Consider the following derivation
There are two reservoirs at temp A & B at temp T1 & T2 .They are connected by a metal rod such that T1 > T2. Q amonut of heat is transfered from T1 to T2.
For reservoir A
entropy change dS=-Q/T1
For reservoir B
entropy change dS=+Q/T2
The net entropy change=-Q/T1 + Q/T2
The above derivation is extracted from "Engineering Thermodynamics" By P.K Nag..But it is also stated that for a heat transfer through a finite temp difference the entropy change > dQ/T..Then how is the entropy change in above process calculated..I am aware of the fact that in a irreversible process we imagine a reversible path between the initial and final state and hence entropy is caculated. But I don't see how that pricipal is applied here..can anyone help me?
1 Answer
In your case, heat conduction is considered to be quasi-static for the subsystems. Let $\Delta Q$ be positive number, energy transferred as heat. Then, change of entropy $S_2$ of the subsystem 2 is $\Delta Q/T_2$, change of entropy $S_1$ of the subsystem 1 is $-\Delta Q/T_1$. For the whole system containing both subsystems, the change of entropy is
$$ \Delta S = \frac{\Delta Q}{T_2} + \frac{- \Delta Q}{T_1}. $$
This is greater than 0, so $\Delta S > 0$. This is consistent with the fact that even quasi-static heat conduction, if it is across finite temp difference, is a non-equilibrium process.
But it is also stated that for a heat transfer through a finite temp difference the entropy change > dQ/T
There is single $T$ in that statement. This is meant to be temperature of reservoir the system is connected to.
We can apply it, say, for subsystem 2. The temperature of the reservoir connected is $T_1$ (there are no other systems considered). The statement then reads
$$ \Delta S_2 \geq \frac{\Delta Q}{T_1}. $$ So $\Delta S_2 = \Delta Q/T_2$ and the above non-equality hold at the same time, because different temperatures are to be used.
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$\begingroup$ Thanks for that..but i still have some doubt. Take a single reservoir . In first case the reservoir gains sone heat Q first through a finite temprature difference . Now in 2nd case it gains the same heat through a qasi static process. Correct me if i am wrong but the increase in entropy due to the same heat is greater in the first case as compared to second case. Does that mean that the reservoir in the second case has reached a different state of heigher entropy than the first but the internal energy is same in both of them? $\endgroup$ Commented May 16, 2015 at 11:54
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$\begingroup$ It is not possible to compare entropy changes if the processes are so vaguely specified. Quasi-static heat transfer and finite temperature difference heat transfer are not mutually exclusive - see your example above, where they are both present. $\endgroup$ Commented May 16, 2015 at 17:48
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$\begingroup$ "There is single T in that statement. This is meant to be temperature of reservoir the system is connected to". So you are saying T is not the temp of the system..I don't think I can agree with that...Because it is clearly stated that T1 is the temp of the reservoir A and T2 the temp of reservoir B $\endgroup$ Commented May 22, 2015 at 11:14
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$\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html..you can check this link $\endgroup$ Commented May 22, 2015 at 14:26
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$\begingroup$ Even if T1 is the temp of the reservoir A and T2 the temp of reservoir B, the Clausius theorem still says that $\int_{cycle} dQ/T_{reservoir} \leq 0$. In differential form, $dS\geq dQ/T_{reservoir}$. In your example, reservoir to the system 2 is the system 1 and the Clausius theorem leads to inequality $dS_2 \geq dQ/T_1$. The hyperphysics explanation of Clausius theorem has many isolated statements that do not follow any line of reasoning, some of them are unclear and some are wrong. I advise to not learn from websites, study good textbooks instead. $\endgroup$ Commented May 22, 2015 at 16:03