Entropy of surroundings and heat

Question

Apparently the entropy change of a system can be found by

$$\mathrm dS=\frac{\mathrm\delta q_\mathrm{rev}}T$$

I am slightly confused about this because I have seen this being used when the temperature of the system and the surroundings are not the same. However for the heat exchange to be reversible between the system and surroundings, need not the temperature of the system and surroundings be the same?

I have a feeling this might have to do with reversibility with respect to the system and the surroundings. I do not know about this and had not heard of it before I read it in a response to this post:

So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir)

Could someone please elaborate on this?

Answer 1

Yes, that is a perfectly acceptable formula for entropy change even outside of thermal equilibrium.

And, as you say, it's related to the irreversibility of heat transfer. The point about ideal thermal reservoirs always exchanging heat reversibly is not necessary to understand this, so let's park it outside for a moment.

Let's imagine a body in an environment. We'll call the temperature of the body $T_b$ and the temperature of the environment $T_e$. If they exchange heat, their respective entropies, $S_b$ and $S_e$, change according to the formula you mentioned:

$dS_b=\cfrac{\delta q}{T_b}$

$dS_e=\cfrac{- \delta q}{T_e}$

Note the sign (the heat flowing into the body is leaving the environment, or vice versa).

Also note, and this is critical, that the magnitude of $\delta q$ is equal in both equations (one receives exactly the same amount of heat as the other loses), but, if $T$ is different, the entropy change won't be the same. The change in entropy (positive or negative) will depend on the temperature of the body or the environment.

Let's have a look at how the total entropy of the system ($S=S_b+S_e$) changes.

If the body and the environment are at thermal equilibrium, in other words, their temperature is the same ($T_b=T_e=T$),

$dS=dS_b+dS_e=\cfrac{\delta q}{T}-\cfrac{\delta q}{T}=0$

In other words, transferring heat between bodies at thermal equilibrium does not change the total entropy of the system.

However, if they are at different temperatures,

$dS=\cfrac{\delta q}{T_1}-\cfrac{\delta q}{T_1}=\delta q \left( \cfrac{1}{T_1}-\cfrac{1}{T_2}\right)$

Now, the sign of the entropy change depends on the relative values of $T_1$ and $T_2$:

If $T_1>T_2 \rightarrow 1/T_1 < 1/T_2 \rightarrow 1/T_1 - 1/T_2 < 0 \rightarrow dS < 0$

If $T_1<T_2 \rightarrow 1/T_1 > 1/T_2 \rightarrow 1/T_1 - 1/T_2 > 0 \rightarrow dS > 0$

Note that I chose the signs of $\delta q$ assuming that the flux of heat went from the environment to the body. So, this means that if the body is hotter than the environment ($T_1 > T_2$), this would imply a reduction in the entropy of the whole system ($dS < 0$) - which the second principle explicitly forbids. And, to the contrary, if the body is colder than the environment, that flux will be spontaneous ($dS > 0$).

So, outside of thermal equilibrium, you still can use that expression to calculate entropy changes - as long as the heat flux goes in the right direction, from hot to cold. It will also apply to heat transfers in equilibrium (and therefore reversible) - although in that case you'll be having some other energy transfer process of equivalent magnitude siphoning that energy off - otherwise the heat would increase the temperature of part of the system, breaking thermal equilibrium (that won't happen).

Answer 2

That formula is used when you have the $Q_{reversible}$ for a system. However, most of the time you don't have $Q_{reversible}$ for the system, but always remember that the $Q$ of the system is actually $-Q_{reversible}$ for the surroundings. So therefore you can use that formula to always find the entropy of the surroundings!