what is wout for the heat engine shown in the figure? This is a topic that many people are looking for. thevoltreport.com is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, thevoltreport.com would like to introduce to you Mastering Physics #11.35 The figure shows three heat engines. For each engine calculate ΔE=. Following along are instructions in the video below:
Figure shows three heat. Engines for each engine. Calculate delta e.
And that qh work work out minus qc. Where qh is the amount of heat transferred from the hot qc is the amount of heat transferred to to the cold reservoir and work out is the energy output of the heat engine qh work out and qc e are all positive quantities. Which if any of the heat engines.
Violates authors first a lot of the thermodynamics then for each engine calculate the theoretical maximum efficiency emax and for each engine calculate the actual efficiency e and then which of any violates the second law of thermodynamics so lets go down. And do part a real quick. So they want us to know.
What is delta e. Delta e. And they say.
Its equal to q. H. Work.
Out minus qc. Its a qh work out minus qc. So the delta e for a.
Is q. H. Is 50.
Joules work out is 30. Joules minus qc. Which is 20 joules and that gives us zero joules.
The delta e for b. Is 10 joules qh work. Which is 4 joules.
Minus qc. Which is seven joules ok and that gives us negative. 1 jewels change in energy for c.
Is q. H. Is.
30. Joules. Work is 10.
Joules minus qc is 20 joules. So that gives us zero joules. So when you put that in you will put zero comma negative one comma zero alright and that is part a okay so now lets move on to part b.
Where they ask which of any of these three violates. The first law of thermodynamics. Which basically says you whatever you put in were conserving energy equals.
What comes out and we cant have more energy out than we put in so for for part b. The one that violates that is engine be the reason is is we have ten joules coming out of stuff that were taking out of the engine. But they say that were getting four that were doing work and seven going to the the cold.
So thats basically 11 joules. So theyre saying. Were using ten joules to get out 11 joules.
Which makes no sense. So that ones gone alright part c. They say calculate the maximum efficiency for all of these engines so for part c well use the equation e max is equal to one minus t c.
Over th so lets go back up and look at the tc and th for all of these. So. If you notice th is all 600.
What i put jewel is on this last. One 600. Kelvin and tc.
Is all 300 kelvin. So we can do this once. And the answer will be the same for all of them.
So. E. Max is equal to one minus 300.
Kelvin over 600 kelvin. Which gives us 1 minus. 3.
Over. 6 which is the same as. One half so we get 1 minus.
05 and so we get 05. Or 50 is our theoretical maximum efficiency for all of those so when you put in well have zero zero point. 5.
Comma zero point. 5. Comma and four engine.
See zero point five all right moving on so for part d. Now they want to know what the actual efficiency is going to be so this one kind of gave me some trouble when i was going through it because we know that the actual efficiency of an engine is e is equal to work out over q h. And so.
I was thinking. Hey will they give us q h. And qc.
So why cant i just change. This to q h q. H.
Minus qc over q h. Which would give us what the actual efficiency is if the most accurate. But what they want to know is do any of these violate the any of the laws of thermodynamics in this case.
The second law of thermodynamics and so in this case. We actually need to just use this one. If you use this equation right here.
Itll tell you youre wrong youll be right in the sense that you did actually calculate the actual efficiency of the engine. But they want to know if any of them violate it so we need to use this one so for engine. A not you sorry little e efficiency for engine.
A is going to be work out. Which is lost my place work out for a is 30 and q h. Is 50.
So we have 30 joules for work out over 50 joules for. Qh which gives us. 06.
For that actual efficiency. The efficiency for b is going to be 40 or 4. Joules for work divided by 10 joules of the qh.
Which gives us zero point four. So this one let me draw it real fast. If we were to have this so we have 10 joules for qh4 joules out and seven going to the cold.
If we were to use this equation. The second one the q h. Minus.
Qc over q h. We say hey we know q. H.
Is. 10. Minus qc.
Which is seven joules over q h. Which is 10. Thats going to give us 3.
Joules over 10. Joules which will give us an actual efficiency of 033. Or 33 percent.
So this is what the engines efficiency actually would be but since we use work were getting more work out than we can so were getting a higher efficiency than as possible so lets go down to engine c. Which is b for c. Is q.
H. Is our work. I mean as 10 joules out for work over q.
H which is 30 joules and that gives us an efficiency of 033. Oh. I said this is.
33 this is third. D if of what it would be 03. Okay.
So now for part d or not part d. Sorry part ii. We want to know.
Which of these breaks. The law of the second left or non thermodynamics. So we figured out that the maximum efficiency for all of these engines is 05.
Or 50. So now that we figured out the actual efficiency. If any of those exceeds 50.
That violates the second law of thermodynamics and for engine. A we got an actual efficiency of 60 so for part ii. The answer is a engine a you you .
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