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.

 

[rezeew]

Your calculations for part a are correct. As for part b, your logic is correct as well. The total energy absorbed from the high-temp reservoir is divided into work and rejected heat according to the efficiency of the engine. So, the work done by the engine would be 1600J and the heat rejected to the low-temp reservoir would be 800J. This is the fundamental concept of the Carnot engine - converting heat into work with maximum efficiency. Your answers seem reasonable and in line with the basic principles of thermodynamics. Keep up the good work!