4 ms·

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 1

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

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.