Carnot Engine Question: Efficiency & Work/Heat Rejected

[1question]

Homework Statement

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?

Homework Equations

A) e = (T₂-T₁)/T₂
ΔU=Q-W

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?

 

[Abhilash H N]

Everything is fine, nothing to worry...
Regards

 

[Simon Bridge]

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

 

[1question]

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

Thanks for the confirmation.