Problem — Efficiency of a real engine (4000-3000 J)

Problem

A real heat engine absorbs $4000\;\text{J}$ from a hot source and releases $3000\;\text{J}$ to a cold source. Calculate the efficiency.

Solution

Net work. In a cycle $\Delta E_\text{int} = 0$, so

$L = Q_H - Q_C = 4000 - 3000 = 1000\;\text{J}$

Efficiency.

$\eta = \frac{L}{Q_H} = \frac{1000}{4000} = 1 - \frac{3000}{4000} = \ev{0{,}25 = 25\%}$

Only a quarter of the absorbed heat becomes useful work; the remaining three quarters are discharged into the cold source.

Links

Topics: Macchine termiche Concepts: Rendimento · Primo principio della termodinamica Skills: Risoluzione di un ciclo termodinamico Objects: Macchina termica