Carnot engine: ΔS of the universe

Problem

A Carnot engine operates between $T_c=500\;\text{K}$ and $T_f=300\;\text{K}$, absorbing $Q_c=1000\;\text{J}$. Calculate: (a) the work $W$, (b) the heat released $Q_f$, (c) the $\Delta S$ of the two reservoirs, (d) the total $\Delta S$.

Solution

(a) The Carnot efficiency depends only on the temperatures: $\eta=1-\frac{T_f}{T_c}=1-0{,}6=0{,}40$. The work is $W=\eta Q_c=0{,}40\cdot 1000=400\;\text{J}$. (b) By the first law $Q_f=Q_c-W=1000-400=600\;\text{J}$. (c) The hot reservoir releases heat, the cold one absorbs it: $\Delta S_c=-\frac{Q_c}{T_c}=-\frac{1000}{500}=-2\;\text{J/K},\qquad \Delta S_f=+\frac{Q_f}{T_f}=+\frac{600}{300}=+2\;\text{J/K}$ (d) Summing the contributions: $\Delta S_\text{univ}=\Delta S_c+\Delta S_f=-2+2=0$ The zero total $\Delta S$ confirms that the Carnot cycle is reversible. $\ev{W=400\;\text{J},\ Q_f=600\;\text{J},\ \Delta S_\text{univ}=0}$

Linked atoms

Topics: Entropia e secondo principio Concepts: Entropia · Ciclo di Carnot · Rendimento · Secondo principio della termodinamica Skills: Risoluzione di un ciclo termodinamico · Bilancio entropico Objects: Macchina termica