An engineer wants to run a heat engine with an efficiency of 40% between a high-temperature reservoir at 310C and a low-temperature reservoir. What is the maximum Celsius temperature of the low-temperature reservoir?
I appreciate the help!
I appreciate the help!
-
1st law of thermodynamics: Q1 = W + Q2 where
Q1 = heat taken from hot reservoir at temp. T1 (per engine cycle)
Q2 = heat expelled to cooler reservoir at temp. T2
Then, definition of efficiency e = W/Q1 = (Q1-Q2)/Q1
2nd law: entropy change of universe cannot be negative:
Entropy lost by hot reservoir = Q1/T1
Entropy gained by cool reservoir = Q2/T2
Entropy change of working substance = 0 since a complete cycle is made and entropy is a state function
So, Q2/T2 = Q1/T1 is the best we can theoretically do (in real life, Q2/T2 > Q1/T1 always).
Therefore, e = [Q1 - Q1*T2/T1]/Q1 = 1 - T2/T1= 0.4
Solve for T2 = 0.6T1 = 0.6*(273 + 310) = 350K = 77C
Q1 = heat taken from hot reservoir at temp. T1 (per engine cycle)
Q2 = heat expelled to cooler reservoir at temp. T2
Then, definition of efficiency e = W/Q1 = (Q1-Q2)/Q1
2nd law: entropy change of universe cannot be negative:
Entropy lost by hot reservoir = Q1/T1
Entropy gained by cool reservoir = Q2/T2
Entropy change of working substance = 0 since a complete cycle is made and entropy is a state function
So, Q2/T2 = Q1/T1 is the best we can theoretically do (in real life, Q2/T2 > Q1/T1 always).
Therefore, e = [Q1 - Q1*T2/T1]/Q1 = 1 - T2/T1= 0.4
Solve for T2 = 0.6T1 = 0.6*(273 + 310) = 350K = 77C