Heat transfer in the isothermal expansion in the Carnot cycle

Question

Problem

In the Carnot cycle there are two adiabatic and two isothermal processes. This question focuses on the isothermal expansion.

In the isothermal expansion books often writes that the system is in thermal equilibrium with a heat reservoir of temperature $T_1$, such that the system is also kept at $T_1$. Furthermore, it is often claimed that heat is being transferred from the reservoir to the system and that this heat is used to perform work, i.e. the gas expands against some piston.

However, how can heat be transferred from the reservoir to the system when they are both at the same temperature $T_1$?

Solution

The solution might be that the temperature of the system is infinitesimally smaller than $T_1$. If that were the case, then I understand that heat can flow from the reservoir to the system.

So why do I think that the system is infinitesimally smaller than $T_1$? I would guess the following:

Consider a reservoir at $T_1$ in thermal contact with a body with temperature $T_2<T_1$. By using Newton's law of cooling we can show that \begin{equation} T(t) = T_1 + (T_2-T_1)e^{-k t} \end{equation} where $k > 0$ is some cooling constant and $T(t)$ is the temperature of the system at time $t$. Note that it actually takes an infinite amount of time before the system is actually at the temperature $T_1$. Hence, the gas in the isothermal expansion is infinitesimally smaller than $T_1$ if we wait long enough before we start the Carnot cycle.

Question

Am I correct in saying that in the isothermal expansion the temperature of the system is actually not equal to $T_1$, it is actually infinitesimally less than $T_1$? Is there a way to derive this without using Newton's law of cooling?

If I am not correct and the system-temperature is actually equal to $T_1$ how can we extract heat from the reservoir and transfer it to our system?

Answer 1

You are exactly correct. To idealize the process as reversible (which is convenient and useful as an example of the highest energy efficiency for a heat engine, one in which no entropy is generated), we cannot have any energy move down a gradient, including a temperature difference. However, heat transfer occurs spontaneously only when there is a temperature difference. These conditions are contradictory. Therefore, as you note, we must assume that the temperature difference is very small—arbitrarily close to zero. And as you note, the consequence is that the power output of the Carnot cycle is zero, as such heat transfer occurs infinitesimally slowly.

Answer 2

If you say that the rate of heat flow is proportional to the temperature difference and that, during the expansion, the combination of heat energy gained from the reservoir and energy loss from doing work on the surroundings results in the temperature difference remaining constant, then $\dot{Q}=k\Delta T=\dot{W}$ and $\dot{U}=0$.