Physics Asked by Khalid T. Salem on July 19, 2021
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;
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?
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)
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.
Correct answer by Mitchell on July 19, 2021
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
Answered by Becharaj on July 19, 2021
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