A Carnot engine takes in $900$ J of heat per cycle from a reservoir at $T_h = 450$ K and rejects heat at $T_c = 150$ K. The work it delivers per cycle is $\_\_\_$ J.

600 (answers within ±0.5 count)

$\eta = 1 - T_c/T_h = 1 - 150/450 = 2/3$, so $W = \eta Q_h = \tfrac{2}{3} \times 900 = 600$ J. Efficiency fixes only the fraction of the intake that can become work; the remaining $300$ J must be dumped to the cold reservoir, and no refinement of the engine removes that debt.