Animated Solution for Physics - Thermodynamics: The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

When dealing with thermodynamic cycles, the Temperature-Entropy ($T-S$) diagram is an incredibly powerful tool. Just as the area under a $P-V$ curve gives us the work done, the area under a $T-S$ curve gives us the heat exchanged during a process. This stems directly from the second law of thermodynamics, where $dQ=TdS$.

In this problem, we are presented with a reversible engine cycle forming a right-angled triangle on the $T-S$ plane. Our goal is to find its efficiency. Let's break down the cycle process by process.

Process 1: The system goes from $(S_0,2T_0)$ to $(2S_0,T_0)$. Notice that the entropy is increasing ($ \Delta S > 0$). This means heat is being absorbed by the system. To find the exact amount of heat absorbed ($Q_1$), we calculate the area under this line down to the entropy axis. This shape is a trapezium.

Process 2: The system moves from $(2S_0,T_0)$ to $(S_0,T_0)$. Here, the entropy is decreasing ($ \Delta S < 0$), indicating that heat is being rejected. The area under this horizontal line is simply a rectangle.

Process 3: The system returns from $(S_0,T_0)$ to $(S_0,2T_0)$. The entropy remains constant ($S_0$), making this an isentropic process. Since $dS=0$, the heat exchanged is zero ($Q_3=0$).

The efficiency $\eta$ of a heat engine is the ratio of the net work done to the total heat absorbed. We can express this as:

Substituting the values we found:

An Elegant Alternative: We could also find the net work done directly. In a $T-S$ diagram, the net work done in a cycle is exactly equal to the area enclosed by the cycle loop. Here, it's the area of the right-angled triangle.

Now, using the fundamental definition of efficiency:

Both paths lead us to the same beautiful result. The efficiency of this reversible engine cycle is $1/3$.