# Entropy and its Differential Treatment in the System vs Surroundings

My question is: Because entropy is a state function, we should be able to calculate $$\Delta S_{surr}$$ along a reversible path in all of these non-limiting cases, such that $$\Delta S_{surr}$$ is always zero. Why is this not done?

I believe the short answer would be, we don’t care.

The choice of the “system” in thermodynamics is completely arbitrary. We define it based on what it is we wish to study, and then focus on that. One person’s system can be another’s surroundings and vice versa. So there is no inherent reason for not calculating the entropy change of the surroundings.

I like to use the following as a definition of thermodynamics (key terms bold faced):

The study of energy transfer, between a system and its surroundings, in relation to the properties of substances.

A key part of this definition is “the properties of substances”, because whatever it is that we wish to study, we need information on its properties. So if we want to change our focus from the currently defined system to its current surroundings, we then need to redefine the surroundings as the system and study the effects of energy transfers to and from it in relation to the properties of its substances. Without knowledge of the properties of the surroundings, we can only come to some general conclusions about the effects of energy transfers to and from it.

For example with, regard to energy, we know that if a system does work on, or transfers heat to, the surroundings, there is an increase in the internal energy of the surroundings equal to the decrease in internal energy of the system. In other words, the first law applies equally to both the system and surroundings. However, without information on the properties of the surroundings we cannot determine the effect of the transfers on, say, the temperature of the surroundings.

With regard to entropy, we know that if heat $$Q$$ is transferred isothermally from the surroundings to the system, there will be a decrease in entropy of the surroundings of

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

And an increase in entropy of the system of

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

For a total entropy change of

$$\Delta S_{Tot}= +\frac{Q}{T_{sys}}-\frac{Q}{T_{surr}}$$

And therefore for all $$T_{surr}>T_{sys}$$, $$\Delta S_{Tot}>0$$. $$\Delta S_{Tot}=0$$ only for the reversible path where $$T_{surr}\to T_{sys}$$. In this example we needed properties of the surroundings, namely its temperature and that it can be considered a thermal reservoir.

Hope this helps.

Because entropy is a state function, we should be able to calculate $$ΔS_{surr}$$ along a reversible path in all of these non-limiting cases, such that $$ΔS_{total}$$ is always zero.

Actually, whenever you're calculating the entropy change in a process you are using a reversible path, since $$S$$ is only defined for quasi-static reversible processes. So in an irreversible process you will calculate the entropy changes of the system and its surroundings assuming each one went in a separate quasi-static reversible process connecting the initial and final states. And if you add the entropy changes for each, it'll not equal zero.

Let's look at an example: suppose body $$(1)$$ is our system with temperature $$T^{(1)}$$ and body $$(2)$$ is the 'surroundings' of the first with temperature $$T^{(2)}$$ such that $$T^{(1)} < T^{(2)}$$. Now we take the two bodies into contact with each other, allowing them to only exchange heat. We know from observation that heat will be transferred from body $$(2)$$ to body $$(1)$$ and not the other way arround, so this process is irreversible.

How can we calculate the entropy change of the two bodies? We assume that body $$(1)$$ went a quasi-static reversible process from it's temperature $$T^{(1)}$$ to the equilibrium temperature $$T_\text{eq}$$ and calculate the entropy along that path. We do the same to body $$(2)$$: take a reversible process connecting $$T^{(2)}$$ and $$T_\text{eq}$$ and calculate the entropy. The change in entropy of the composite system $$(1) + (2)$$ in an infinitesimal step will be $$dS = dS^{(1)} + dS^{(2)} = \frac{\delta Q^{(1)}}{T^{(1)}} + \frac{\delta Q^{(2)}}{T^{(2)}}$$ but since energy is conserved, $$\delta Q^{(2)} + \delta Q^{(1)} = 0$$, so $$dS = \delta Q^{(1)} \left (\frac{1}{T^{(1)}} - \frac{1}{T^{(2)}} \right ) > 0$$ for the whole process, since $$T^{(2)}$$ will always be bigger than $$T^{(1)}$$ until $$T_\text{eq}$$ is reached and $$\delta Q^{(1)}$$ is positive. So integrating yields $$\Delta S > 0$$ for the whole process.