# Carnot Engine Question

1. Oct 7, 2013

### 1question

1. The problem statement, all variables and given/known data

A Carnot heat engine absorbs heat form a high-temp reservoir at 1200K and concerts some of it into useful work and rejects the rest to a low-temp reservoir at 400K. A) What is the efficiency of the engine? b) If the engine absorbs 2400J per cycle from the high-temp reservoir at 1200K, what is the amount of work done, and how much heat is rejected to the low-temp reservoir at 400K?

2. Relevant equations

A) e = (T2-T1)/T2
ΔU=Q-W

3. The attempt at a solution

For part a, I got (1200-400)/1200= 0.6666...

For part b, my logic is that Total energy = work + heat rejected, based on ΔU formula above.
Therefore, work done should be 2400J*0.66 = 1600 J and heat rejected should be the rest, aka 2400*0.33 = 800J.

These answers seems to easy, especially for part b. Am I doing something wrong?

2. Oct 7, 2013

### Abhilash H N

Everything is fine, nothing to worry.....
Regards

3. Oct 7, 2013

### Simon Bridge

Well done. Sometimes physics is easy - with that reasoning it sounds like you have understood the material.

4. Oct 7, 2013

### 1question

Thanks for the confirmation.