The Efficiency of Heat Engines: Calculating Work Output from 150J Heat Input

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The discussion revolves around calculating the work output of a heat engine with an efficiency of 35% and a heat input of 150J. The initial calculation of 52.5J for work output is presented but questioned for accuracy. Participants confirm the formula for efficiency as work output divided by heat input, leading to the same result of 52.5J. There is some confusion about whether the calculation is correct, but it is ultimately validated by another participant. The consensus indicates that the calculation method is sound, despite initial doubts.
Dx
A heat engine has efficiency of 35% and receives 150J heat per cycle. how much work dos it perfrom in each cycle?

I said 52.5J

w=e*q_h =
.35 * 150 =
52.5
its wrong

what did i do wrong with this calculation. unless it mean 150J of put in. can anyone please explain this to me.
TY!
dx :wink:
 
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Can you post the definition of efficiency from your book?
 
efficiency(e) = W(work out) / Q_H(work in)
 
Originally posted by Dx
efficiency(e) = W(work out) / Q_H(work in)

That's what I thought. OK, just plug-n-chug.

(You know, it's always possible that the book is wrong!)
 
Originally posted by Dx
A heat engine has efficiency of 35% and receives 150J heat per cycle. how much work dos it perfrom in each cycle?

I said 52.5J

w=e*q_h =
.35 * 150 =
52.5

I thought i did that like shown but aparently its not right. I just wanted to know does your answer match mine?
Dx:wink:
 
Yes, it does.
 
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