This may sound trivial but I am having a hard time linking different statements of second law. What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy. Also from second law, entropy of any isolated system never decreases. How are the two statements similar? Please use simple layman terms, I am simply trying to grasp the idea.
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1$\begingroup$ It is not to circumvent your question but Wikipedia has an article going through the different formulations providing also an historical overview. It seems sufficiently on layman terms and would be somehow redundant to re-do it here. Still you could get a nice summary from a willing user. en.m.wikipedia.org/wiki/Second_law_of_thermodynamics $\endgroup$– AlchimistaMay 27, 2021 at 10:19
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Please use simple layman terms, I am simply trying to grasp the idea.
Thermodynamics is a rigorous field, which is one of the things that makes it so powerful. Explaining something in simple layman terms will make it less powerful and more mysterious. One way around this is to pick and choose some parts that are necessary to answer the question and ask you to suspend your disbelief while following the argument. In this case, we need to know that in a ideal heat engine, the change in entropy of the reservoirs is:
$$ \Delta S_\mathrm{reservoir} = \frac{Q_\mathrm{rev}}{T}$$
Also, the overall change in entropy will never be negative (and for an ideal heat engine, it will be zero).
What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy.
You can figure that out because if you extract heat from a reservoir, the entropy of it increases. Your process has to increase the entropy of something else by at least the same amount, so you can't just turn thermal energy into work.
Also from second law, entropy of any isolated system never decreases. How are the two statements similar?
If it never decreases, it is zero at best. You can apply this general statement to the specific case of an ideal heat engine. In this process, you have to heat up the cold reservoir to match the entropy decrease of taking heat out of the hot reservoir. Because we are dividing by $T$, and it is different for the two reservoirs, we can extract some of the energy in form of work without breaking the second law. The higher the temperature difference, the more work we can extract for a given amount of heat transfer.
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$\begingroup$ $\Delta S_\mathrm{reservoir} = \frac{Q}{\mathrm{T}}$; the reversible restriction is unnecessary because, from the (infinite) reservoir's perspective, all heat flow is reversible. But you know that, so I'm curious what was your reason for including it? $\endgroup$– theoristMay 27, 2021 at 13:20
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$\begingroup$ ∆S = q/T, does this equation hold for irreversible? @theorist $\endgroup$ May 27, 2021 at 13:24
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$\begingroup$ @pravakarwang Only for the reservoir. The reservoir is assumed to be so large that heat flow from the system only has an infinitesimal effect on the reservoir, and thus is effectively reversible. $\endgroup$– theoristMay 27, 2021 at 13:25
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$\begingroup$ @theorist got it. I also wanted to ask why that equation holds only for reversible cases in general? Will be great if u can clarify without creating another post. Thanks $\endgroup$ May 27, 2021 at 13:28
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1$\begingroup$ @theorist Not everyone knows the implicit assumption that reservoir means infinite well-stirred reservoir. In a real heat engine, we would also talk of reservoirs, but these are neither infinite nor well-stirred. Also, it is hard in a simple equation to make the link that the equality holds only if we are talking about a reservoir, so folks jump to incorrect stuff like $$\Delta H_{rx} / T = \Delta S_{rx} $$ $\endgroup$– Karsten ♦May 27, 2021 at 13:40