# Why are source and system at same temperature in a Carnot cycle

I am learning Carnot cycle and Carnot theorem from my textbook. I am also familiar with the graph of Carnot cycle featuring a pair of adiabatic and isothermal processes.

Say that the source is at a temperature $$T_2$$ and the sink at a temperature $$T_1$$. My query is only regarding the isothermal processes in it so I would take only isothermal compression or expansion.

It undergoes isothermal expansion from $$V_1$$ to $$V_2$$ being in contact with the sink at temperature $$T_2$$. Then the work done ($$n=1$$) would be $$w =-RT_2 ln\frac{V_2}{V_1}$$ Since the process is isothermal and the work done is negative due to expansion, a $$q$$ amount of heat is absorbed by the system from the source. We also know that heat flows only due to a difference in temperature, therefore the system must be at some temperature lower than that of the source.

But the formula above asserts the temperature of the system as $$T_2$$ which is same as source's temperature. This means that the source and the system must be at the same temperature $$T_2$$ which shouldn't be right either as in this case there would be no flow of heat and the process will be no longer isothermal. Could anyone figure out what has gone wrong?

• One ancient Greek dude said something along the lines that he's not going to be present at his own death, so why bother about it. Same thing here. When the system is at $T_2$, there can't be and isn't any process anymore. The process was before that. Aug 10 at 19:47

We depict Carnot's process in the $$T-S$$ diagram: Steps $$2$$ and $$4$$ are adiabatically reversible, also called isentropic, so we don't need to worry about those two. We focus in steps $$1$$ and $$3$$. Since they are isothermal, for an ideal gas, there is no change in the internal energy. Thus, $$q = -w$$ or \begin{align} w_{12} = -RT_1\ln\left(\frac{v_2}{v_1}\right)<0 \qquad q_{12} = RT_1\ln\left(\frac{v_2}{v_1}\right)>0 \tag{1/2} \\ w_{34} = -RT_2\ln\left(\frac{v_4}{v_3}\right)>0 \qquad q_{34} = RT_2\ln\left(\frac{v_4}{v_3}\right)<0 \tag{3/4} \\ \end{align}

For the surroundings, Carnot introduces the concept of a heat reservoir. This means "something" that can exchange an infinite amount of heat without a change of temperature. You may laugh at this, but they exist for practical purposes on earth. The most simple example is the ocean. We can dump a lot of high-energy toxic residuals.... Do you think the temperature of the ocean is going to change? No.

The entropy change for the a reservoir, since $$T$$ is constant, is just the famous formula $$\Delta S = Q/T$$.

We also remember the entropy change for an ideal gas as a function of temperature and volume $$\Delta S = \int_{T_1}^{T_2} \frac{C_V^\mathrm{ig}(T)}{T} \, \mathrm{d}T + R \ln\left(\frac{v_2}{v_1}\right) \tag{5}$$

First we will prove that the best outcome for the step $$1-2$$ is that the temperature of the system is equal to that of the surroundings. We will name this unknown temperature $$\color{magenta}{T_\mathrm{sur,I}}$$, as in the picture, in pink. Then we will discuss it.

Step $$1-2$$ Here, the fluid expands so the system does work. In order to maintain the temperature constant, the surroundings must give heat to the system. The entropy change of the system, by Eq. (5) since $$T$$ is constant, yields $$\begin{equation} \Delta S_\mathrm{sys,12} = R\ln\left(\frac{v_2}{v_1}\right) > 0 \tag{6} \end{equation}$$ For the entropy change of the surroundings, the heat that it loses is exactly Eq. (2), but with its sign reversed $$\begin{equation} \Delta S_\mathrm{sur,12} = \frac{-q_{12}}{T_\mathrm{sur,I}} \rightarrow \Delta S_\mathrm{sur,12} = -\frac{RT_1\ln(v_2/v_1)}{T_\mathrm{sur,I}} \tag{7} \end{equation}$$ Thus, the total entropy change in this step is the sum of Eq. (6) and (7) \begin{align} \Delta S_\mathrm{12} &= \Delta S_\mathrm{sys,12} + \Delta S_\mathrm{sur,12} \\ &= R\ln\left(\frac{v_2}{v_1}\right) - \frac{RT_1\ln(v_2/v_1)}{T_\mathrm{surr,I}} \\ &= R\ln\left(\frac{v_2}{v_1}\right) \left(1 - \frac{T_1}{T_\mathrm{surr,I}}\right) \tag{8} \end{align} What can be the best possible outcome for the heat engine? We know the answer, and is that outcome that minimizes the entropy. Not only that, we can achieve its limit given by the second law, and by Eq. (8) this means that $$\begin{equation} R\ln\left(\frac{v_2}{v_1}\right)\left(1 - \frac{T_1}{T_\mathrm{sur,I}}\right) = 0 \therefore \boxed{T_1 = T_\mathrm{sur,I}} \tag{9} \end{equation}$$ I leave to you the proof for step $$3-4$$, which is analogous, and derive that $$T_2 = T_\mathrm{sur,II}$$.

The idealization of Carnot engine stems from equaling Eq. (8) to zero. There is no need to do that, in fact, we can perfectly use a reservoir that is at a higher temperature than $$T_1$$. However, this inevitably brings some increase of entropy.

You are correct to affirm that to bring a change of temperature we need a driving force. This is what every phenomenological law about a flux tells us. Fourier's law states that a temperature gradient gives rise to a heat flux, Ohm's law that a density current flux gives rise to an electric potential difference, etc. Smaller is the temperature gradient, smaller will be the flux, and more time we will need to bring desired temperature changes to the system. This is an inevitable consequence of nature. We would like things to behave like Eq. (9) states, but things cannot move with Eq. (9). In real life, we need finite fluxes in order to move the system out of equilibrium, with the cost of an increase of entropy.

Eq. (9) represents a limit, in the sense that it is the best step for this heat engine. For practical purposes, to approach this in real life is not that crazy. Imagine your fluid in step $$1-2$$ is at $$\pu{99.8 °C}$$. It needs to recieve heat from a heat reservoir at a higher temperature. Boling water at $$\pu{100 °C}$$ can be chosen, and this is very near an isentropic operation, by applying Eq. (8). We will need a lot of boiling water, because as the reservoir gives heat, its temperature will go down. If the temperature of the system and the surroundings are equaled, beyond here, no temperature changes can be obtained.

• Hmm, the arrows or the legends of the diagram must be reversed, as reversible adiabatic compression/expansion does not cause temperature drop/raise, but vice versa. Aug 11 at 11:14
• @Poutnik Hello! Wow, yes, totally correct. Thank you very much for the comment, I have changed the labels now. Aug 11 at 13:21