Same temperature, different heat

Problem

Two students are arguing. One says: “if $T_A = T_B$ then the gas has received no heat.” The other says: “not true, it depends on how the transformation went.” Who is right? Give an example with $T_A = T_B$ but $Q \neq 0$, and one with $T_A = T_B$ and $Q = 0$.

Solution

The SECOND student is right. Temperature fixes the internal energy, not the heat: $T_A = T_B$ only implies $\Delta E_\text{int} = 0$, but $Q$ depends on the path.

Example with $Q \neq 0$ — reversible isotherm. The gas expands at constant temperature: it receives heat $Q = n R T \ln\!\frac{V_B}{V_A} > 0$ and performs an equal amount of work on the surroundings. The internal energy does not change ($T_A = T_B$), but $Q \neq 0$.

Example with $Q = 0$ — free expansion into a vacuum. The gas is isolated ($Q = 0$) and expands into a vacuum ($L = 0$): $\Delta E_\text{int} = 0 \Rightarrow T_B = T_A$, and here $Q = 0$.

Conclusion. $T_A = T_B$ fixes $\Delta E_\text{int} = 0$, but does NOT fix $Q$: that depends on the path followed.

$\ev{\text{The second student is right}}$

Links

Topics: Thermodynamics Concepts: First law of thermodynamics · Thermodynamic transformations Skills: Limiting-case and thought-experiment analysis