Answer:
The correct option is (e) η = 0.66
Explanation:
given data:
temperature at entry is 627 degree C
temperature at exit is 27 degree C
the efficiency of a engine is given as η
[tex]\eta = (1-\frac{T_{L}}{T_{H}})*100[/tex]
where [tex]T_{L}[/tex] is temperature in kelvin at exit point
[tex]T_{H}[/tex] is temperature in kelvin at entry point
[tex]\eta = (1-\frac{27+273}{627+273})*100[/tex]
η = 66.66 %
