Problem — Entropy in a Carnot cycle

Problem

A Carnot engine absorbs $1000\;\text{J}$ at $500\;\text{K}$ and releases $600\;\text{J}$ at $300\;\text{K}$. Calculate: (a) $\Delta S_c$ of the hot reservoir, (b) $\Delta S_f$ of the cold reservoir, (c) the total $\Delta S$ of the universe.

Solution

Entropy change of a reservoir. A reservoir at constant temperature $T$ exchanging a quantity of heat $Q$ (with sign) undergoes $\Delta S = Q/T$. The reservoir releases heat to the engine, so for it $Q<0$; the cold reservoir receives it, so $Q>0$.

(a) Hot reservoir. Loses $1000\;\text{J}$ at $500\;\text{K}$:

$\Delta S_c = \frac{-1000\;\text{J}}{500\;\text{K}} = \ev{-2{,}0\;\text{J/K}}$

(b) Cold reservoir. Receives $600\;\text{J}$ at $300\;\text{K}$:

$\Delta S_f = \frac{+600\;\text{J}}{300\;\text{K}} = \ev{+2{,}0\;\text{J/K}}$

(c) Universe. The working fluid returns to its initial state ($\Delta S_\text{fluid}=0$ over a cycle), so the change of the universe is the sum of the two reservoirs:

$\Delta S_\text{univ} = \Delta S_c + \Delta S_f = -2{,}0 + 2{,}0 = \ev{0\;\text{J/K}}$

The zero result is the signature of reversibility: the ideal Carnot cycle generates no entropy. Any real engine would have $\Delta S_\text{univ} > 0$.

Collegamenti

Argomenti: Entropy and the second law · Heat engines Concetti: Entropy · Carnot cycle · Second law of thermodynamics Competenze: Entropy balance Oggetti: Heat engine