Carnot cycle with efficiency > 1

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SUMMARY

The discussion centers on the theoretical possibility of constructing an engine using the Carnot cycle with two reservoirs, one at a negative temperature (T_1 < 0) and the other at a positive temperature (T_2). Participants conclude that such a configuration cannot yield an efficiency greater than 1, as negative temperatures, while numerically lower, represent a state that is hotter than infinite temperature. The consensus is that the Carnot cycle's efficiency equation would result in a negative efficiency, contradicting the laws of thermodynamics.

PREREQUISITES
  • Understanding of the Carnot cycle and its efficiency equation
  • Familiarity with thermodynamic temperature scales, particularly Kelvin
  • Knowledge of two-state paramagnet systems and their implications on temperature
  • Basic principles of energy conservation in thermodynamic processes
NEXT STEPS
  • Research the implications of negative temperatures in thermodynamics
  • Study the Carnot cycle and its efficiency calculations in detail
  • Explore the concept of two-state paramagnets and their thermal properties
  • Investigate the relationship between temperature scales and thermodynamic laws
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Students and professionals in physics, particularly those focused on thermodynamics, as well as engineers interested in the theoretical limits of heat engines.

paweld
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Is it possible (at least theoretically) to construct engine which uses Carnot cycle
and two reservoirs with temperatures T_1&lt;0&lt;T_2 (one of them has
negative temperature which is possible e.g. in case of two-state paramagnet).
Such cycle would have efficiency greater then 1.

(of course the reservoir with negative temperature is "hotter")
 
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No. Please see earlier threads.
 
paweld said:
Is it possible (at least theoretically) to construct engine which uses Carnot cycle
and two reservoirs with temperatures T_1&lt;0&lt;T_2 (one of them has
negative temperature which is possible e.g. in case of two-state paramagnet).
Such cycle would have efficiency greater then 1.

(of course the reservoir with negative temperature is "hotter")
I'm not following - you want the cold reservoir to be hot and the hot reservoir to be cold? If you plug that into the efficiency equation, you get an efficiency below zero! In other words, nothing.

And a temperature below zero...you mean below absolute zero? :confused:

http://en.wikipedia.org/wiki/Carnot_cycle
 
paweld said:
Is it possible (at least theoretically) to construct engine which uses Carnot cycle
and two reservoirs with temperatures T_1&lt;0&lt;T_2 (one of them has
negative temperature which is possible e.g. in case of two-state paramagnet).
Such cycle would have efficiency greater then 1.

(of course the reservoir with negative temperature is "hotter")

Temperature conventions that use negative values do not do so for any absolute physical reasons. They are more designed around human convenience (water's state changes for Celsius and I think human comfort for Farenheit).

That is why we have the Kelvin scale. You can't go below 0 Kelvin because the Kelvin temperature relates directly to energy, which can't be negative. Kelvin best represents the actual thermodynamics, which doesn't allow for a negative number.
 
Sounds like meaningless twaddle to me.
 
paweld said:
Is it possible (at least theoretically) to construct engine which uses Carnot cycle
and two reservoirs with temperatures T_1&lt;0&lt;T_2 (one of them has
negative temperature which is possible e.g. in case of two-state paramagnet).
Such cycle would have efficiency greater then 1.

(of course the reservoir with negative temperature is "hotter")

No. The reason is that negative temperatures, although numerically lower than zero, are higher than infinite temperature. From http://en.wikipedia.org/wiki/Negative_temperature#Heat_and_molecular_energy_distribution":

"The temperature scale from cold to hot runs:

+0 K, . . . , +300 K, . . . , +∞ K, −∞ K, . . . , −300 K, . . . , −0 K."

On the "other side", everything works in reverse: when you add energy, the temperature decreases; when you remove energy, the temperature increases. So let's assume you have a Thot that is negative and a Tcold that is positive. Going from Thot to −∞ K, you will need to add energy. Then going from +∞ K to some temperature Tx you will recover the amount of energy you spent "cooling" your negative temperature, making your net energy balance = 0. Finally going from Tx to Tcold will represent the actual energy you will recover. So the actual Carnot cycle have to be calculated with Tx as the hotter temperature.

Not a pro on the subject, that is just my understanding.
 
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