 5.6703e-8 x 293 x 0.1 for a GoPro camera. That's 1.66e-6 Watts, or 1.66 microwatts. That's a pretty damn low transfer rate: we are talking about roughly 3 mJ of energy lost over 30 minutes. Now, the delta T here is equal to Q/(mC), where Q is our 0.003 Joules, m is the mass, 0.135 kilograms, and C is the specific heat. The specific heat is tricky but let's assume this thing is made out of copper: 390 J/(K kg). That gives us a delta T of, drum roll please, 5.7e-5 degrees. So assuming we are in the shade and are not touching cold objects, the GoPro will come back at the exact temperature it left after a 30 minute space walk. We can of course figure out what it will be in the sun, where I expect it'll heat up quite a bit, which is probably the more dangerous thing for a piece of equipment. The equations are nearly the same.
  - lutusp 13y ago
    > Let's phrase the question this way: if I have a 1 kg mass of plastic and metal in some proportion and I throw it out of the airlock with the initial temperature of 293 K, what will its temperature be 30 minutes later if it was (a) in the shadow and (b) in the sunlight? To answer, I have to say first that radiation is a much more efficient way to transfer heat energy than convection, and in some contexts it rivals conduction. If it's in shadow, exposed to space and not the sun, the object in your thought experiment will radiate most of its heat energy rather quickly, and will ultimately fall to a temperature near the CMB, i.e. 2.7 Kelvins. > Now, the radiation heat transfer out of a black body will be: q = σ T^4 A, where q is the Watts of heat transferred, σ is the Boltzman constant, T is the temperature in Kelvin, and A is the surface area. Roughly: 5.6703e-8 x 293 x 0.1 for a GoPro camera. That's 1.66e-6 Watts. In your calculation, you failed to take the fourth power of the temperature. http://en.wikipedia.org/wiki/Black-body_radiation#Stefan.E2.80.93Boltzmann_law http://en.wikipedia.org/wiki/Black-body_radiation#Stefan.E2.... j = σ T^4 (j = radiated power watts) = 5.67 * 10^−8 * 295^4 Watts from a "unit square", presumably a square meter, or about 430 watts. Remember that three-dimensional objects lose their heat energy more quickly as they become smaller in size (because their dimensions decrease proportional to the square of their dimensions, but their volume declines as the cube). This means the GoPro camera, initially at room temperature, exposed to space would radiate away its heat energy very quickly. The linked article suggests that a person (at body temperature and in a normal earthly environment) radiates away a net power of about 100 watts. Remember about this figure that is is a net (radiation minus absorption) for a person of about 2 square meters surface area at room temperature. If we calculate the example of a person exposed to space (in shadow) directly without any heat inflow, the radiated power would be about 860 watts. If a person were shaped like a GoPro camera (with the same density) and we scaled it down proportionally, the rate of heat loss would increase (even though the power would decrease) for reasons given above. That consumer camera would not be long for this world -- in fact, it would probably expire faster than the hapless human in the above example. > So assuming we are in the shade and are not touching cold objects, the GoPro will come back at the exact temperature it left after a 30 minute space walk. Do take the fourth power of temperature. See how that turns out. :)