Question

Consider the atmosphere to be a heat engine, which converts absorbed radiation to kinetic energy of winds. Let the global mean radiation absorbed be 200 Wm$^{-2}$. In steady-state, if the global mean kinetic energy dissipation is 10 Wm$^{-2}$, then the efficiency of the atmospheric heat engine is _________% (round off to one decimal place).

Hint:

The efficiency of a heat engine can be calculated by dividing the work output (or useful energy dissipation) by the total heat input.

Answer 1

Correct Answer: 4.9

The efficiency of a heat engine is given by the formula: \[ \eta = \frac{\text{Work Output}}{\text{Heat Input}} \times 100 \] Here, the work output is the global mean kinetic energy dissipation, and the heat input is the global mean radiation absorbed: \[ \eta = \frac{10}{200} \times 100 = 5% \] Thus, the efficiency of the atmospheric heat engine is: \[ \boxed{5.0%} \]