Problem — Efficiency from exchanged heats (800-600 J)

Problem

A heat engine absorbs $Q_c = 800\;\text{J}$ from the hot reservoir and expels $Q_f = 600\;\text{J}$ to the cold reservoir per cycle. Calculate the efficiency.

Solution

Setup. In a cycle $\Delta E_\text{int} = 0$, so the net work is the difference between absorbed and released heat:

$L = Q_c - Q_f = 800 - 600 = 200\;\text{J}$

Efficiency. By definition $\eta = L/Q_c = 1 - Q_f/Q_c$:

$\eta = \frac{L}{Q_c} = \frac{200}{800} = 1 - \frac{600}{800} = \ev{0{,}25 = 25\%}$

Only a quarter of the absorbed heat becomes useful work; the rest ($600\;\text{J}$) is dumped into the cold reservoir, as required by the second law.

Collegamenti

Argomenti: Heat engines Concetti: Efficiency · First law of thermodynamics Competenze: Solving a thermodynamic cycle Oggetti: Heat engine