A heat engine with a thermal efficiency of 35% produces 750kJ of work. Heat adde
ID: 1818858 • Letter: A
Question
A heat engine with a thermal efficiency of 35% produces 750kJ of work. Heat added to the engine is transferred from a reservoir at 550 Kelvin, and the engine rejects heat to the surrounding air, which is at 300 Kelvin. Determine the amount of heat added to the engine and the amount of heat rejected by the engine.
I used a variation of the efficiency equation: Qsupplied = Wperformed / efficiency
where Qsupplied was found to be 2142.86 kJ
Then I used Qrejected = Qsupplied - Wperformed
where Qrejected was found to be 1392. 86 kJ
Would this be the correct procedure?
Explanation / Answer
It seems that you have the correct answer. The only thing i can think to mention is the carnot efficiency! this is given by: n_max = 1 - T_C/T_H Where: T_C = 300K T_H = 550K so n_max = 1 - 300/550 = 45.45% This means that the max efficiency is 45.45% and the actual efficiency is 35%. This means that the numbers you are working with are for a legitimate engine. Other than that I don't know what else to tell you. Cheers, Ian