Entropy in an irreversible cycle

Problem

A real heat engine operates between the same temperatures $T_c=500\;\text{K}$ and $T_f=300\;\text{K}$ as a Carnot engine, but with efficiency $\eta=0{,}25$ (rather than Carnot’s $0{,}40$). In one cycle it absorbs $Q_c=1000\;\text{J}$ from the hot reservoir. Calculate the entropy change of the universe $\Delta S_\text{univ}$ produced in one cycle and verify that it is positive.

Solution

Setup: the engine performs work $W$, releases $Q_f$ to the cold reservoir, and over one cycle its own entropy does not change (it returns to the initial state). The entropy of the universe is the sum of the changes of the two reservoirs.

Work produced from the real efficiency: $W=\eta\,Q_c=0{,}25\cdot 1000=250\;\text{J}$

Heat released to the cold reservoir (first law over the cycle): $Q_f=Q_c-W=1000-250=750\;\text{J}$

Entropy change of the hot reservoir (releases $Q_c$ at $T_c$): $\Delta S_c=-\frac{Q_c}{T_c}=-\frac{1000}{500}=-2\;\text{J/K}$

Entropy change of the cold reservoir (receives $Q_f$ at $T_f$): $\Delta S_f=+\frac{Q_f}{T_f}=+\frac{750}{300}=+2{,}5\;\text{J/K}$

Entropy balance of the universe: $\Delta S_\text{univ}=\Delta S_c+\Delta S_f=-2+2{,}5=+0{,}5\;\text{J/K}$

The result is positive, as it must be for an irreversible cycle (an ideal Carnot cycle would instead give $\Delta S_\text{univ}=0$). $\ev{\Delta S_\text{univ}=+0{,}5\;\text{J/K}>0}$

Linked atoms

Topics: Entropia e secondo principio Concepts: Entropia · Secondo principio della termodinamica · Rendimento Skills: Bilancio entropico Objects: Macchina termica