this vid and the one about PV diagrams and internal energy (about 4h long) summed up my entire thermodynamics in physics degree im so glad this vid exists
@@ahmadsleiman37 Probably thinking of thermochemistry. If not, chemistry and physics do tend to have many overlaps in some areas so im not too surprised.
This video may be 7 years old, but DAMN it still holds up to it's quality. My professor is horrible at lecturing but this video clears up so many confusions and doubts. Thank you Julio. I don't know what I'd do without your videos.
Absolute king, all your videos are so helpful! The lecture notes on thermodynamics were hard to understand but you just simplified the whole process thanks! Keep doing what you're doing
thank you so much! with your videos I'm learning more than I have ever learned at school! i started to understand physics normally, all with the help of your explanation.
I watched so many videos and looked up explanations on google to try to understand these topics but less than a couple minutes after watching your videos, everything just makes sense. You really break things down so simply. Thank you!! D:
Many Thanks for your time and this helpful video. My roommates said Thermodynamics is hard. however, in 1h :18 min and 25 second. I understand the HARD THERMODYNAMICS. May my almighty God bless you and give more energy to pu on the Helpful videos like this one . Again, thank you so every much for time and Bless you hear.
After this course unit and the results are out I will send this video to my lecturer to see how things are supposed to be done and explained not just to brag around. Thank you Sir
Hello... sir... Really really really.. Your videos are awesome.. I loved it, the way of your explanations.. Like a practical one.. Its amazing.. I dont want to memorize any formule after watching your videos. the thermodynamics and HMT is the most dangerous subjects in Mechanical Engineering.. But you made that things very easy for me..
For temperature differences and weighted averaging, you don't need to convert to Kelvin. You ultimately need Kelvin before you plug it in to the rest of the equation, but it isn't necessary to convert in advance of averaging. (T1 + T2)/2 + K will give the same outcome as: (T1 + K + T2 + K)/2 Where the two T's give temperature in Celsius, and the K is the Celsius to Kelvin conversion adder of 273.15 Kelvin.
First of all, WOW!! So helpful! I'm a little confused about the ideal gas problem - So in my ChemE thermo class we talk about the entropy change of an ideal gas as delta S = integral of Cp*dT/T - ln(P/P0)*R. Now this would equal zero for a reversible process, but nowhere in the problem does this say that it is reversible and the only ispentropic processes are reversible and adiabatic. I'm not sure what to think because your reasoning makes sense here but I'm sure if you have an irreversible adiabatic process there's an entropy change, as all irreversible processes have entropy changes. This is quite a thinker...
1:07:05 In that problem, why dS = 0 when Q = 0? Although it is an adiabatic process (Q = 0), Qr (reversible) is not 0 as far as I know. Since Q and Qr (reversible) are not the same, so maybe we can calculate change in entropy dS. Can anybody explain about that? Thanks ^^
one question 1:10:00 , isn't the net change in entropy of a heat engine affected by the work done by the engine? Is there a differnce between a carnot engine and a regular heat engine?
what is meant by a jet engine cycle? Is that a measurement over time? by the same token, does a typical reciprocating or piston engine cycle refer to a single intake combustion exhaust cycle? please advise if you know
That's dT in the numerator, rather than deltaT. That indicates that the integral is with respect to T. When you integrate, the integrand is the part of the integral excluding the d-whatever term. So you are looking for the integral of 1/T, from your look-up table, which will be ln(T) + C
@The Organic Chemistry Tutor Hello Sir, I am a little bit confused at 25:11 min (Carnot engine). You said heat is absorbed here and every time gas expands the engine does work hence Work is positive. But in your Internal Energy you said system does work so work is negative. and work is done on system, work is positive. I am finding these two contradicting and I am confused. This is what I think when you talking about work in the Cornot engine, Please correct me if I am correct on these two concepts. W= -PdeltaV. This formula is relative to the Work of the system not the surrounding right? In the Carnot engine, when you mentioned Work is positive when gas expands, you are talking about the work of the surrounding system not the work that the engine gains from the surrounding system right? and when the gas compress you said work is negative, you mean the work of the surrounding is negative? Please help me clarify. Thank you very much, and I am highly appreciate your time.
I think in this video he’s using W(by system)=P*dV as physicists use, and that's what you use with the graphs because you have P on y axis and V on x axis. Which would mean when gas expands, work done by the system is positive. So when he says work is positive, he means work done by the system, or as you say "work of the surrounding system" or work gained by the surrounding. If he said work is negative ("But in your Internal Energy you said system does work so work is negative. and work is done on system, work is positive"), that would mean either he meant when work is done by system (e.g expansion), work done on the system is negative. Or he was using the chemistry convention W(by system) = -P*dV. (Chemistry convention). In chemistry, it's the opposite, when gas expands, work done by system is negative. But work done on the system is positive.
There are different sign conventions. The heat engine sign convention assigns work as positive when the system provides the work, and heat as positive when it is added to the system.
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this vid and the one about PV diagrams and internal energy (about 4h long) summed up my entire thermodynamics in physics degree
im so glad this vid exists
I'm confused, I am given to study thermodynamics as in chemistry and everyone is saying its physics tf?
@@ahmadsleiman37 Probably thinking of thermochemistry. If not, chemistry and physics do tend to have many overlaps in some areas so im not too surprised.
@@ahmadsleiman37 u teach thermodynamics?
thats a very simple and light thermodynamics mod you have there then
@@ahmadsleiman37 in my classes, thermodynamics is in both subjects.
messy room is the best analogy I have heard for entropy 10/10 stars!
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This video may be 7 years old, but DAMN it still holds up to it's quality. My professor is horrible at lecturing but this video clears up so many confusions and doubts. Thank you Julio. I don't know what I'd do without your videos.
hey, mechanical engineering bouncing back to get his F.E. i had a bad teacher 3 years ago in college and have to say this is a life saver. thank you!
Absolute king, all your videos are so helpful! The lecture notes on thermodynamics were hard to understand but you just simplified the whole process thanks! Keep doing what you're doing
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thank you so much! with your videos I'm learning more than I have ever learned at school! i started to understand physics normally, all with the help of your explanation.
You real make me to succeed since am going for exams tomorrow , thank
God bless you
I watched so many videos and looked up explanations on google to try to understand these topics but less than a couple minutes after watching your videos, everything just makes sense. You really break things down so simply. Thank you!! D:
Many Thanks for your time and this helpful video. My roommates said Thermodynamics is hard. however, in 1h :18 min and 25 second. I understand the HARD THERMODYNAMICS. May my almighty God bless you and give more energy to pu on the Helpful videos like this one . Again, thank you so every much for time and Bless you hear.
You didn't really understand it but whatever
as jee aspiring student i find it quite simple solving sum before you did it is good for understanding the concept
Why did you kill konan tho?
Entropy starts at 56:17
Thank you!!
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After this course unit and the results are out I will send this video to my lecturer to see how things are supposed to be done and explained not just to brag around. Thank you Sir
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In 19:19, why the unit is J?
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Great video, but I'm a bit confused at 19:14 . It's 154 kW so shouldn't the answer be 13.3 GW rather than MW?
Yeah, it should.
Yes, you are right the answer should be given in Kj's.
yup! Or 1.33*10^10 J
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Hello... sir... Really really really.. Your videos are awesome.. I loved it, the way of your explanations.. Like a practical one.. Its amazing.. I dont want to memorize any formule after watching your videos. the thermodynamics and HMT is the most dangerous subjects in Mechanical Engineering.. But you made that things very easy for me..
true bro
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at 19:20, werent we suppossed to write the answer in KJ??
1:13:06 Shouldn't you convert to Kelvin first, and then take the average? 323K vs 338K
For temperature differences and weighted averaging, you don't need to convert to Kelvin. You ultimately need Kelvin before you plug it in to the rest of the equation, but it isn't necessary to convert in advance of averaging.
(T1 + T2)/2 + K
will give the same outcome as:
(T1 + K + T2 + K)/2
Where the two T's give temperature in Celsius, and the K is the Celsius to Kelvin conversion adder of 273.15 Kelvin.
Great sir wonderful experience
Finally understood these concepts
you are the best, thank you very much for such an amazing lecture. great of all time in Thermodynamic
First of all, WOW!! So helpful! I'm a little confused about the ideal gas problem - So in my ChemE thermo class we talk about the entropy change of an ideal gas as delta S = integral of Cp*dT/T - ln(P/P0)*R. Now this would equal zero for a reversible process, but nowhere in the problem does this say that it is reversible and the only ispentropic processes are reversible and adiabatic. I'm not sure what to think because your reasoning makes sense here but I'm sure if you have an irreversible adiabatic process there's an entropy change, as all irreversible processes have entropy changes. This is quite a thinker...
Mans doing God’s work here.
1:07:05 In that problem, why dS = 0 when Q = 0? Although it is an adiabatic process (Q = 0), Qr (reversible) is not 0 as far as I know. Since Q and Qr (reversible) are not the same, so maybe we can calculate change in entropy dS.
Can anybody explain about that? Thanks ^^
one question 1:10:00 , isn't the net change in entropy of a heat engine affected by the work done by the engine? Is there a differnce between a carnot engine and a regular heat engine?
Really really nice vids my dude, thanks man
Thanks
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was the equation at 43:50 wrong? I thought K=Qc/W or I was wrong? could someone explain that part for me
@19:19, is that Joules or is it supposed to be KJ since we were dealing with KW?
Yeah I think it should be KJ
Very perfect, thanks alot 👍👍👍
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1:00:42 Isn't the specific heat capacity for water 4.184 J and not 4184 J?
Nvm I forgot all about the kilogram conversion.
This is Amazing! What’s the difference between Ql and Qc please?
Ql means heat LOST , Qc means heat from the COLD reservoir or to the COLD reservoir
Lmk if u understand
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please do something on rankine cycle
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what is meant by a jet engine cycle? Is that a measurement over time? by the same token, does a typical reciprocating or piston engine cycle refer to a single intake combustion exhaust cycle? please advise if you know
@ 1:02:50 why did DeltaT/ T become 1/T?
What happened to the change in temp in the numerator?
That's dT in the numerator, rather than deltaT. That indicates that the integral is with respect to T.
When you integrate, the integrand is the part of the integral excluding the d-whatever term. So you are looking for the integral of 1/T, from your look-up table, which will be ln(T) + C
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At 1:13:53 there is an error...the negative 30°c should be in kelvin...so it make the math after that wrong
At 19:17, I think the final answer should be 13305600kJ not 13305600J
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@The Organic Chemistry Tutor
Hello Sir, I am a little bit confused at 25:11 min (Carnot engine). You said heat is absorbed here and every time gas expands the engine does work hence Work is positive. But in your Internal Energy you said system does work so work is negative. and work is done on system, work is positive. I am finding these two contradicting and I am confused. This is what I think when you talking about work in the Cornot engine, Please correct me if I am correct on these two concepts.
W= -PdeltaV. This formula is relative to the Work of the system not the surrounding right?
In the Carnot engine, when you mentioned Work is positive when gas expands, you are talking about the work of the surrounding system not the work that the engine gains from the surrounding system right? and when the gas compress you said work is negative, you mean the work of the surrounding is negative?
Please help me clarify. Thank you very much, and I am highly appreciate your time.
I think in this video he’s using W(by system)=P*dV as physicists use, and that's what you use with the graphs because you have P on y axis and V on x axis. Which would mean when gas expands, work done by the system is positive. So when he says work is positive, he means work done by the system, or as you say "work of the surrounding system" or work gained by the surrounding.
If he said work is negative ("But in your Internal Energy you said system does work so work is negative. and work is done on system, work is positive"), that would mean either he meant when work is done by system (e.g expansion), work done on the system is negative. Or he was using the chemistry convention W(by system) = -P*dV. (Chemistry convention). In chemistry, it's the opposite, when gas expands, work done by system is negative. But work done on the system is positive.
thnkyou so muchhhhhhhhhhhh. now i know thermo is easy thankyou bro
In carnot cycle, the sign of W is reversed , no? 26:24
Thank you very much!!! this is very helpful to me
How much work is required to explain the university of minnesota to the university of cincinnati?
Thank you brah for this perfect explination.
When work is done on the system, to compress the gas, the work is positive because it is added into the system.
There are different sign conventions. The heat engine sign convention assigns work as positive when the system provides the work, and heat as positive when it is added to the system.
again thank you very much
19:15 13.3millionKJ or 13.3GJ
I was searching for this comment
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*Thank you so much* 🌹
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