The Zeroth Law of Thermodynamics — Question 1

Question 1

Equilibrium comparisons form connected temperature classes

Unchanged equilibrium bodies A, B, C, D and E are tested in pairs. A is in thermal equilibrium with B, B with C, and D with E. Separately, A is found hotter than D. Write $X\sim Y$ for thermal equilibrium. Assume each comparison tests the same retained body state.

1. List all pairs whose equilibrium follows from the zeroth law. How many distinct temperature classes are established?

2. Infer the ordering of C and E. Which extra comparison could merge the classes, and why would that contradict one of the original observations?

Question 1 - Solution

Equilibrium comparisons form connected temperature classes

(a) Method and calculation

Transitivity gives$$A\sim B,\ B\sim C\ \Longrightarrow\ A\sim C.$$$$\mathcal C_1=\{A,B,C\},\qquad\mathcal C_2=\{D,E\}.$$The unordered equal-temperature pairs are AB, AC, BC and DE; the hotter-than observation makes these two classes distinct.

(b) Interpretation and further calculation

$$T_C=T_A>T_D=T_E\ \Longrightarrow\ T_C>T_E.$$An equilibrium result such as $C\sim E$ would imply$$T_A=T_C=T_E=T_D,$$contradicting $T_A>T_D$. At least one result, retained-state assumption or comparison method would then require revision.

Physical check

Equilibrium is transitive for unchanged equilibrium states, not for labels attached to bodies whose temperatures change during testing.