The Zeroth Law of Thermodynamics — Question 2

Question 2

An insulating partition makes no equilibrium test

A and B are separated by an ideal adiabatic partition. Their temperatures remain $350$ and $290\,\mathrm K$ while no energy crosses it. Replacing it by a conducting link gives the supplied initial transfer law $P_{A\to B}=G(T_A-T_B)$ with $G=.40\,\mathrm{W/K}$.

1. Calculate the initial rate after replacement. Why was the earlier zero transfer not evidence of thermal equilibrium?

2. With the conducting link, a detector reports $|P|\le.020\,\mathrm W$. Bound $|T_A-T_B|$. State why the result depends on knowing that $G$ is nonzero.

Question 2 - Solution

An insulating partition makes no equilibrium test

(a) Method and calculation

$$\Delta T=350-290=60\,\mathrm K,\qquad P_{A\to B}=.40(60)=24\,\mathrm W.$$The insulating case has$$G=0\ \Longrightarrow\ P=0\quad\text{for every }\Delta T.$$Preventing interaction does not demonstrate equality of temperatures.

(b) Interpretation and further calculation

$$|\Delta T|=\frac{|P|}{G}\le\frac{.020}{.40}=.050\,\mathrm K.$$This is a finite-resolution equality test. If the link were broken, $G=0$ and division would be invalid; no temperature bound would follow.

Physical check

The zeroth law concerns mutual thermal equilibrium when thermal comparison is possible, not mere absence of observed transfer through an isolator.