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

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.

• The second law simply states that, in absence of any other effect, heat only moves from hot to cold. That's it. Why you would not want to use that is beyond me. The second law of thermodynamics is essentially the definition of temperature (find me another, if you can). There is nothing more exciting behind it. I would, by the way, refrain from using the word "universe" in anything relating to thermodynamics. Use "closed system", instead. That way you are not fooling yourself into believing that the observed universe, at the current level of our knowledge, is actually a closed system. – CuriousOne Dec 26 '14 at 5:03
• The second law is never violated, by the way. You don't need a criterion for it to be true. The only question is if your choice of system boundaries is correct, but that has nothing to do with the second law. – CuriousOne Dec 26 '14 at 5:05
• Thanks for your reply it is good to hear your point of view. To address your second comment I understand that the second law can never be violated in the real world. If not it would not be a law! I am trying to evaluate proposed processes to see if they can really happen in the real world. For example, "An inventor claims he can increase the heat recovery of your powerplant by ...". – Appguy1 Dec 26 '14 at 5:24
• As for the question of why I would want to have an analytical way of determining whether a claimed process violates the second law, you are right that for the two body system it is not necessary, and we can tell just by looking. However, if I want to analyse more complex systems (eg 3 + bodies, work etc) it may not be as obvious when a violation has occurred. If my only option is to use my derivative method then so be it, but I am just wondering if a simpler method is already established. – Appguy1 Dec 26 '14 at 5:28
• If you want to check a technical process, there is software for that which actually uses the correct thermodynamic data for working media (like steam) and common chemical reactions. It also has the advantage of being able to handle problems that can only be solved numerically. At best you are trying to reinvent the wheel. – CuriousOne Dec 26 '14 at 5:29

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.

• Thank you for your answer, however Carnot's theorem does not apply to my system, because it is not a heat engine. A heat engine is defined as a device that: "Uses energy provided in the form of heat to do work and then exhausts the heat which cannot be used to do work." Reference: hyperphysics.phy-astr.gsu.edu/hbase/thermo/heaeng.html My scenario does not extract work from the reservoir, it simply consists of one body exchanging heat with another. So when I try to use Carnot's theorem there is no value for Q2. I will try to add a diagram to make my question clearer. – Appguy1 Dec 27 '14 at 1:50
• @Appguy1: "one body exchanging heat with another" is an engine, regardless if work is performed by your system or onto your system. – Geremia Dec 27 '14 at 2:27
• @Appguy1: Why does $Q_2=0$ in your situation? How are you planning to measure and ensure that, experimentally? – Geremia Dec 27 '14 at 2:28
• The scientific definition of a heat engine does not include two bodies that exchange heat without generating work. A second university level references is as follows: "A heat engine is a cyclic process that absorbs heat and does work on the surroundings." Reference: chem.arizona.edu/~salzmanr/480a/480ants/carnot/carnot.html Also I do not agree with your statement that if work is performed onto a system that transfers heat it would still be an engine. In this case it would be a heat pump (provided the energy is transferred from cold to hot). – Appguy1 Dec 27 '14 at 2:59
• As for how we could set up this system experimentally, we could achieve this in the following way. Place a large mass of temperature $T_R$ and a small mass of temperature $T_R - 20K$ into a well insulated container. This will ensure negligible heat transfer to the surroundings, and can be measured by measuring the temperature of the outer surface of the container to ensure that no temperature change occurs. To be clear I am not suggesting this would result in the final temperatures in my question, as I know this to be impossible. My question is regarding the simplest way to prove this. – Appguy1 Dec 27 '14 at 3:03

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.

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 $\Delta\,S<0$.