Asked by MZAOEM

A heat engine operating between 100 degree celcius and 700 degree celcius has an efficiency equal to 40% of the maximum theoretical efficiency. How much energy does this engine extract from the hot reservoir in order to do 5000 J oe mechanical work?
Answered by bobpursley
So 40 percent of Energy equals 5000?

Energy= 5000/.4
Answered by tekalign
1005
Answered by chaltu
A heat engine operating b/n 100°c and 700°c has efficiency equal to 40% the maximum theoretical efficiency.how muh does this extract from the hot reservoir in order to do 500J of mechanical work?
Answered by Barii
202.6
Answered by Galma
Process of this answer?
Answered by Yonas workneh
Emax(carnot engine)=1-(Tc/Th)=1-(373/973)~0. 617
Actual efficiency is 30%of max efficiency so 30%(0. 617)~0. 185 which means. 18. 5 % effiecient u take that and plug it in the normal effieciency formula for engine which is (Qh-Qc)/Qh we have been given the value of work output which in other words mean Qh-Qc= 5000 plug that and find the value for Qh , Qh=5000/0. 185~ 27027. 8J
Answered by Yonas workneh
My bad make it 40% and the answer becomes 20270. 8J
Answered by Firex
Natural
Answered by Firex
Heat engine
Answered by Naol
The actual efficiency is 40% of
maximum theoretical efficiency w/c is [1-(373K/973K)]=0.62. Actual efficiency =0.62*40% => 24.8% now we can find the heat of the reservoir!
0.248=5000J/Qh
Qh =5000J/0.248
QH =20161.3J
The energy required to extracted from the hot reservoir is equal to QC!
QH=QC+W
QC=QH-W
QC=20161.3-5000J
QC=15161.3J.
So the answer will be 15161.3J!
Answered by Wakjira
Qc-QH=by what formula
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