The Zeroth Law of Thermodynamics — Question 10

Question 10

A loop of comparison offsets tests internal consistency

Three calibrated differential comparisons report $d_{AB}=T_A-T_B=+2.0\,\mathrm K$, $d_{BC}=+3.0\,\mathrm K$ and $d_{CA}=-4.6\,\mathrm K$. First treat the reports as exact for unchanged states.

1. Derive the loop closure condition and decide whether one set of temperatures can satisfy all three.

2. Now allow each report an independent hard error of $.20\,\mathrm K$. Can a consistent assignment exist? Give one set of corrected differences and temperatures, taking $T_C=300\,\mathrm K$.

Question 10 - Solution

A loop of comparison offsets tests internal consistency

(a) Method and calculation

Every scalar temperature assignment satisfies$$(T_A-T_B)+(T_B-T_C)+(T_C-T_A)=0.$$The reports instead sum to$$2.0+3.0-4.6=.40\,\mathrm K,$$so they cannot all be exact descriptions of unchanged states.

(b) Interpretation and further calculation

The sum uncertainty allows $[-.20,1.00]$ K, which includes zero. One permitted correction is$$(d_{AB},d_{BC},d_{CA})=(1.8,2.8,-4.6)\,\mathrm K.$$$$(T_A,T_B,T_C)=(304.6,302.8,300.0)\,\mathrm K.$$All three corrected relations hold. Compatibility does not identify which measurements were in error.

Physical check

Adding the same constant to all temperatures leaves every comparison unchanged; differences alone do not determine the absolute origin.