# 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