(a) 0.249 (24.9 %)
The maximum efficiency of a heat engine is given by
[tex]\eta = 1-\frac{T_C}{T_H}[/tex]
where
Tc is the low-temperature reservoir
Th is the high-temperature reservoir
For the engine in this problem,
[tex]T_C = 270^{\circ}C+273=543 K[/tex]
[tex]T_H = 450^{\circ}C+273=723 K[/tex]
Therefore the maximum efficiency is
[tex]\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249[/tex]
(b-c) 0.221 (22.1 %)
The second steam engine operates using the exhaust of the first. So we have:
[tex]T_H = 270^{\circ}C+273=543 K[/tex] is the high-temperature reservoir
[tex]T_C = 150^{\circ}C+273=423 K[/tex] is the low-temperature reservoir
If we apply again the formula of the efficiency
[tex]\eta = 1-\frac{T_C}{T_H}[/tex]
The maximum efficiency of the second engine is
[tex]\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221[/tex]