> 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.
= 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. :)
Ha! You are totally right. Yes, at 430 Watts we are talking about substantial heat transfer. Without getting into the calculus, if we approximate that the heat transfer will stay constant for a short period of time, then in the first second, we will lose roughly 8.2 degrees. Yup, you are correct, the thing will be damn cold.
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....
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. :)