Answer
The new efficiency of the composite system is the same as the original engine. The new arrangement of engines is:
Work Step by Step
In the composite system, we have two Carnot engines with working systems $A$ and $B$. Substance $A$ accepts heat $Q_H$ heat at temperature $T_H$. It does $W_1$ work and rejects the remaining heat to a reservoir at temperature $T'$. Working substance $B $ accepts the remaining heat ($Q_H-W_1$) at temperature $T'$. $B$ does $W_2$ work and rejects the remaining heat to a reservoir at temperature $T_C$. We use the formula of the efficiency of the Carnot engine.
Efficiency of engine with substance $A=\dfrac{W_1}{Q_H}=1-\dfrac{T'}{T_H}$
Total efficiency of the composite system $=\dfrac{W_1+W_2}{Q_H}$
Efficiency of engine with substance $B$ is:
$\hspace{9mm} \dfrac{W_2}{Q_H-W_1}=1-\dfrac{T_C}{T'}$
$\Longrightarrow W_2=(Q_H-W_1)(1-\dfrac{T_C}{T'})$
$\Longrightarrow W_2=Q_H-\dfrac{T_C}{T'}Q_H-W_1+\dfrac{T_C}{T'}W_1$
Now by dividing both sides by $Q_H$:
$\hspace{9mm} \dfrac{W_2}{Q_H}=1-\dfrac{T_C}{T'}-\dfrac{W_1}{Q_H}+\dfrac{T_C}{T'} \dfrac{W_1}{Q_H}$
$\Longrightarrow \dfrac{W_2+W_1}{Q_H}=1-\dfrac{T_C}{T'}+\dfrac{T_C}{T'}\dfrac{W_1}{Q_H}$
$\Longrightarrow \dfrac{W_1+W_2}{Q_H}=1-\dfrac{T_C}{T'}+\dfrac{T_C}{T'}(1-\dfrac{T'}{T_H})$
$\Longrightarrow \dfrac{W_1+W_2}{Q_H}=1-\dfrac{T_C}{T'}+\dfrac{T_C}{T'}-\dfrac{T_C}{T_H}=1-\dfrac{T_C}{T_H}$
With just one working substance between temperatures $T_C$ and $T_H$, efficiency of the engine is $1-\dfrac{T_C}{T_H}$
Thus, the efficiency of the composite system is the same as that of the original engine.
