These lecture notes covering week 8 of https://paradigms.oregonstate.edu/courses/ph441 include a small group activity in which students derive the Carnot efficiency.
These lecture notes are week 8 of https://paradigms.oregonstate.edu/courses/ph441.
Note that the figures come from https://paradigms.oregonstate.edu/media/figures/sankey-engines.py
This week we will be zooming through chapters 8 of Kittel and Kroemer. Chapter 8 covers heat and work, which you learned about during Energy and Entropy. Hopefully this will be a bit of review and catch-up time, before we move on to phase transitions.
As we reviewd in week 1, heat and work for a quasistatic process are given by \begin{align} Q &= \int TdS \\ W &= -\int pdV \end{align} But we can often make use of the First Law in order to avoid computing both of these (if we know how to find the internal energy): \begin{align} \Delta U &= Q + W \end{align}
We have a monatomic ideal gas, and you can use any of its properties that we have worked out in class. We can begin with what you saw in Energy and Entropy \begin{align} pV &= NkT \\ U &= \frac32 NkT \end{align} and we can add to that the results from this class: \begin{align} S&= Nk\left(\ln\left(\frac{n_Q}{n}\right) + \frac52\right) \\ F &= NkT\left(\ln\left(\frac{n}{n_Q}\right) -1\right) \\ n &= n_Q e^{-\beta\mu} \\ n_Q &\equiv\left(\frac{mkT}{2\pi \hbar^2}\right)^{\frac32} \\ \end{align}
Let us consider a simple cycle in which we start with the gas at temperature \(T_C\).
Putting these all together, the total work done is \begin{align} W &= NkT_H\ln 2 - NkT_C\ln 2 \\ &= \ln 2Nk (T_H-T_C) \end{align}
If we are interested in this as a heat engine, we have to ask what we put into it. This diagram shows where energy and entropy go. The engine itself (our ideal gas in this case) returns to its original state after one cycle, so it doesn't have any changes. However, we have a hot place (where the temperature is \(T_H\), which has lost energy due to heating our engine as it expanded in step 2), and a cool place at \(T_C\), which got heated up when we compressed our gas at step 4. In addition, over the entire cycle some work was done.
The energy we put in is all the energy needed to keep the hot side hot, which is the \(Q\) for step 2. \begin{align} Q_H &= NkT_H\ln 2 \end{align} The efficiency is the ratio of what we get out to what we put in, which gives us \begin{align} \varepsilon &= \frac{W}{Q_H} \\ &= \frac{\ln 2Nk (T_H-T_C)}{NkT_H\ln 2} \\ &= 1 - \frac{T_C}{T_H} \end{align}' and this is just the famous Carnot efficiency.
We could also have run this whole cycle in reverse. That would look like the next figure. This is how a refridgerator works. If you had an ideal refridgerator and an ideal engine with equal capacity, you could operate them both between the inside and outside of a room to acheive nothing. The engine could precisely power the refridgerator such that no net heat is exchanged between the room and its environment.
Naturally, we cannot create an ideal Carnot engine or an ideal Carnot refridgerator, because in practice a truly reversible engine would never move. However, it is also very useful to know these fundamental limits, which can guide real heat engines (e.g. coal or nuclear power plants, some solar power plands) and refridgerators or air conditioners. Another use of this ideal picture is that of a heat pump, which is a refridgerator in which you cool the outside in order to heat your house (or anything else). A heat pump can thus be more efficient than an ordinary heater. Just looking at the diagram for a Carnot fridge, you can see that the heat in the hot location exceeds the work done, preciesly because it also cools down the cold place.