Asked by anonymous

If a gasoline engine has an efficiency of 21 percent and loses 780 J to the cooling system and exhaust during each cycle, how much work is done by the engine?

Is it 3709.5???

Chemical energy in (as fuel and oxygen) = Work out + Heat out
21% is the ratio of Work out to Chemical energy in. That means that 79% is the ratio (Heat)out/(Chemical E)in.

The ratio of (work)out to (heat)out is 21/79

W/Q = .21/.79 = 21/79 = W/780 J

W = (780/79)*21 = ?
Answered by Anonymous
207.34
Answered by Henry
1. Work done = Energy output(Eo).
Eff. = Eo/Ein = Eo/(Eo+780) = 0.21
Eo = 0.21(Eo+780)
Eo = 0.21Eo + 163.8
0.79Eo = 163.8
Eo = 207.3 J.
Answered by someone
i don't understand henry's equation, maybe use Qh and Qc next time?
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