The Zeroth Law of Thermodynamics — Question 3

Question 3

A drifting reference breaks a transitivity argument

A negligibly disturbing thermometer reaches equilibrium with body A and reads $310\,\mathrm K$. It later reaches equilibrium with B and reads $315\,\mathrm K$. The operator reports that both were in equilibrium with the same thermometer and therefore A and B were in equilibrium. Both bodies retained their original states.

1. Identify the two thermometer states and evaluate $T_B-T_A$. Explain precisely why the inference fails.

2. A stable reference R is instead maintained at $312\,\mathrm K$. Define $d_X=T_X-T_R$. Find $d_A,d_B$ and a condition involving these differences for A and B to be in equilibrium.

Question 3 - Solution

A drifting reference breaks a transitivity argument

(a) Method and calculation

The physical thermometer occupies different states, $R_1$ and $R_2$:$$A\sim R_1,\qquad B\sim R_2,\qquad T_{R_1}=310\ne315=T_{R_2}.$$$$T_B-T_A=315-310=5\,\mathrm K.$$The zeroth-law premise needs the same reference state, not merely the same object.

(b) Interpretation and further calculation

$$d_A=310-312=-2\,\mathrm K,\qquad d_B=315-312=3\,\mathrm K.$$$$T_A=T_B\ \Longleftrightarrow\ d_A=d_B.$$They need not both equal zero, but their differences from one unchanged reference must match.

Physical check

An equilibrium relation compares states; reusing an instrument does not freeze its thermodynamic condition.