An ideal Carnot engine is composed of two reservoirs and a working fluid. The hot reservoir and the cold reservoir have temperatures $T_1$ and $T_2$ respectively, with $T_1>T_2$. The working fluid is in a phase transition and has temperature $T_1$ at the start of the Carnot cycle. It undergoes another phase transition at $T_2$ at the end of the cycle to return to its original state.

This is a P-V diagram of the Carnot cycle which proceeds in four steps:

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I'm particularly interested in the two stages (from 1 to 2) and (from 3 to 4). They can be described as follows:

1) Stage (from 1 to 2) is a reversible isothermal expansion of the working fluid to transform from the liquid state to the gaseous one. The working fluid is at $T_1$ and it happens to have boiling point at $T_1$. Hence, heat $Q_1$ is supplied to the fluid from the hot reservoir until it transforms to a gas keeping its temperature constant along the whole process. (That the fluid's temperature is constant during the whole process is owing to it being in a phase transition.)

2) Stage (from 3 to 4) is a reversible isothermal compression, and it's similar to what we have just described, with the difference being in this case, heat $Q_2$ is drawn out of the fluid and transfers to the cold reservoir, and the fluid transforms from gas to liquid retaining a constant temperature of $T_2$ throughout the whole process.

I'm puzzled by the mechanism by which the working fluid undergoes phase transition. So, at stage (from 1 to 2), both the fluid and the hot reservoir have the exact same temperature, so that they're in a thermal equilibrium. Hence, there should be no heat or energy exchange between the two bodies. The same can be said of stage (from 3 to 4).

So how is it possible for heat to flow from two bodies having the exact same temperature?

  • $\begingroup$ You're also assuming that the liquid phase is compressible (from point 4 to 1)? $\endgroup$ – Soba noodles Sep 28 '15 at 15:25
  • $\begingroup$ @Sobanoodles Yes. $\endgroup$ – Omar Nagib Sep 28 '15 at 17:21

The only reason the reservoir and the fluid in 1-2 are at the same temperature is that the fluid is boiling. Heat is being absorbed by the fluid, and providing the energy required to cause a phase change until all the liquid has been converted to vapor. If the working fluid were not undergoing a phase change, the process would not be isothermal: both the reservoir and the fluid would be increasing in temperature.

Likewise, in 3-4, the reservoir and fluid are losing heat to the environment, and the temperature would be falling if not for the release of energy of fusion.

  • $\begingroup$ I know that the fluid is boiling, I already stated that in my question(that the working fluid has a boiling point at $T_1$ and it's in a phase change), my question is: how it is possible for heat to be absorbed by the fluid when it has the same temperature as that of the hot reservoir? $\endgroup$ – Omar Nagib Sep 28 '15 at 17:19
  • $\begingroup$ for the heat exchange to take place you only need infinitesimally different temperatures between the bodies, the heat exchange will be reversible but very slow. $\endgroup$ – hyportnex Sep 30 '15 at 13:41
  • $\begingroup$ there is always heat exchange..that is the point...even when the temperature is the same. With different temperatures this exchange is more on one side than the other. Isothermal process is one that is very slow, unlike adiabatic, where we assume very quick, so quick so no heat can be transfered.. $\endgroup$ – Žarko Tomičić Jun 24 '18 at 9:13

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