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I quote from Zemansky's "Heat & Thermodynamics";

"Imagine two systems A and B separated from each other by an adiabatic wall but each in contact with a third system C through diathermic walls, the whole assembly being surrounded by an adiabatic wall as shown in Fig. 1-2a. Experiment shows that the two systems will come to thermal equilibrium with the third and that no further change will occur if the adiabatic wall separating A & B is then replaced by a diathermic wall (Fig. 1-2b). If, instead of allowing both systems A & B to come to equilibrium with C ar the same time, we first have equilibrium between A & C and then equilibrium between B & C (the state of system C being the same in both cases), then, when A & B are brought into communication through a diathermic wall, they will be found to be in thermal equilibrium."

My question is;

  1. What does he exactly mean by "the state of system C being the same in both cases"? Does C get connected to A first and then after reaching thermal equilibrium with A, gets connected to B? Or do we have like 2 identical systems to C and we connect A to one and B to the other?

  2. If it means that C is just one system and we connect A first and then to B ( without C being in its initial condition before it was connected to A), then what I understand is that A & C will reach thermal equilibrium and will have same "temperature" (I know we still didnt define temperature yet but at least based on how it "feels") so afterwards when B is connected to C, C being at the same temp as A now, the temp of C will change to the equilibrium temp with B. So A and B will have different temperatures, so how come will they be at thermal equilibrium when connected? (No change will occur in either A or B).

This is the figure he refers to (https://i.stack.imgur.com/iiAe5.jpg)

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  • $\begingroup$ C could be a thermometer. Its state the same would mean "indicating the same value on its readout". $\endgroup$
    – user137289
    Jun 15, 2017 at 23:30
  • $\begingroup$ So he means the state of C after reaching thermal equilibrium with each of them is the same? $\endgroup$ Jun 15, 2017 at 23:32
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    $\begingroup$ It means that one has two identical systems identified as C. Equivalently one can consider just one system C but it has to be a thermal reservoir so A would not change the state of B. $\endgroup$
    – Diracology
    Jun 15, 2017 at 23:40
  • $\begingroup$ As a side comment, I've found Zemansky particularly unreadable, although I don't have a good suggestion for an alternative book for thermodynamics at that level. (Maybe Schroeder's book?) $\endgroup$ Jun 16, 2017 at 1:08
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    $\begingroup$ Zemansky is tough sledding at times, but there is useful insight in that thicket that I didn't get from easier texts. That said, I prefer a statistically motivated approach to thermodynamics over the historical development. And like @Diracology I think that bringing in the notions of heat reservoir (a system with effectively infinite heat capacity) and thermometer (a system with effectively zero heat capacity) are useful for these kinds of arguments. $\endgroup$ Jun 16, 2017 at 15:04

2 Answers 2

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Your arguments are correct.

If $A$ and $C$ are brought in equilibrium initially, then for $A$ and $B$ to be in equilibrium, B has to be in equilibrium with that state of $C$ which was initially in equilibrium with $A$, without any transfer of heat energy.

In other words, $B$ and $C$ must be at the same temperature before they are brought in contact (Here, the system $C$ is the one in equilibrium with $A$).

Therefore, the state of system $C$ ,which the same in both cases, refers to that state of C which is in equilibrium with $A$ $\space$ and with $B$ (without any heat transfer). Only then $A$ and $B$ will be in thermal equilibrium.

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  • $\begingroup$ Why can't there be any heat transfer? Couldn't both A and B change C's observable characteristics in the same way? (Same effect due to heat transfer) but that wouldn't ensure that they'll be in thermal equilibrium? In other words, can A and B change C in the same way while not being at the same "hotness"? $\endgroup$ Jun 16, 2017 at 11:25
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    $\begingroup$ There can be heat transfer,of course. Two systems brought in contact will always try to attain equilibrium through heat transfer. After A and C get the same temperature, if B and C dont have the same temperature (without contact) then the common temperature of B and C (after contact) will not be equal to A's. Which means that A and B aren't at the same temperature. The best way to state the zeroth law is by saying that when A and B are simultaneously in equilibrium with C then it is certain that A and B have the same temperature (even if A and B are separated by an adiabatic wall). $\endgroup$
    – Mitchell
    Jun 16, 2017 at 11:47
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your analysis is correct The way the sentence in Case2 is written,there will be no thermal equilibrium between A, B and C The sentence in the second case should read instead: If, instead of allowing both systems A & B to come to equilibrium with C at the same time, we first bring A in thermal contact with C and then B with C (while keeping A in contact with C)), then, when A & B are brought into communication through a diathermic wall, they will be found to be in thermal equilibrium

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