3 ms·

> A perfect refrigerator will use vastly more than 1J to remove 1J from a 4.2K device. While your actual point is true, your example is too loosely specified -

by a1369209993 6y ago

> A perfect refrigerator will use vastly more than 1J to remove 1J from a 4.2K device. While your actual point is true, your example is too loosely specified - the energy used by a perfect (or even imperfect) refrigerator depends on the temperature of the environment it's dumping the heat into. In the extreme (and also engineering-wise preferred) case, with a heat sink at cosmic microwave background temperatures of ~2.7K, a perfect refrigerator (rather, heat exchanger) would actually gain energy. (Of course, since the literally-glowing-hot sun takes up some portion of the sky, actually getting a <4.2K environment would likely require siting your computer in the Oort Cloud, if not outside the galaxy entirely, hence why your actual point is true.)

deleted 6y ago
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amluto 6y ago
Touché. If you want to go down that road, I believe you’ll find that the maximum amount of heat that any machine can exhaust into the CMB (or, more generally, radiatively transfer into a medium at an effective temperature T_h) is precisely the amount of heat it would emit as a blackbody at T_h. IOW sending entropy into space requires generating a uniformly distributed random distribution of something and sending it into space and, if you want to use light, you end up with delta S = Q / T, where T is the temperature of the light. So your device will need a silly surface area, in addition to needing to be very well shielded from the sun. Now maybe you can cheat because, at the wavelengths in question, the effective surface area of an antenna is quite large, but I doubt this buys you much. But in some moral sense, you are entirely correct. If you put your computer on a cold moon with no view of the sun, you can sink quite a lot of heat at a low temperature. (A really good spectrally and directionally selective filter means you may not need to be quite as far from the sun as the Oort Cloud. Even just pointing an IR thermometer into a clear night sky gives a nice low temperature.)
MertsA 6y ago
> actually getting a <4.2K environment would likely require siting your computer in the Oort Cloud, Well actually you can just use multilayer sun shades made of sheets of aluminized kapton like the James Webb Telescope. It wouldn't get it down to CMB temperature but supposedly some of the Oort cloud is already down to 4K so it should be possible to reach <4.2K with enough attenuation by a sun shade.

deleted 6y ago

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amluto 6y ago

Touché. If you want to go down that road, I believe you’ll find that the maximum amount of heat that any machine can exhaust into the CMB (or, more generally, radiatively transfer into a medium at an effective temperature T_h) is precisely the amount of heat it would emit as a blackbody at T_h. IOW sending entropy into space requires generating a uniformly distributed random distribution of something and sending it into space and, if you want to use light, you end up with delta S = Q / T, where T is the temperature of the light. So your device will need a silly surface area, in addition to needing to be very well shielded from the sun. Now maybe you can cheat because, at the wavelengths in question, the effective surface area of an antenna is quite large, but I doubt this buys you much. But in some moral sense, you are entirely correct. If you put your computer on a cold moon with no view of the sun, you can sink quite a lot of heat at a low temperature. (A really good spectrally and directionally selective filter means you may not need to be quite as far from the sun as the Oort Cloud. Even just pointing an IR thermometer into a clear night sky gives a nice low temperature.)

MertsA 6y ago

> actually getting a <4.2K environment would likely require siting your computer in the Oort Cloud, Well actually you can just use multilayer sun shades made of sheets of aluminized kapton like the James Webb Telescope. It wouldn't get it down to CMB temperature but supposedly some of the Oort cloud is already down to 4K so it should be possible to reach <4.2K with enough attenuation by a sun shade.