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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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2 Answers2

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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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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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