1. **State the problem:** A heat engine performs 100 J of work per cycle with an efficiency of 25%. We need to find the heat absorbed ($Q_{in}$) and heat rejected ($Q_{out}$) in each cycle. 2. **Formula and explanation:** Efficiency ($\eta$) of a heat engine is defined as the ratio of work done ($W$) to heat absorbed ($Q_{in}$): $$\eta = \frac{W}{Q_{in}}$$ Given efficiency $\eta = 0.25$ and work done $W = 100$ J. 3. **Calculate heat absorbed:** Rearranging the formula: $$Q_{in} = \frac{W}{\eta} = \frac{100}{0.25} = 400 \text{ J}$$ 4. **Calculate heat rejected:** Heat rejected is the difference between heat absorbed and work done: $$Q_{out} = Q_{in} - W = 400 - 100 = 300 \text{ J}$$ 5. **Summary:** - Heat absorbed per cycle: 400 J - Heat rejected per cycle: 300 J This means the engine absorbs 400 J of heat, does 100 J of work, and rejects 300 J of heat each cycle.