Ranking the Q/T ratio

Problem

Four systems undergo reversible processes. In each case the heat $Q$ absorbed and the temperature $T$ at which the exchange occurs are given: (A) $Q=100$ J, $T=500$ K. (B) $Q=200$ J, $T=500$ K. (C) $Q=500$ J, $T=250$ K. (D) $Q=400$ J, $T=400$ K. Rank the entropy changes $\Delta S = Q/T$ from smallest to largest.

Solution

For a reversible exchange at constant temperature $\Delta S = Q/T$. We compute the four values: $\begin{aligned} \Delta S_A &= \frac{100}{500} = 0{,}20\;\text{J/K} \\ \Delta S_B &= \frac{200}{500} = 0{,}40\;\text{J/K} \\ \Delta S_D &= \frac{400}{400} = 1{,}00\;\text{J/K} \\ \Delta S_C &= \frac{500}{250} = 2{,}00\;\text{J/K} \end{aligned}$

Ordering from smallest to largest: $\ev{A < B < D < C}$

Collegamenti

Argomenti: Entropy and the second law Concetti: Entropy · Temperature Competenze: Ranking reasoning · Entropy balance