A heat engine operating between energy reservoirs at 20C and 600C has 30% of the
ID: 2202579 • Letter: A
Question
A heat engine operating between energy reservoirs at 20C and 600C has 30% of the maximum possible efficiency. How much energy must this engine extract from the hot reservoir to do 1000 J of work? (Please explain how to do it)Explanation / Answer
First question Efficiency of an ideal Carnot engine is given by: ? = 1 - Tc/Th => Th = Tc / (1 - ?) at 30% efficiency Th = 283K / ( 1 - 0.3) = 404K at 70% efficiency Th = 283K / ( 1 - 0.7) = 943K => ? T = 943K - 404K = 539K You have to increase hot reservoir temperature by about 540°C to raise efficiency to the desired level. Second question Maximum possible efficiency is the efficiency of an ideal Carnot engine: ?_c = 1 - Tc/Th = 1 - 293K / 873K = 0.6644 So the actual efficiency of the engine is: ? = 0.3·?_c = 0.1993 Efficiency is the ratio of (useful) work done divided by heat extracted from hot reservoir ? = W/Qh Hence: Qh = W/? = 1000J / 0.1993 = 5017J