Problem — Carnot efficiency and work (500-300 K)

Problem

A Carnot engine operates between $T_c = 500$ K and $T_f = 300$ K. Calculate the efficiency. How much work does it produce for every $1000$ J absorbed from the hot reservoir?

Solution

Carnot efficiency.

$\eta = 1 - \frac{T_f}{T_c} = 1 - \frac{300}{500} = 1 - 0{,}60 = 0{,}40$

Work produced. The efficiency links work and absorbed heat via $\eta = W/Q_c$:

$W = \eta \, Q_c = 0{,}40 \times 1000 = 400 \ \text{J}$

Heat released. From the first law on the cycle, $Q_f = Q_c - W$:

$Q_f = 1000 - 400 = 600 \ \text{J}$

$\ev{\eta = 40\%, \qquad W = 400 \ \text{J}, \qquad Q_f = 600 \ \text{J}}$

Of the $1000$ J drawn at $500$ K, $400$ J become work and $600$ J are dumped at $300$ K: this is the theoretical maximum allowed between these two temperatures.

Collegamenti

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