An engine does 10 J of work and exhausts 25 J of waste heat during each cycle.
what is the thermal efficiency?

I did it like this...
eff = \frac{work}{Qh}= \frac{10J}{15J} = 0.67 = 67%

But my homework said it was incorrect

So i tried it this way
eff = \frac{work}{Qh}= \frac{10J}{25J} = 0.4 = 40%

But that is also incorrect, so what am i doing wrong?

Question

An engine does 10 J of work and exhausts 25 J of waste heat during each cycle.
what is the thermal efficiency?

I did it like this...
eff = \frac{work}{Qh}= \frac{10J}{15J} = 0.67 = 67%

But my homework said it was incorrect

So i tried it this way
eff = \frac{work}{Qh}= \frac{10J}{25J} = 0.4 = 40%

But that is also incorrect, so what am i doing wrong?

irishboy123 · Answer

try 10/35 or ask @osprey

osprey · Answer

@IrishBoy123 I'd go for 10/35 also. 35 is total energy, implied by having 25 and 10 "accounted" for, and using conservation of energy. 10J out of the total of 35 is used. so 100 percent it ...
Mind you if we're both wrong ...

savy · Answer

I did it backwards, it is supposed to be 35, thank you

osprey · Answer

@Savy It is very very easy to get into a totally brain numbing muddle when doing physics, as I know at the moment to my frustration with endless bits of algebra that I seem to need to hack through. So, "doing it backwards" is probably par for the course, esp if you don't know what you're doing.