# Homework Help: The Carnot Engine Problem

1. Sep 22, 2010

### JSGhost

1. The problem statement, all variables and given/known data
A Carnot heat engine operates between temperatures Th and Tc.
(a) If Th = 515 K and Tc = 363 K, what is the efficiency of the engine?

(b) What is the change in its efficiency for each degree of increase in Th above 515 K?
? K-1

(c) What is the change in its efficiency for each degree of change in Tc?
? K-1

(d) Does the answer to part (c) depend on Tc?
Yes or No?

Explain.

2. Relevant equations
e_c = 1 - T_c/T_h

3. The attempt at a solution

(a)
T_h = 515k
T_c = 363k
e_c = 1 - T_c/T_h = 1 -363k/515k = 0.2951456311 = 0.30

I'm don't know how to start part (b). Do I use the same equation or a different one?

2. Sep 22, 2010

### kuruman

You want to find the rate of change of the efficiency e with respect to Th. Rate of change is also known as ...? (Hint: It's the "D" word.)

3. Sep 22, 2010

### JSGhost

So do the derivative of e = 1 - T_c/T_h ?

4. Sep 22, 2010

### kuruman

Yes.

5. Sep 22, 2010

### JSGhost

e = 1 - T_c/T_h

How do you do the derivative of that equation?

dE = -ln(T_h/T_c) ?

6. Sep 23, 2010

### kuruman

Would it help if you said let x = TC and wrote

e = 1 - x/TH?

Can you find de/dx ?

7. Sep 23, 2010

### JSGhost

Actually I did not have to do the derivative. It was so simple roflol.

b) Increase T_h by one and calculate the efficiency

1 - 363/516= 0.2965

calculate difference from the original efficiency

0.2965-0.2951=.0014

c) Decrease T_c by one and calculate

1 - 362/515= 0.2971

Now find the difference

.2951 - .2971= 0.002

d) No