Carnot Cycle & Heat Engines, Maximum Efficiency, & Energy Flow Diagrams Thermodynamics & Physics

[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/

[@wolvsongthechosen]

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[@colestinson8564]

13:14 for Carnot Cycle

[@abdinhuda3007]

Thanks a bunch

[@MuhammadMohad]

Thank you

[@charlesmutuku9345]

you made the carnot cycle easy for me to understand,i passed in my thermodynamics exam.thank you

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absolute chad, thanks mr. organic chemistry tutor. If I pass my thermodynamics class, It's mostly thanks to you!

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did you pass?

[@guguruseto]

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[@rheaplawat]

i am so grateful for this elegant and simple explanation of the carnot cycle. it saved me from losing so many marks in my exams!

[@ClydeBuka]

You explained better than my lecturer thankyou

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Last week our lecture gave us an assignment to explain about the Carnot cycle and I took note of exactly what our organic tutor is explaining guess what,,? I got an 100 %

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[@srw2289]

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[@chakeol4776]

thank you so much for the video. self studying thermo dynamics and these videos really help explain the concepts and equations

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I love u dude I was dealing with a hard problem with carnot graph I was trying the question for 2 days and now after ur video 🤗🤗🤗I came up with the answer😍😘😘😘

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[@cryptobossz]

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[@KG_Studios]

Thanks man Very thorough

[@johns228]

Hi, at 18:08, I think you missed a negative sign. Q(c)=-nRT(c)ln(Vd/Vc) Your videos really help a lot!!

[@HassanmohamudIdleSenior]

Thank you teacher 🎉🎉

[@ELIUDKIPCHIRCHIR-ll3xf]

Great job professor,,grateful for your good job

[@workerpowernow]

i've watched your videos from my freshman year of undergrad through my physics PhD courses when I get confused lol. Thanks

[@felipegarcia2269]

On Question 2, why did you use a different variant for the efficiency, and when do you know to use which one?

[@serious-power3085]

this is 2 years late but if helps anyone they are equivalent

[@serious-power3085]

*4 months late

[@serious-power3085]

divide the numerator by the denominator and it goes from (Th - Tc)/Th to 1 - (Tc/Th)

[@geo-f2i]

am really loving the way he explains. i do understand

[@shizulislam9908]

Saved my day

[@benjaminamirkhosravi3686]

This guy is much better than our mad teacher

[@domd1268]

Hello, Can I clarify if in 18:08, should the it be Q=-W? since the Work is(-) from the compression giving you delta U = Q - (-W)? and also in 19:12, isnt the delta U = +W because Work is (-) from the compression giving you delta U = Q - (-W)? Or am i wrong?

[@bhoumik911]

im a little bit confused about that one too

[@David-gn5rp]

6:38 Wouldn't Qc be -6000 J?

[@ratzax1120]

thank you very much once again

[@NK-iy6if]

God Bless You! Thank You!

[@fuge74]

thank-you, studying for thermal physics and we are using Daveil V. shreoders book, this is a great complement to it.

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I am very interested in physics and related subjects. This video is simple for me to understand

[@mjzotv7353]

Thank you lecture

[@frankchothefact6502]

Helpful video

[@ninobettencorti6701]

Very clear, thanks man

[@andrewdawe9610]

Near the end of the video when you are calculating the work done by the adiabatic expansion between B and C you use the constant Cv. Isn't Cv a constant used when the volume is constant? In the same way Cp is used in a process where the pressure is constant?

[@carultch]

You don't necessarily need constant volume for Cv to apply, or constant pressure for Cp to apply. They relate to the special case where these specific heat capacities are the constants of proportionality between change in temperature and heat added to the system, but they apply outside this context as well. Cv is the proportionality constant between temperature an internal energy, while Cp is the proportionality constant between temperature and enthalpy. And this is true, regardless of whether volume or pressure is constant in any situation. And since both enthalpy and internal energy are state functions, it doesn't matter what path we take. The difference in these thermodynamic energy values still applies. Cv and Cp do have some dependence on temperature, so it isn't an exact proportionality. But it is close, and these values aren't very temperature sensitive, so you still get meaningful results by assuming they are constant with temperature. In this particular application, U=Cv*T is the way we calculate internal energy at each point of interest. We then take changes in internal energy, and equate them to account for work done by or on the system, associated with each adiabatic process.

[@tochukwutk5190]

Wherever you got your examples from please keep it coming my test was the exact same thing no difference ❤️❤️❤️ I love you 😭❤️🔥😂😂

[@puritymutheu4892]

U are a blessing

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Superb ❤️its very valuble for every time for any undergraduates 🥳✨️

[@priyanshi9909]

Thank you ☺️

[@michaelbrohier4075]

Great video - thanks

[@sallymarie1665]

Who else is here few hours to exam. Thank you O.C❤️😩

[@justinlaberge2522]

In Step 2 of the Carnot Cycle, the Equation for Change in internal energy used is dU = n(Cv)dT, why do we use Cv (molar heat capacity at constant volume) if we are talking about an adiabatic expansion where the volume is not constant?

[@leesnotbritish5386]

Exam time in 18 minutes, let’s see what we get

[@douhanezar4912]

Thank you so much I really appreciate it.

[@vasanthkumarcroos8605]

very much appreciate sir

[@yigitcan824]

Professor I have a question here. Is Carnot cycle totally isothermal? I mean there are two isothermal processes, which are isothermal compression and isothermal expansion. Others are adiabatic compression and adiabatic expansion. These adiabatics are also isothermal, at the same time ?

[@tamukamaungwa3392]

Well explained. Thank you very much

[@WorldofGossiptv]

Responsible for my graduation🎉

[@mohamadhanan5706]

if any expanssion gives a +w ... then how the work will be positive in second step of carnot cycle ?? because the work= - ncv deltaT

[@fruit_punch_samurai7217]

∆T is negative beacuse temperature is reduced in the second step hence the work is positive.

[@mohamadhanan5706]

shahzan ahmad ohhh! Now it makes sense, thanks a lot.

[@bessil7955]

my glorious king

[@tahmiddaniel9241]

😄 Thank You!

[@iarenubie78]

Umm as a chemist I would prefer the sign convention (Delta U = Q + W ) but the rest are fine

[@gustavopaz5453]

Why do chemists want to impose a sign convention that only meshes well with chemistry problems? The laws of thermodynamics are supposed to be universal. Why do some people want to tailor them for specialized cases?

[@yashyelmame1460]

@@gustavopaz5453 the sign convention of chemistry can be used in physics and vice versa Both of them are universal

[@Laws2]

Ah yes, Physical Chemistry. The bane of my undergraduate GPA

[@peterfaloni5058]

I have my chemistry exam tomorrow I hope it helps 😩

[@menglimarrero4296]

Sir at problem 4 What is the exhaust temp? Is it the Tc=266.7? not the Th=666.7?

[@ianfama462]

1:36 how'd you solve the 0.26?

[@UnknownUser049]

I am also questioning that

[@UnknownUser049]

I get it now

[@UnknownUser049]

You need to minus Tc and TH after that Minus the Answer to TC again to get the answer and multiply by 100 to get the percentage

[@lasarith2]

@@UnknownUser049 just think of 1.00 as 100% .

[@xaviiifrances]

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[@davidjnrananiawusah118]

POWERFUL EXPLANATION BUT PLS CAN U ANSWER FELIPE GARCIA'S QUESTION. i am really interested in it

[@serious-power3085]

They are equivalent

[@serious-power3085]

divide the numerator by the denominator and it goes from (Th - Tc)/Th to 1 - (Tc/Th)

[@ram_pande8039]

Thanks man

[@julianalexisumanabernal2281]

where is the entropy do you have anything with that??

[@MrYahya0101]

I don't understand why you're using dU = n*Cv*dT where Cv is used for constant volume but volume is not constant during adiabatic expension/compression?

[@carultch]

I responded to this same question to a user named Andrew Dave on this video.

[@daminkim0703]

Why is the work done by the system in question number 2 not minus ?

[@Gil_FR]

김다민 it should be minus, since it’s leaving the system

[@sezerkoyun07]

Eyvallah hocam

[@jithinajith4295]

Which textbook is this?

[@LionnelMebenga]

Please for help in homogeneity

[@HarunOrRashid-pp5ju]

Please sir upload the diesel cycle curve.

[@post-centrist666]

Carnot Engine eh? More like a Cannot Engine am I right! Cuz it Cannot exist ha!……crickets

[@happyfrezar2813]

🤣🤣🤣WHAT A WORD PLAY

[@Oscar-gx2yf]

I love you man

[@MrKaje72]

What does n represent?

[@ppgsheinagracemayordobsabe3860]

I think there's something wrong in number 3. 🤔

[@aheer8687]

Awat kat slide kami takdak, nanti time explain memang aku jawab according to youtube

[@thepinkdude1854]

isn't delta u = w+q not q-w?

[@ZizoZizo-fp6js]

u=q-w in physics u=q+w in chimestry

[@carultch]

Sign convention preferences vary, depending on application and context.

[@minjinseo3681]

When is engine an ideal one?

[@frostbite07]

Carnot engine is an ideal engine, its efficiency is 1. It’s impossible for a real engine to have efficiency equal to 1.

[@TavartDukod]

@@frostbite07 Carnot engine doesn't have the efficiency equal to one. Before proceeding remember that "thermodynamic temperature" just means we're using a temperature scale with 0 being the absolute zero like Kelvin or Rankine. Let T_h be the thermodynamic temperature of the heat source, and T_c be the temperature of the heat sink. Then, the thermal efficiency of Carnot engine equals 1 - T_c/T_h. So even the perfect thermal engine cannot have the thermal efficiency of one. Imagine that your engine takes heat from the source at 100°C (373.15 K) and dispenses of it into a sink at 0°C (273.15 K). Than your engine can never be more efficient than (1 - 273.15/373.15) * 100% ≈ 27%.

[@abdulkadiribrahim9640]

A mistake if Qh - Qc and Qc= -8000 then it should be 14000 - (-8000) = 14,000 + 8000 =22,000 not 14,000 - 800

[@frogger6478]

What’s delta U

[@carultch]

The change in internal energy. In other words, the change in the sum total of all the modes of internal kinetic energy of gas molecules, in translation, rotation and vibration. Don't ask why it's called U, it just is.

[@johnkelly3886]

Thermodynamics gives you bounds on real engines. Moreover, the explosion of the fuel mixture is not even approximately reversible. How can thermodynamics be of any use in designing real world internal combustion engines?

[@carultch]

It gives you an idea for what efficiency you can expect, in a given engine design. It also gives you guidance for how to fine-tune the geometry of the engine for the greatest efficiency. This way, given a target power rating of the engine, you can determine the stroke, the bore, and the clearance volume, from the theoretical efficiency of the Otto cycle. And when you monitor the pressure and volume at every point along the way in a real engine, it does closely follow the theoretical curve of the Otto cycle.

[@mec1789]

the fact that you must write down basic algebra for americans to understand is hilarious for me not gonna lie

[@frankchothefact6502]

If W=/QH/-/QC/ Then W= /14000/- /-8000/ Which is W= 22000j

[@anarosedacutanan5997]

Why don't you include the negative sign of - 26 %

[@tinalgh4636]

کدوم دانشگاه درس خوندی

[@janekwicher]

why TH is divided by TC?

[@yalesbin]

I’d probably be better off watching your videos rather than reviewing my notes

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i love you

[@魏鑫林]

mercy！

[@georgeseamans768]

Mark Wahlberg?

[@puritymutheu4892]

Thank you so so much 😊