Question

A heat engine operating between 30°C and 450°C has 40% of maximum possible efficiency. If its power output is 125kW, at what rate does the engine absorb heat?

Answer 1

Maximum efficiency is given by:1-Tl/Th, where Tl is lower temperature and Th is higher temperature(in kelvins).

Here,Tl=30 C = 30+273.16 K = 303.16 K.

Similarly,Th=450 C=450+273.16 = 723.16 K.

So,maximum efficiency = 1-303.16/723.16 = 0.580784335

So, actual efficiency =40% of maximum efficiency = 0.4*0.580784335=0.2323137.

Now, efficiency = (power output)/(rate of heat absorption)

So, 0.2323137 = 125/(rate of heat absorption)

=>rate of heat absorption = 125/0.2323137 = 538.07 kW.