Topic: Thermodynamics
A heat engine working with a thermal efficiency of 35% receives 2 kW of heat from a furnace. The waste heat rejected from the engine is
- A. 0.7 kW
- B. 0.35 kW
- C. 1.3 kW
- D. 1.65 kW
Answer: C. 1.3 kW
Explanation: Thermal efficiency ($\eta$) is given by $1 - (Q_R / Q_S)$, where $Q_S$ is the heat supplied and $Q_R$ is the waste heat rejected. Given $\eta = 0.35$ and $Q_S = 2$ kW, we can write $0.35 = 1 - (Q_R / 2)$. Solving for $Q_R$ gives $Q_R / 2 = 1 - 0.35 = 0.65$, therefore $Q_R = 0.65 \times 2 = 1.3$ kW.