Calculate the ideal (Carnot) efficiency of a heat engine

1. Dec 11, 2005

lilkrazyrae

Calculate the ideal (Carnot) efficiency of a heat engine operating between 23.0 degrees and 515 degrees. How much heat would be rejected by the engine if 1.00 x 10^6 calories were taken from the high temperature resivoir.

Would you just use e=1-(T(c)/T(H))?
And for the second part i don't have any clue how to get started.

2. Dec 11, 2005

lightgrav

You have to use Kelvin ... .

efficiency is not "1" if your Energy input is split between 2 (or more) outputs. You have a million cal input, and only "e %" of them will be used for what you had intended. Once you get Tc/Th , you know what fraction are wasted.

3. Dec 11, 2005

Andrew Mason

The efficiency is defined as the ratio of work output to heat input:

Efficiency$$= W/Q_H = (Q_H-Q_C)/Q_H = 1 - Q_C/Q_H$$

It can be shown that for a Carnot engine (no entropy change), $Q_C/Q_H$ is equal to $T_C/T_H$.

That should enable you to answer the question.

AM

Last edited: Dec 11, 2005
4. Dec 11, 2005

lilkrazyrae

So for efficiency e =1-T(c)/T(h)=1-(298.18/788.15)=62.171%
And for the heat rejected .62171=x/1.00*10^6=6.22*10^5 cal.

Is this right

5. Dec 11, 2005

lilkrazyrae

Or would you have to include the 1 in the efficiency equation again?

6. Dec 12, 2005

lilkrazyrae

No Body knows?!?

7. Dec 13, 2005

Andrew Mason

You are trying to find the heat rejected. You have found the work done: W = 6.22 x 10^5 cal. The heat rejected (heat flow out to cold reservoir) plus the work done has to be equal to the heat flow into the engine.

$$Q_C = Q_H - W = 1 \times 10^6 - 6.22 \times 10^5 = 3.78 \times 10^5 cal.$$

AM