Carnot cycle and Carnot engine | Thermodynamics | Physics | Khan Academy

[@musicforever1486]

This video was released when I was in the kindergarten

[@naveedthalappil7976]

Music Forever ! Same

[@darealpoopster]

Same, I was 5!

[@datsmydab-minecraft-and-mo5666]

same!

[@joyeetaroy9305]

I was learning how to talk than

[@nihilisticalbino]

I was 8 xD

[@ninjabomberboy]

Haven't started watching this explanation yet, but I already know its going to be a shitload better then my lecturers

[@TurdFurgeson571]

Don't discount the fact that by now you're hearing this information for at least the second time. Try doing things the other way next time. Watch Khan first, then go to lecture. I promise you will enjoy lecture more than you do now. You will be prepared to work with this information on a new level in class and you will likely be able to ask better questions, which will only help you understand this stuff. Perhaps that may be the source of your trouble, too, that you don't ask questions. I don't know you, but I do know that most students remain quiet and confused, and thus frustrated at their professors or the content. It seems like most don't want to look stupid in front of peers. It's okay to look stupid when learning things you've never learned before. (You're all stupid! Why wouldn't you be? You've never done this before.) Regardless of the expectations you have for yourself, I promise you your instructor is not expecting you to completely understand this stuff after hearing it _one_ time. No one expects that. For the love of who or whatever you believe in, ask questions. If you're afraid to speak up in class, send an email, go to office hours, hell do your homework at office hours. "That's so awkward!" Oh well. This is why you pay tuition, not to sit in a class and hear something once, but to actually learn the material; learning requires this work on your part though, it's an active process. That's how to use your professors, and get the most out of a mortgage's worth of student loans you're going to have. Khan is great, but realize that you had to take time out of your day to participate in your learning. Your professor and Khan are both available, but it's up to you to actually utilize the resource.

[@motopilot322]

You're animations are just fine man.. I've never taken formal physics courses, but for 10+ years now I've done my own research via books,magazines etc... Your videos by far are the most clear and to the point..your analogies are tough sometimes..but het..that's what pause and repay are for! Keep it up brotha! This is much appreciated!!

[@khletus9061]

10 more years have passed since your comment, how is your research going ?

[@juniorphysicist4055]

The day after tomorrow is my physics exam. While revising thermodynamics , i was facing some difficulties to understand this topic. Thanks to Sal for explaining the concept in a easy way. Love from Bangladesh❤️❤️

[@rubaiyattasnim1297]

For some reason I kept getting confused about how Q1 was coming into the system and Q2 was going out. This video helped so much!

[@TET2005]

Thank you sir. Your video is worth a million dollar instead of reading the boring physical textbook. As long the concepts are grasp properly, it will stay with you forever.

[@amassad127]

All of these videos are really just incredible. I'm a sophomore engineer and I still find these videos really helpful! I wish there was some way to help you put out these videos. Really, really admirable--thank you

[@annym.8170]

amassad127 donate khanacademy.org

[@jiayu7597]

You are so much better than my thermo professor....

[@omarkhanlilcurry]

why do i pay tuition

[@cbear9355]

You summed up college for me.

[@amazingworld9463]

Paagal

[@SaeedAcronia]

@@audacioustux Jesus. Poor guy just asked a question man.

[@shubhaprada9423]

but I feel why do I pay even to college?

[@chrisscott3725]

For that piece of paper called a degree

[@liaoyuqian4172]

It’s kinda crazy when I fully understood the mechanism behind it. It gave me that”aha!” moment. Again, I am thoroughly impressed by Khan academy

[@yutongyang-b2e]

I love captions so much...especially when I find that I can open a Chinese captions.感觉世界充满关怀

[@lilcote93]

im taking this right now with a prof that graduated from MIT as well ;) thanks for all the help sal

[@73Datsun180B]

I want to know what drugs carnot was doing, they seem pretty good!

[@shohanthegreat9857]

Lmfao

[@ok_computer12]

Existential crisis.

[@SabbirAhmed-nw2cv]

I was struggling to understand this. Thank you for this!

[@MalikRizwanElahi]

all lectures deliverd by this HONOURABLE professor are soooooooooo nice. i like all the lectures and enjoyed. i realy appreciate wow. keep it up sir. May God give u long life to educate us

[@ahobimo732]

Such a great explanation. I'd forgotten just how incredible Khan Academy was.

[@anandapatmanabhansu]

Teach forever sir

[@raiedahmednishat8883]

I can't thank you enough!!! SAL!!! MA MAN!!!! seriously, I was really troubled by this... I'm so grateful that youtube exists and there's a channel called khan academy

[@jamiegollings]

Fell much more comfortable about my Therm exam on Monday now, thanks!

[@arjuncalidas2736]

10:13 I think as pressure is decreased, the volume is increased. You said volume goes down. But awesome work here.

[@zpiazza]

these lectures should be put into number order

[@RKYT0]

not finished repeating it but already IMMENSLY helpful

[@OpsOfCod]

wow i have to say this is awesome. I didnt think i would find a video that would explain carnot cycle to a point that i would understand. Thanks man !

[@williamjpiano]

Very comprehensive video on Thermal cycles, nice work!

[@abhisekupadhyaya11]

Great Video.... I would have failed my physic xm if I wasn't there....

[@alijaved3356]

+Abhisek Upadhyaya no shit sherlock

[@janmay6349]

+Ali Javed Give him some ice for that burn

[@alijaved3356]

Janmay Patel lol :P i had forgotten about this

[@anandapatmanabhansu]

Thankyou sir

[@Quran722.]

Simply made too simple by khan

[@Gahstupidcomments]

chemistry textbooks u = q + w physics textbooks u = q - w chemistry defines w as work done on the system physics defines w as work done by the system same equation MIND YOUR SIGNS

[@taltheking100]

Thank you for the video! You made it all so simple to understand

[@isadora13correa]

THANK YOU!

[@kaif2562]

Thanks for helping me in learning it.

[@beyzayigit1912]

ı wish ı could like it for billion times hahhah

[@huzaifa129]

i would like to see such a talented professer!!!!!!!

[@jdfab_ulous]

From B to C (adiabatic) process, why does Volume keep increasing since there's no change in heat transfer and there's no source / reservoir?

[@arnatri1503]

Fabrice Vieillesse when adiabatic expansion is taking place, the piston is just allowed move out. Therefore the gas expands increasing the volume.

[@iknownothing35]

Shouldn't delta-U have a specific value considering for an adiabatic process Q = 0 always? In that example, W(AB) = 0; no work has been done. But the gas did do work considering the change in Volume... idk maybe I'm not seeing it.

[@s0m0c]

at 10:11 Salman Could have meaned "my pressure would have kept going down and my volume would have kept going up"

[@cyrilsmith6396]

Thank u so much! U r doing a GREAT work there...

[@mikeb5174]

also, at 7:00 : im taking pchem I right now, and we do define deltaU=q+w. so qab=-wab.

[@matthewheide6816]

you deserve a Nobel prize..nuf sayed

[@ptcala99]

shitty microsoft paint-esque program + 20 minutes = lucid explanation of the Carnot cycle....amazing!!! Thank you so much!!! I feel more comfortable with thermo than ever!

[@Anujsingh-mo1vd]

I understood, work done by the system is transferred the heat from hot reservoir to cold reservoir.

[@jeanbernard3469]

Thank you dear!

[@johnpartridge7052]

He sounds like Tom Hanks at 1.25x speed lol

[@wings4994]

Yeah lol

[@s0m0c]

adiabatic process means isolated, so no heat transfer can ocurr,the system travels trough an isoterm and if the temperature doesn`t change for our system, neither do the Internal energy is going to change, so is constant. But in and adiabatic process the temperature of the system is not constant is not travelling along the isoterm anymore.

[@ok_computer12]

During the isothermal expansion process, does the system simultaneously absorb heat while doing work?

[@Mijal15]

From B to C where is the heat flow? The temperature is changing from T1 to T2 therefore its not isothermal. Once the reservoir is gone the system is completely isolated so the heat exchange would stop.

[@s0m0c]

in both process the volume is changing going up during expansions and going down in a compresion

[@khellilg]

Haaa if i had access to internet when i was a student....

[@amritmcamc]

I tripped on your shit. I understood it all. thanks.

[@glimpseofanpilgrimage5323]

hey why the slop in the graph is more steaper at BC

[@KVI3I]

One Question: By removing the Rocks, you have to use Work from the outside - would'nt that make the process irreversible?

[@Vasu-qn6kj]

Wouldn't that also make it redundant? I mean Carnot cycle is utilised in an engine right.. you made an engine because you wanted to do work on something.. but if you have to do some outside work to make that happen.. then you're basically doing work to get work. Doesn't make sense to me. But ofcourse I'm obviously missing something. I just don't know what that is.

[@phantomj8449]

They were actually adapted from the rotation of a train wheel, when as the rocks decrease, it indicates the axes start to have less push on the wheel (or backward resistance), closing in on 90 degrees to the forward motion of the train. Keep on going, and the wheel’s spin pushes the piston in the opposite direction, compressing the gas and then there’s the next 2 stages, completing the cycle.

[@vijgenboom2843]

Thank you ❤️

[@pyg1116]

I thought you couldn't draw the line from A to B because we can't measure the system while it's in flux...or do we always assume it's a quasistatic system...or can we draw the path because the system is kept at a constant temperature?

[@sukritmanikandan3184]

Yeah you assume quasistatic

[@amritbasavaraj9146]

But how does this cycle interact with Kevin-Planck's statement that any system cannot absorb heat from a reservoir at a fixed temperature and convert it wholly into work?

[@orangeinfotainment620]

TYSM 🥰

[@mikeandino81]

So A to B is adiabatic and so is B to C? I'm sorry but isn't T2 going into the system? Or is this a system definition issue? Thanks for the video, though. Very good!

[@biswajitmarndi9146]

wow!!! Thanks Sal.

[@Anujsingh-mo1vd]

I understand. Work done went

[@clementkoshy5438]

If the internal energy of the ideal gas from 1-2 was zero ( the work done (1-2) was to increase the volume and decrease the pressure ) , then what is the internal energy of the ideal gas during adiabatic expansion ( process 2-3 )

[@Oh4Chrissake]

How do we know that Q2 < Q1? That is, how do we know that a smaller amount of heat is transferred from the system, between C and D, than to it, between A and B?

[@theblackknight4295]

I just love your videos vai,love from Bangladesh.

[@draegernaut]

THANK YOU, i understand now

[@bindelapranay7921]

its awesome i can learn phy very easily

[@DonGuzmann]

first 2 mins ----> instant subscribe

[@jasonpunza]

all your videos are very helpful. thank you!

[@BANKO007]

Why does the gas continue to expand during the adiabatic expansion after the isothermic expansion? What determines when the heat should be removed? Heat can be turned into mechanical energy without the adiabatic elements. Something is missing. "Remove some rocks" doesn't explain anything.

[@phantomj8449]

Because the rocks are being removed. You are right in a slow reversible process, the weight of the rocks should only be marginally lighter than the gas pressure under reversible heating, and as soon as heating stops, expansion should stop. They were actually adapted from the rotation of a train wheel, when as the rocks decrease, it indicates the axes start to have less push on the wheel (or backward resistance), closing in on 90 degrees to the forward motion of the train. Keep on going, and the wheel’s spin pushes the piston in the opposite direction, compressing the gas and then there’s the next 2 stages, completing the cycle.

[@sinankuniyilathoot5188]

thAnks

[@sgsg5903]

Very good thanks Very much!!

[@ahmedelbanna5156]

How is adiabatic process digrammed with isothermic process in 2D and in isothermic process T is const while in adiabatic process T is variable so there must be 3D diagram For 3 variables P&V&T

[@adityakhanna113]

It's a projection. The third variable is just mentioned. it's a bit messy to do 3d and 2d is understandable

[@tejasgajra2731]

A graph is always drawn between independent variables P,V&T altogether are not independent i.e. if two are known ,we can find the value of the third.

[@christiantischer632]

I think something's missing in the explanation: In the end, the system is at the exact same state as in the beginning: the same amount of pebbles at the same height. So how is there a net work done by the system?

[@Thaumius]

because work is a path function, not a state function.

[@gazhill89]

Thanks

[@子阳-h7x]

great video! it help me a lot

[@mohiuddinkhan4069]

from 2021

[@therealjordiano]

useful engine ... .. childhood comes to mind D,:

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Thanks khan academy👌👌All ur lactrs r v Funnt and amazing

[@themeharsingh]

Why temperature should goes down in 1 st and 2nd step if pressure is decreased and volume incresed ( if reservoir wasn't there)

[@DMZ509]

you need to watch the previous vids, you can't understand a subject without knowing the sub-subjects, especially in chemistry

[@Growriginator]

Thank's from Germany!

[@kinzaarif8053]

Great explanation!!keep it up

[@gauravsaimaddipati8356]

I was two years old when this video came out

[@welcomewehavecookies]

Thank you so much for your helpful tutorial! :D

[@CreQ]

Our internal energy did not change? I thought pressure was kinetic energy, and if the pressure went down the internal kinetic energy went down? Just like if you removed pressure from a Co2 cartridge by firing a pellet gun, the internal energy changes and each successive pellet will fire at a slower velocity.

[@M_0892]

Thanx so much!

[@tig8med]

May allah (God) give u long life to educate us

[@internationalgolfconstruction]

Can we say that the work done to the system from A to C is equal to mgh of the stones where h is the distance traveled by the piston?

[@shootshoot]

man you're really cranking these out, it's hard to keep up

[@arjundenath8535]

11 years, How those videos affected you?

[@DMZ509]

Same thing. Thermodynamics is one of the subsubjects of chemistry, even physics.

[@bscutajar]

What I don't get is that you can easily devise a system where the piston does work when kt moves both ways. So in that case both increasing and decreasing volumes would mean the piston is doing work. So you can just add and remove the same qty of heat and have the system do work, whicb would violate the first law.

[@sergiolozavillarroel3784]

8:43 "I should be talking about thermodinamics not drawing" lol

[@doganguler1]

there is no reservoir from D to A as no heat is exchanged in a adiabatic process.

[@s0m0c]

Gracias.

[@mitch.4615]

Brilliant video, you have helped me so much! Thank you :)

[@Dan.U.]

Appreciate ur work!

[@kashiii2011]

what do you do if it isn't a mono atomic ideal gas and if we are dealing with steam...?

[@hobinsmith2611]

bcos the volume expanded that's why the temperature decreases.

[@mcgregor107]

Dude, this is an iso-thermal expansion, not adiabatic. No heat flow and a change in temperature for adiabatic, no temperature change (and heat flow required to compensate) for isothermal.

[@UnitedPebbles]

Can't believe I actually wrote an essay about this, a long time ago...yet I don't understand the model most people explaining...wth.

[@amritas2400]

Thanks a million 🙏❤❤❤❤❤

[@ronaldbadic9828]

Good work.

[@theincapable]

@TheMIkex24 He accidentally took the wrong word. The pressure is going down and the volume is going up, as you can can see in the figure.