A reversible heat engine absorbs 40 kJ of heat at 500 K and performs 10 kJ of work rejecting the rem — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta S_{\text{univ}} = \Delta S_{\text{source}} + \Delta S_{\text{sink}} + \Delta S_{\text{engine}}$. Since the engine operates in a cycle, $\Delta S_{\text{engine}} = 0$. The source loses 40 kJ at 500 K: $\Delta S_{\text{source}} = -\frac{40000}{500} = -80 \text{ J/K}$. The sink gains 30 kJ at 300 K: $\Delta S_{\text{sink}} = +\frac{30000}{300} = +100 \text{ J/K}$. $\Delta S_{\text{univ}} = -80 + 100 = 20 \text{ J/K}$.