Discussion
A substance experiences an entropy change of \(\Delta s \gt 0\) in a quasi-steady process. The rise in temperature (corresponding to the entropy change \(\Delta s\)) is highest for the following process:
Answer: C
The temperature rise is highest for the isochoric process.
For a reversible process \(T\,ds = \delta q\). At constant volume \(\delta q = c_{v}\,dT\), and at constant pressure \(\delta q = c_{p}\,dT\), so for the same entropy change
\[ \begin{aligned} dT_{v} &= \frac{T\,ds}{c_{v}} \\ dT_{p} &= \frac{T\,ds}{c_{p}} \end{aligned} \]
Since \(c_{v} \lt c_{p}\), the isochoric rise is larger, by the factor \(\gamma = c_{p}/c_{v}\). In an isothermal process the temperature does not rise at all, and an isenthalpic process of an ideal gas also keeps the temperature constant, since enthalpy depends only on temperature.
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