Problem — Balances of a cycle

Problem

A heat engine runs a cycle: it absorbs heat $Q_c = 800$ J from the hot reservoir and releases $Q_f = 600$ J to the cold one. What is the net work and the efficiency?

Solution

First law on the cycle. In a cycle the engine returns to the initial state, so $\Delta U = 0$ and all the net heat becomes work:

$W = Q_c - Q_f$

Net work.

$W = 800 - 600 = 200 \ \text{J}$

Efficiency. This is the fraction of absorbed heat converted into useful work:

$\eta = \frac{W}{Q_c} = \frac{200}{800} = 0{,}25$

$\ev{W = 200 \ \text{J}, \qquad \eta = 0{,}25 = 25\%}$

Of the $800$ J drawn from the hot reservoir, only $200$ J become work; the remaining $600$ J are 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 · Conservation of energy Oggetti: Heat engine