7. A Carnot engine has an efficiency of 0.45%. If the operating temperature in Celsius is such that Ti-0.30TH, what is the operating temperatures of this engine? Solution given e = 0.45% = 0.0045 let T1 and T2 cold and hotreservir temperatures respectively we know, e = 1 - (T1 + 273)/(T2 + 273) = 0.0045 = 1 - (0.30*T2 + 273)/(T2 + 273) ==> T2 = 1.77 C (TH) <<<<<-----------Answer T1 = 0.03*T2 = 0.3*1.77 = 0.531 C (TL) <<<<<-----------Answer .