The temperature-entropy (T-S) diagram of a re | vedclass

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(C) The heat exchanged in a process is given by $Q = \int T \, dS$,which represents the area under the curve in a $T-S$ diagram.$1$. Heat absorbed $(Q

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$0.33$

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(C) The heat exchanged in a process is given by $Q = \int T \, dS$,which represents the area under the curve in a $T-S$ diagram.$1$. Heat absorbed $(Q_1)$ from $A$ to $B$: The process $A ightarrow B$ is a straight line. The area under this line is a trapezoid with parallel sides $2T_0$ and $T_0$ and height $(2S_0 - S_0) = S_0$.$Q_1 = 	ext{Area under } AB = \frac{1}{2} (2T_0 + T_0) (2S_0 - S_0) = \frac{3}{2} T_0 S_0$.$2$. Heat rejected $(Q_2)$ from $B$ to $C$: The process $B ightarrow C$ is at constant temperature $T_0$. The area under this line is a rectangle.$Q_2 = 	ext{Area under } BC = T_0 (2S_0 - S_0) = T_0 S_0$.$3$. Efficiency $(\eta)$: The efficiency of a heat engine is given by $\eta = 1 - \frac{Q_2}{Q_1}$.$\eta = 1 - \frac{T_0 S_0}{\frac{3}{2} T_0 S_0} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.

The temperature-entropy $(T-S)$ diagram of a reversible engine cycle is given in the figure. Its efficiency is

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