1. **State the problem:** A Carnot engine operates between a hot reservoir at temperature $T_H = 600K$ and a cold reservoir at temperature $T_C = 200K$. It produces work $W = 1000J$ per cycle. We need to find the heat rejected $Q_C$ to the cold reservoir per cycle. 2. **Formula and important rules:** For a Carnot engine, the efficiency $\eta$ is given by: $$\eta = 1 - \frac{T_C}{T_H}$$ The efficiency is also the ratio of work output to heat absorbed from the hot reservoir: $$\eta = \frac{W}{Q_H}$$ where $Q_H$ is the heat absorbed from the hot reservoir. 3. **Calculate efficiency:** $$\eta = 1 - \frac{200}{600} = 1 - \frac{1}{3} = \frac{2}{3}$$ 4. **Find heat absorbed from hot reservoir $Q_H$:** Using $\eta = \frac{W}{Q_H}$, rearranged: $$Q_H = \frac{W}{\eta} = \frac{1000}{\frac{2}{3}} = 1000 \times \frac{3}{2} = 1500J$$ 5. **Calculate heat rejected $Q_C$:** Energy conservation for one cycle states: $$Q_H = W + Q_C$$ Rearranged: $$Q_C = Q_H - W = 1500 - 1000 = 500J$$ **Final answer:** The heat rejected by the cold reservoir per cycle is $500J$.