# Heat: heat input

Dx
A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

is 1.8Kw right?

e=W/q_h
.30 * 600

TY
dx

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Tom Mattson
Staff Emeritus
Gold Member
No, that would correspond to an efficiency of 0.33...

Dx
Originally posted by Tom
No, that would correspond to an efficiency of 0.33...
Is my equation wrong, what am i doing incorrectly tom?
Dx

Tom Mattson
Staff Emeritus
Gold Member
Originally posted by Dx
Is my equation wrong, what am i doing incorrectly tom?
Dx
I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input.

Dx
Originally posted by Dx
A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

is 1.8Kw right?

e=W/q_h
.30 * 600
yourt the man Tom! Thats what my symbols mean. e= efficiency, W=work ouot/Q_h=work in. so why did i get this wrong, tom.
Im lost in the math and dont know where? plz help?
Dx

HallsofIvy
Homework Helper
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

Tom Mattson
Staff Emeritus
Gold Member
Originally posted by HallsofIvy
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.
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.

Originally posted by HallsofIvy
You might be better off to take Tom's answers with a grain of salt and check them carefully.
Speaking of grains of salt,
600 watts/30% = 2,000 watts, not joules.

A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

is 1.8Kw right?

e=W/q_h
.30 * 600

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