Physics I / Second Law of Thermodynamics and Entropy
Practice question · Estimate

A heat engine runs between a 600 K flame and a 300 K environment. Estimate the maximum possible efficiency, in percent.

Estimate on a scale from 0 % (0) to 100 % (100).

Hints
  1. Carnot’s limit: ηmax=1Tcold/Thot\eta_{max} = 1 - T_{cold}/T_{hot} (temperatures in kelvin).
  2. 1300/6001 - 300/600.
Show the answer

50 (answers within ±5 count)

Why

ηmax=1300/600=50%\eta_{max} = 1 - 300/600 = 50\%, and no engineering cleverness can beat it, because it follows from the second law itself. Real engines between these temperatures manage perhaps 30%. Waste heat is not a design flaw; it is physics rent.

Read the lesson: Second Law of Thermodynamics and Entropy →

Practise Second Law of Thermodynamics and Entropy

The app has 6 more questions on this lesson, and keeps your place in the course. Physics I is free to start.

More questions on Second Law of Thermodynamics and Entropy