Homework Help: Heat: heat input

No, that would correspond to an efficiency of 0.33...

 

I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input.

 

You might be better off to take Tom's answers with a grain of salt and check them carefully. He seems to be doing the same thing you are: taking the numbers given and putting them together in the simplest way without regard for the correct formula.


In this case, the formula you give is: e=W/q_h and the values are efficiency= e= 0.30, Work out= W= 600.

In other words, 0.30= 600/q_h. Solving that, 0.30*q_h= 600 so
q_h= 600/0.30= 2000 Joules.

It should occur to you that, since efficiency is always less than 1, there must be more energy put in that you get out.

.30 * 600

 

No, Dx was quoting himself (not me) in his post above. The only "answer" I gave was...

"I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input."

...which is no different from what you posted.

 

Speaking of grains of salt,
600 watts/30% = 2,000 watts, not joules.

 

can anybody please tell me the formula for heat rate for fuels?