# When doing entropy calculations, why do you use the reversible heat for the system but the actual heat for the surroundings?

Let there be a system and surroundings. Let the system undergo an irreversible process. I understand that to calculate the entropy change of the system, one can use a reversible path between the initial and final states instead because entropy is a state function.

What I don't understand is why you must use the actual quantity of heat that flowed when calculating the entropy change for the surroundings.

I'm looking for either a mathematical view or an intuitive view.

To make my questions mathematically precise....

Why is

$\Delta S_{system}= \int _1^2 \frac{dQ_{reversible}}{T}$

but

$\Delta S_{surroundings}= \int _1^2 \frac{dQ_{irreversible}}{T}$

• For calculating the entropy of the surroundings you should use a reversible processes as well. Besides, what is the physical difference between the system and the surroundings? It is just a matter of convenience what names you give. – Diracology Jan 16 '18 at 11:21
• @Diracology You are correct however what is the reversible heat change in the surroundings? Generally the surroundings are taken to be large enough that any changes in there state are negligible, however the reversible heat needed to bring about no change in state is clearly 0, which is clearly wrong (though unsurprising as having no change in state clearly implies no change in entropy). It is not obvious to me how to find the appropriate reversible path to calculate the change in entropy of the surroundings. – By Symmetry Jan 16 '18 at 11:43
• The second equation doesn't sound right. Where did you find it? – valerio Jan 16 '18 at 11:49
• @BySymmetry I don't think there is a general procedure to find such paths. You shall use the same tricks you use for the actual systems. – Diracology Jan 16 '18 at 12:00

It is a standard practice in textbook thermodynamics analyses to tacitly equip the surroundings with equipment that does not generate entropy during a process. One such piece of equipment is the so-called ideal constant temperature reservoir. The fluid in such a reservoir is assumed to have an infinite mass times heat capacity so that the temperature of the fluid remains virtually constant when it exchanges heat with the system. For such an ideal reservoir, the key operating characteristic is that its entropy change is equal to the heat it receives from the system $Q_{res}$ divided by its absolute temperature (irrespective of any irreversible entropy generation within the system):$$\Delta S_{res}=\frac{Q_{res}}{T_{res}}$$

The "ideal constant temperature reservoir" describes the limiting behavior of any finite thermal capacity reservoir, for which $$\Delta S=\int_{T_i}^{T_f}{mC\frac{dT}{T}}=mC\ln{(T_f/T_i)}$$ with $$Q=mC(T_f-T_i)$$where $T_i$ and $T_f$ are the temperatures in the initial and final thermodynamic equilibrium states of the reservoir and Q is the actual amount of heat received by the reservoir in the irreversible process.

If we combine these two equations, we obtain: $$\Delta S=mC\ln{\left(1+\frac{Q}{mCT_i}\right)}$$In the limit of mC becoming infinite, this reduces to:$$\Delta S=\frac{Q}{T_i}$$

I should also mention that the fluid within the ideal constant temperature reservoir is assumed to have either an infinite thermal conductivity or is extremely well-mixed, such that the temperature throughout the reservoir is uniform spatially at $T_{res}$, including (and specifically) at the interface with the "system."

• Thank you. I find it odd that none of my texts write this out explicitly. – Ben Jan 16 '18 at 15:16
• So do I. It would take so little to do and would clear up so much confusion among students. – Chet Miller Jan 16 '18 at 15:29
• I know this next question deviates from the original question, but is it a "reasonable" approximation to take mC to be essentially infinite? Doing some math showed me that the difference even at mC=20,000 (~5L of water) gives a less than 2% difference for entropy changes up to around 500 J/K (from a chemists perspective that's a just a bit bigger than standard changes). – Ben Jan 17 '18 at 16:35
• Hi. I really don't follow this question. – Chet Miller Jan 17 '18 at 17:07
• That's fine, don't worry about it. – Ben Jan 17 '18 at 17:08