Answer:c>a>b
Explanation:
Given
All the engine extracts same amount of heat(Q) from High-temperature reservoir
For a) 400 and 500 K
[tex]\eta _{engine}=1-\frac{T_L}{T_H}[/tex]
[tex]\eta _{engine}=\frac{work\ supplied}{heat\ supplied}[/tex]
[tex]1-\frac{400}{500}=\frac{W_a}{Q}[/tex]
[tex]W_a=\frac{Q}{5}[/tex]
For b)500 K and 600K
[tex]1-\frac{500}{600}=\frac{W_b}{Q}[/tex]
[tex]W-b=\frac{Q}{6}[/tex]
For c) 400 K and 600 K
[tex]1-\frac{400}{600}=\frac{W_c}{Q}[/tex]
[tex]W_c=\frac{2Q}{3}[/tex]
So c will give the highest amount of work
c>a>b