A Carnot engine operates between and . If it absorbs 120 cal heat per cycle from the source, the hea — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$T_1 = 327 + 273 = 600 \text{ K}$. $T_2 = 117 + 273 = 390 \text{ K}$. For a Carnot engine, $\frac{Q_1}{T_1} = \frac{Q_2}{T_2}$. Heat rejected $Q_2 = Q_1 \left(\frac{T_2}{T_1}\right) = 120 \times \left(\frac{390}{600}\right) = 120 \times 0.65 = 78 \text{ cal}$.