Carnot Heat Engines, Efficiency, Refrigerators, Pumps, Entropy, Thermodynamics - Second Law, Physics

This physics tutorial video shows you how to solve problems associated with heat engines, carnot engines, efficiency, work, heat, refrigerators, heat pumps and entropy. It discusses the concepts of the first and second law of thermodynamics. This video contains plenty of examples and practice problems.
Carnot Cycle:
ttps:// • Carnot Cycle & Heat En...
Otto Cycle:
• Otto Cycle of Internal...
Refrigerators and Heat Pumps:
• Refrigerators, Heat Pu...
Entropy:
• Entropy Change For Mel...
Physics 1C Final Exam Review:
• Physics 1C Final Exam ...
_______________________________
Physics 2 - Basic Introduction:
• Physics 2 - Basic Intr...
Electric Charge - Physics:
• Electric Charge - Physics
Conservation of Electric Charge:
• Law of Conservation of...
Coulomb's Law and Electric Force:
• Coulomb's Law - Net El...
Electric Fields:
• Electric Field Due To ...
Electric Dipole Moment:
• Electric Dipole Moment...
_______________________________
Electric Flux:
• Electric Flux, Gauss's...
Gauss Law Problems:
• Gauss Law Problems, In...
Electric Potential:
• Electric Potential
Equipotential Lines:
• Equipotential Lines & ...
The Electron Volt:
• Electron Volt Explaine...
Electric Potential Energy:
• Electric Potential Energy
Physics PDF Worksheets:
www.video-tutor.net/physics-b...

[@TheOrganicChemistryTutor]

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

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Entropy starts at 56:17

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[@Soor101]

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...

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[@user-iz2et4kj9y]

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?

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Can you make a playlist for thermodynamics

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[@tartsart2828]

In 19:19, why the unit is J?

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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

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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

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[@h_ym_o]

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 ^^

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[@scifi_the_dev]

@19:19, is that Joules or is it supposed to be KJ since we were dealing with KW?

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Yeah I think it should be KJ

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[@muskaanmemon8475]

Can someone explain why the heat of combustion is equal to QH and not QL @ 21:54 ?

[@BoBo121213]

Muskaan Memon Combustion is basically the burning of the added fuel that gives the QH to the engine aaaand I just realized that the comment is 2 months old so no point

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[@asandasiphesihle4771]

at 19:20, werent we suppossed to write the answer in KJ??

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[@boomertaste3404]

@ 1:02:50 why did DeltaT/ T become 1/T? What happened to the change in temp in the numerator?

[@carultch]

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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IN 17.57 ,IS THIS RATE OR SIMPLY HEAT DISCARDED?

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I want to know whether this is university level. I am first year mechanical engineering student. The questions that I see in the exam are more complicated than these.

[@Akshay-bc1pv]

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?

[@21area21]

Yeah, it should.

[@farlyn3]

Yes, you are right the answer should be given in Kj's.

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yup! Or 1.33*10^10 J

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[@valcheung5747]

@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.

[@MrYahya0101]

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.

[@mrknowmyself]

In carnot cycle, the sign of W is reversed , no? 26:24

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