Question

A heat engine operating between 40°C and 450°C has 21 % of maximum possible efficiency. If its power output is 125 kW, 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=40 C = 40+273.16 K = 313.16 K.

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

So,maximum efficiency = 1-313.16/723.16 = 0.566956

So, actual efficiency = 21% of maximum efficiency = 0.21*0.566956=0.11906.

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

So, 0.11906 = 125/(rate of heat absorption)

=>rate of heat absorption = 125/0.11906 = 1049.891 kW.