3 ms·

Theoretically keeping something at the same temperature is at most going to double the energy requirements, though there's likely to be some additional practica

Theoretically keeping something at the same temperature is at most going to double the energy requirements, though there's likely to be some additional practical issues with maintaining such a low temperature.

amluto 6y ago
Not even close. A perfect refrigerator will use vastly more than 1J to remove 1J from a 4.2K device. Look up Carnot’s theorem.
- 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
    [deleted]
  - 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.
- contravariant 6y ago
  By Carnot's theorem an ideal engine will convert convert 1J heat into less than 1J of work and this process is invertible. So as a corollary you can transfer 1J of heat using less than 1J of work. Since the temperature difference is so large the theoretical limit is really close to 1J of work though.
  - amluto 6y ago
    Something went wrong in your calculation. An ideal heat engine operating between reservoirs at T_h and T_c has efficiency 1 - T_c/T_h. This means, when operating reversibly, it will send Q_c heat to the cold side, remove Q_h from the hot side, and do W work, where Q_c = Q_h - W and W = (1 - T_c/T_h)Q_h. If T_c = T_h/10, then 1 - T_c/T_h = 90% (very efficient!) and Q_c = 0.1*Q_h (not much waste heat!). Now turn this around, because it’s reversible. A heat pump removes Q_c from the cold side and exhausts Q_h to the hot side. If you remove 1J from the cold side, you exhaust 10J to the hot side, and that 9J difference is the work done. That is, it costs 9J to remove 1J of heat from a 27.3K refrigerator if the exhaust is at 273K. It’s worse at 4.2K. (You can do an equivalent calculation a little more tidily by considering the entropy change of each side. To remove a given amount of entropy from the cold side, you must add at least as much entropy to the hot side. Plug in the usual formula for isothermal entropy change, and you get the answer.)
    - contravariant 6y ago
      A I see where my mistake is now, thanks! I suppose it makes sense that reversing the process would also meant that the 'heat intake' is equivalent to the exhaust heat, which for an efficient engine wouldn't be all that much. I suppose that means that the cooling requirement will increase the energy requirement by nearly 2 orders of magnitude.
  - HPsquared 6y ago
    This is only true for heat pumps where the useful heat transfer occurs at the 'hot' side. In the ideal reversible case, COP_heating = T_h / (T_h - T_c), and is always greater than 1 (as the heat from work done is added to the heat transferred from the cold side). For refrigeration the situation is different: we are interested in the amount of heat _removed_ from the cold side. In this case, for a Carnot cycle COP_cooling = T_c / (T_h - T_c) and is NOT always greater than 1.

amluto 6y ago

Not even close. A perfect refrigerator will use vastly more than 1J to remove 1J from a 4.2K device. Look up Carnot’s theorem.

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
  [deleted]
- 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.
contravariant 6y ago
By Carnot's theorem an ideal engine will convert convert 1J heat into less than 1J of work and this process is invertible. So as a corollary you can transfer 1J of heat using less than 1J of work. Since the temperature difference is so large the theoretical limit is really close to 1J of work though.
- amluto 6y ago
  Something went wrong in your calculation. An ideal heat engine operating between reservoirs at T_h and T_c has efficiency 1 - T_c/T_h. This means, when operating reversibly, it will send Q_c heat to the cold side, remove Q_h from the hot side, and do W work, where Q_c = Q_h - W and W = (1 - T_c/T_h)Q_h. If T_c = T_h/10, then 1 - T_c/T_h = 90% (very efficient!) and Q_c = 0.1*Q_h (not much waste heat!). Now turn this around, because it’s reversible. A heat pump removes Q_c from the cold side and exhausts Q_h to the hot side. If you remove 1J from the cold side, you exhaust 10J to the hot side, and that 9J difference is the work done. That is, it costs 9J to remove 1J of heat from a 27.3K refrigerator if the exhaust is at 273K. It’s worse at 4.2K. (You can do an equivalent calculation a little more tidily by considering the entropy change of each side. To remove a given amount of entropy from the cold side, you must add at least as much entropy to the hot side. Plug in the usual formula for isothermal entropy change, and you get the answer.)
  - contravariant 6y ago
    A I see where my mistake is now, thanks! I suppose it makes sense that reversing the process would also meant that the 'heat intake' is equivalent to the exhaust heat, which for an efficient engine wouldn't be all that much. I suppose that means that the cooling requirement will increase the energy requirement by nearly 2 orders of magnitude.
- HPsquared 6y ago
  This is only true for heat pumps where the useful heat transfer occurs at the 'hot' side. In the ideal reversible case, COP_heating = T_h / (T_h - T_c), and is always greater than 1 (as the heat from work done is added to the heat transferred from the cold side). For refrigeration the situation is different: we are interested in the amount of heat _removed_ from the cold side. In this case, for a Carnot cycle COP_cooling = T_c / (T_h - T_c) and is NOT always greater than 1.