The First Law of Thermodynamics — Question 9

Question 9

A network of state differences exposes an inconsistent measurement

Use $\Delta U=Q-W$, where $Q$ is heat into the system and $W$ is work done by it. Measurements give, from A to B, $Q_{AB}=600$ J and $W_{AB}=200$ J; from B to C, $Q_{BC}=-100$ J and $W_{BC}=150$ J. A separate C-to-A run reports $Q_{CA}=-200$ J and $W_{CA}=-100$ J.

1. Calculate the implied three internal-energy changes and test their sum around A–B–C–A.

2. If both first two runs and $W_{CA}$ are correct, infer the corrected $Q_{CA}$. State why the inconsistency cannot be repaired by choosing a different zero for internal energy.

Question 9 - Solution

A network of state differences exposes an inconsistent measurement

(a) Method and calculation

$$\Delta U_{AB}=400\,\mathrm J,\quad\Delta U_{BC}=-250\,\mathrm J,\quad\Delta U_{CA}=-100\,\mathrm J.$$$$\sum_{\mathrm{loop}}\Delta U=400-250-100=50\,\mathrm J\ne0.$$The reported state differences cannot all be correct.

(b) Interpretation and further calculation

$$\Delta U_{CA}=-(400-250)=-150\,\mathrm J.$$$$Q_{CA}=\Delta U_{CA}+W_{CA}=-150-100=-250\,\mathrm J.$$Replacing $U$ by $U+U_0$ leaves every difference unchanged, so the 50 J loop residual cannot be removed by a reference shift.

Physical check

State-function differences telescope to zero over a closed loop independently of path-dependent heat and work.