A Carnot engine operates between a hot source and a cold sink, initially having a thermal efficiency — Thermodynamics and Thermochemistry Chemistry Question
Question
A Carnot engine operates between a hot source and a cold sink, initially having a thermal efficiency of $40\%$. When the temperature of the sink is subsequently reduced by $60^\circ\text{C}$, the efficiency increases to $55\%$. What are the original absolute temperatures of the source ($T_H$) and sink ($T_C$) respectively?
💡 Solution & Explanation
Carnot Efficiency $\eta_1 = 1 - \frac{T_C}{T_H} = 0.4 \Rightarrow \frac{T_C}{T_H} = 0.6$. After reducing the sink temperature, $\eta_2 = 1 - \frac{T_C - 60}{T_H} = 0.55 \Rightarrow \frac{T_C - 60}{T_H} = 0.45$. Expanding the second equation yields: $\frac{T_C}{T_H} - \frac{60}{T_H} = 0.45$. Substitute $0.6$ for $\frac{T_C}{T_H}$: $0.6 - \frac{60}{T_H} = 0.45 \Rightarrow \frac{60}{T_H} = 0.15 \Rightarrow T_H = 400 \text{ K}$. The initial sink temperature is $T_C = 0.6 \times 400 = 240 \text{ K}$.