Simplest way to analytically determine whether a claimed heat transfer process obeys the second law of thermodynamics?

Question

I want to find the simplest method to determine whether a proposed heat transfer process violates the second law of thermodynamics. Specifically I am looking for a method that meets the following needs:

• A general method that can be used to analyse any heat transfer process.
• I want to do this without resorting to word definitions such as the Clausius statement of the second law.

To explain my progress so far, a few days ago the "obvious answer" that I would have given was that $\Delta S_{Universe}$ for the process must be $>=0$.

However, on closer inspection $\Delta S_{Universe}>=0$ does not always mean a process is possible. An example of a process that satisfies this criterion, but is clearly impossible, is using a thermal reservoir to heat a body from $T_R-20K$ to $T_R + 10K$.

The reason that this process is impossible is that the part of the process that involves heating from $T_R$ to $T_R+10K$ involves heat transfer from a cooler body to a hotter body, which results in a decrease of $S_{Universe}$.

Based on this insight I came up with the following rule: Every stage of the process must result in a positive $\Delta S_{Universe}$.

The way we can test for this condition is as follows:

1. Formulate an equation for $\Delta S_{Universe}$
2. Differentiate the above equation with respect to the state variable of interest (in this case the temperature of the body)
3. Set $ \displaystyle\frac{d\Delta S_{Universe}}{dT_{Body}} = 0$ in the above equation
4. Solve for $T_{Body}$, which will henceforth be referred to as $T_{MaxS}$
5. Check whether either of the following conditions are true: $T_{BodyFinal} < T_{MaxS} < T_{BodyInitial}$ or $T_{BodyInitial} < T_{MaxS} < T_{BodyFinal}$
6. If either of the above conditions are true, then by the mean value theorem, some part of the process must have resulted in a negative $\Delta S_{Universe}$. If not, then the process obeys the second law of thermodynamics.

The above method appears to be applicable to all processes, however the issue is that even for a simple two body system this method is difficult and time consuming to carry out. For a worked example of this method see the previous thread that I started:

Can calculations find positive entropy change for heat transfer from cold reservoir to hot body?

Therefore my question is: Does anyone know of a simpler method to analytically determine whether a proposed process obeys the second law of thermodynamics? This must also meet the criteria mentioned above.

I appreciate anyone's time and thank you in advance.

Answer 1

The simplest way would be to measure efficiencies. Carnot's theorem says that the maximal efficiency of a heat engine (or, as you call it, a "heat transfer process") is:$$\eta = 1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}.$$Clausius's based his concept of entropy, that$$\oint\frac{dQ}{T}\le0,$$on Carnot's theorem, where $dQ>0$ means heat is absorbed by the system and "$=$" means an irreversible cycle.

Thus, if Carnot's theorem is violated, so is the 2nd Law.

Answer 2

There is an error in your thinking. The appropriate condition for determining whether a process is possible is $\Delta S_\text{Universe} \ge 0$. Your mistake is in believing that the process you mention is impossible.

Let us imagine that in addition to the body (whose initial temperature is $T_0 = T_R-20K$) and the heat reservoir (with constant temperature $T_R$), have a reversible Carnot engine and a sealed piston of ideal gas. To heat the body up to $T_1 = T_R + 10K$ we have to perform three stages:

Stage 1: first we connect the Carnot engine up as a heat engine, taking its heat from the reservoir, dumping its waste heat into the body, and we use its work output to compress the gas piston. This is a reversible process with $\Delta S_\text{Universe}=0$. We let the engine run until the resulting waste heat brings the body up to $T_R$, at which point we can no longer extract any more work.

Stage 2: Now the body is at $T_R$, but we have some work stored in the compressed gas. In stage 2 we set the Carnot engine up to use this work to pump heat from the reservoir into the body, which we do until it reaches $T_0$. This is a reversible process with $\Delta S_\text{Universe}=0$.

Stage 3: We have now accomplished our objective, but a pedant might claim that we didn't leave the system in the same state it started in, because the gas in the piston is (probably) more compressed than it was initially. So in stage 3 we simply release the piston, letting it return to its initial state. This is an irreversible process with $\Delta S_\text{Universe} \ge 0$.

Whether this is possible or not will depend on how the body's heat capacity depends on its temperature. (For example, if the heat reservoir is at $-5^\circ\,\mathrm{C}$ and the body is a container of water, it will not be possible - the work stored in the piston will be used up while melting the water, so we can't heat it to $5^\circ\,\mathrm{C}$.) But the condition for whether it's possible or not just boils down to $\Delta S_\text{Universe}\ge 0$.

This is always the condition for whether a process is thermodynamically possible. If a process with $\Delta S_\text{Universe}\ge 0$ seems impossible then you've either failed to imagine the right way to do it (as in this case) or it's impossible for some other reason besides thermodynamics.

Answer 3

I think you should accept Nathaniel's answer, because the process you cite does indeed fulfill $\Delta\,S<0$, thus by the second law is impossible.

Another way to show this (less general and elegant than Nathaniel's answer) is to put definite heat capacities on your body and reservoir and do a simple calculation for the closed system comprising only body and reservoir of $\Delta S$ assuming constant heat capacities. You'll find you always have $\Delta\,S<0$ for this special system.

Now you can apply this argument to a sequence of arbitrarily small temperature changes making up a nonzero temperature change in the direction you state. Smaller and smaller temperature changes make the constant heat capacity assumption in each step more and more valid. Passing to the limit, by the argument in the foregoing paragraph, the integral that results has an integrand that is always negative throughout the noninfinitessimal process, thus showing that $\Delta\,S<0$.