A heat engine is operating in between 500 K and 400 K. If the engine absorbs 100 J heat per cycle, t — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Maximum possible efficiency (Carnot): eta_max = 1 - T_cold/T_hot = 1 - 400/500 = 0.20 (20%). Since η = 1 - Q_out/Q_in: Q_out >= 80 J (second law requirement). Heat rejected less than 80 J is thermodynamically impossible. Among the options: 75 J (A) and 70 J (D) are less than 80 J, so both are impossible.