Relating Entropy and the 2nd law

[guyvsdcsniper]

So I am midway through my Thermodynamics course in college and still feel a bit unsure about the 2nd law and entropy.

I've learned that the 2nd law was states 3 different ways, and by contrapositive proofs we can determine they are all equivalent. What we end up getting for the 2nd law is basically, "A cooler object never heats a hotter object". Thats as simple as I believe I could state it with where I am at in my course at the moment.

Then I learned about entropy, which is a bit hard to grasp intuitively. My professor told me we can't truly define entropy until we get to Statistical Mechanics. I may butcher his words a bit, but he stated for now we can just think of entropy as a state variable, and we can use it to find other variables. We worked through an example where a heat reservoir and a Joule apparatus were connected and we found only the reservoir had an increase of entropy equal to mgh/T. I believe he also said for now we can think of it as a transfer of thermal energy. Again I may be butchering his words and he probably stated this a lot more precisely.

Again I am probably messing up somethings my professor said but I believe theyre pretty close and I am only halfway through my Thermo course.

So how do the 2nd law and entropy relate? When you ask someone what's the 2nd law of thermodynamics, they say Entropy. But according to Clausius its "A cooler object never heats a hotter object".

 

[Chestermiller]

Check out this link to my Physics Forums Insight Article on Entropy and the 2nd law:
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

Also consider these basics:

1. Just like internal energy U, entropy S is a physical (state) property of the material(s) comprising a system at thermodynamic equilibrium, and the entropy change between two thermodynamics equilibrium states of a system depends only on the two end states (and not on any specific process path between the two end states).

2. For a closed system, there are only two ways that the entropy of the system can change:

(a) by heat flow across the system boundary with its surroundings at the temperature present at the boundary$T_B$; this is equal to the integral from the initial state to the final state of $dQ/T_B$ along whatever path is taken between the two end states.

(b) by entropy generation within the system as a result of irreversibility. The mechanisms for entropy generation within a system experiencing an irreversible process include (a) heat conduction with finite temperature gradient, (b) viscous dissipation of mechanical energy to internal energy resulting from finite velocity gradients within the system, and (c) molecular diffusion with finite concentration gradients.

3. Contribution (a) is present both for reversible and irreversible paths. Contribution (b) is positive for irreversible paths and approaches zero for reversible paths. For any arbitrary path between the two end states, the two contributions add linearly: $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$where $\sigma$ is the amount of entropy generated within the system as a results of irreversibility. This is the Clausius inequality in equation form, and constitutes a mathematical statement of the 2nd law of thermodynamics. It is compatible with all "word statements" of the 2nd law.

4. Determining the amount of entropy generation along an irreversible path is very complicated so, to determine the entropy change for a system between any initial and final thermodynamic equilibrium states, we are forced to choose only from the set of possible paths that are reversible in applying our equation (i.e., paths for which $\sigma$ is zero). The reversible path we choose does not have to bear any resemblance to the actual path for the process of interest. All reversible paths with give the same result, and will also provide the entropy change for any of the irreversible paths.

5. One can determine the change in entropy for a closed system experiencing any process, provided that application of the 1st law of thermodynamics is sufficient to establish the final thermodynamic equilibrium state.

Questions?

 

[andresB]

"A cooler object never heats a hotter object".

"A cooler object never spontaneously heats a hotter object".