A heat engine operating between energy reservoirs at 20C and 600C has 30% of the max possible efficiency. How much energy does this engine extract from the hot reservoir to do 1000J of work? Looking for guidance to solve.
by 
Anonymous
ANSWERS: 2
-
0.3= 1000/Heat extracted. =1000/0.3 =3333.33 joules
-
You need to know what the max possible efficiency is, before you can take 30% of it. The max possible efficiency is 1 - cold_temperature_in_kelvin/hot_temperature_in_kelvin = 1 - 293/873 = about 0.66 and 30% of 0.66 is about 20% efficiency = work/(heat extracted from hot reservoir) 20% = 1000J/ QH QH = 1000/.2 = 5000 J
RELATED QUESTIONS
Copyright 2023, Wired Ivy, LLC