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"The entropy of the cold source in decreases when energy leaves it as heat, but when that heat enters the hot sink the rise in entropy is not as great (because the temperature is higher). Overall there is a decrease in entropy and so the transfer of heat from a cold source to a hot sink is not spontaneous."

This is the reason given in my physical chemistry book(Atkins,Paula) but doesn't the second law state that the process is spontaneous when change in entropy of the system and surrounding has to be greater than 0. Here the author only calculates the entropy change of the system. Why ?

Ali
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    Did try say energy is being transferred to/from the surrounding? It looks like they are considering an isolated system. – BioPhysicist Jan 22 '22 at 20:42
  • @BioPhysicist I want to ask something about this question asked on this site : https://physics.stackexchange.com/q/193500/232914. Will it be okay if I ask it in the comments ? And if not then how do I ask it ? – Ali Jan 22 '22 at 21:18
  • New questions should be asked in a new post. Not in the comments of another post – BioPhysicist Jan 22 '22 at 21:31

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We're considering the transfer of heat between two bodies. The two bodies make up the system. We assume that (perhaps because of insulation) no heat is exchanged between the system and anything else. Therefore there are no relevant surroundings to consider!

Philip Wood
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  • Welp, I guess I should have just answered this instead of asking the OP for clarification. – BioPhysicist Jan 22 '22 at 21:13
  • @Philip Wood can you help me with another unrelated question ? https://physics.stackexchange.com/q/193500/232914 Why does the OP say that the change in entropy of the surrounding is negative of the change of entropy of the system. I know that Q(surrounding) will be equal to -nRTln(V2/V1) but it has to be divided by the temperature of the surrounding not the temperature of the system ! Please inform me if asking this in the comments goes against the policies of the site. – Ali Jan 22 '22 at 21:26
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    @BioPhysicist In my experience this sort of thing happens quite often. Incidentally I hadn't seen your comment before posting my answer. Keep up your excellent posts. – Philip Wood Jan 22 '22 at 21:32
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    @Ali (a) I'm not good on site policies, but think that in general you should ask a separate question. You might get better answers than my simple response, which is ... (b) The entropy decrease of the gas's surroundings should indeed be calculated using the temperature of the surroundings. But for a reversible isothermal change (an idealised change) the temperature of the surroundings should be only infinitesimally higher than that of the expanding gas. The change will then, of course, be extremely slow! – Philip Wood Jan 22 '22 at 22:03
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Here the author only calculates the entropy change of the system. Why ?

The author is simply calculating the total change in entropy for a transfer of heat between a cold and hot source which, together, can be considered the system plus surroundings = the universe. Which is the system and which is the surroundings is irrelevant. The point is the spontaneous transfer of heat from cold to hot violates the second law since the entropy change of the system plus surroundings has to be $\ge0$. If heat transferred spontaneously from cold to hot the change would be $\lt 0$.

This can easily be seen by assuming the two sources are thermal reservoirs (constant temperature sources). Then, if heat transferred from cold to hot you would have for the cold reservoir

$$\Delta S_{C}=-\frac {Q}{T_{C}}$$

And for the hot reservoir

$$\Delta S_{H}=+\frac {Q}{T_{H}}$$

For a total entropy change of

$$\Delta S_{tot}=+\frac {Q}{T_{H}}-\frac {Q}{T_{C}}$$

Since $T_{C}<T_{H}$, $\Delta S_{tot}\lt 0$

Hope this helps.

Bob D
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