|Ch 01: Units & Vectors||2hrs & 22mins||0% complete||WorksheetStart|
|Ch 02: 1D Motion (Kinematics)||3hrs & 11mins||0% complete||WorksheetStart|
|Ch 03: 2D Motion (Projectile Motion)||3hrs & 8mins||0% complete||WorksheetStart|
|Ch 04: Intro to Forces (Dynamics)||3hrs & 42mins||0% complete||WorksheetStart|
|Ch 05: Friction, Inclines, Systems||4hrs & 32mins||0% complete||WorksheetStart|
|Ch 06: Centripetal Forces & Gravitation||3hrs & 51mins||0% complete||WorksheetStart|
|Ch 07: Work & Energy||3hrs & 55mins||0% complete||WorksheetStart|
|Ch 08: Conservation of Energy||6hrs & 54mins||0% complete||WorksheetStart|
|Ch 09: Momentum & Impulse||5hrs & 35mins||0% complete||WorksheetStart|
|Ch 10: Rotational Kinematics||3hrs & 4mins||0% complete||WorksheetStart|
|Ch 11: Rotational Inertia & Energy||7hrs & 7mins||0% complete||WorksheetStart|
|Ch 12: Torque & Rotational Dynamics||2hrs & 9mins||0% complete||WorksheetStart|
|Ch 13: Rotational Equilibrium||4hrs & 10mins||0% complete||WorksheetStart|
|Ch 14: Angular Momentum||3hrs & 6mins||0% complete||WorksheetStart|
|Ch 15: Periodic Motion (NEW)||2hrs & 17mins||0% complete||WorksheetStart|
|Ch 15: Periodic Motion (Oscillations)||3hrs & 16mins||0% complete||WorksheetStart|
|Ch 16: Waves & Sound||3hrs & 25mins||0% complete||WorksheetStart|
|Ch 17: Fluid Mechanics||4hrs & 39mins||0% complete||WorksheetStart|
|Ch 18: Heat and Temperature||4hrs & 9mins||0% complete||WorksheetStart|
|Ch 19: Kinetic Theory of Ideal Gasses||1hr & 40mins||0% complete||WorksheetStart|
|Ch 20: The First Law of Thermodynamics||1hr & 49mins||0% complete||WorksheetStart|
|Ch 21: The Second Law of Thermodynamics||4hrs & 56mins||0% complete||WorksheetStart|
|Ch 22: Electric Force & Field; Gauss' Law||3hrs & 32mins||0% complete||WorksheetStart|
|Ch 23: Electric Potential||1hr & 55mins||0% complete||WorksheetStart|
|Ch 24: Capacitors & Dielectrics||2hrs & 2mins||0% complete||WorksheetStart|
|Ch 25: Resistors & DC Circuits||3hrs & 20mins||0% complete||WorksheetStart|
|Ch 26: Magnetic Fields and Forces||2hrs & 25mins||0% complete||WorksheetStart|
|Ch 27: Sources of Magnetic Field||2hrs & 30mins||0% complete||WorksheetStart|
|Ch 28: Induction and Inductance||3hrs & 38mins||0% complete||WorksheetStart|
|Ch 29: Alternating Current||2hrs & 37mins||0% complete||WorksheetStart|
|Ch 30: Electromagnetic Waves||1hr & 12mins||0% complete||WorksheetStart|
|Ch 31: Geometric Optics||3hrs||0% complete||WorksheetStart|
|Ch 32: Wave Optics||1hr & 15mins||0% complete||WorksheetStart|
|Ch 34: Special Relativity||2hrs & 10mins||0% complete||WorksheetStart|
|Ch 35: Particle-Wave Duality||Not available yet|
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|Intro to Heat Engines||12 mins||0 completed|
|Efficiency of Heat Engines||14 mins||0 completed|
|Heat Engines & PV Diagrams||19 mins||0 completed|
|Four Stroke Piston Engine||44 mins||0 completed|
|Carnot Cycle||8 mins||0 completed|
|Refrigerators||54 mins||0 completed|
|Intro to Refrigerators||27 mins||0 completed|
|Refrigerators & PV Diagrams||27 mins||0 completed|
|Entropy and the Second Law||46 mins||0 completed|
|Entropy & The Second Law||27 mins||0 completed|
|Statistical Interpretation of Entropy||12 mins||0 completed|
|Second Law for Ideal Gas||7 mins||0 completed|
Concept #1: Refrigerators
Hey guys, we're going to start talking about refrigerators now let's get to it. A refrigerator can be thought of as the opposite of a heat engine right a heat engine uses heat flow from a hot reservoir to a cold reservoir to produce work right heat naturally wants to flow from hot to cold and an engine uses that fact to convert some of it into work a refrigerator does the opposite a refrigerator uses work in order to move heat in the opposite direction from the cold to the hot against the direction that it wants to flow. So that's the basic idea with the refrigerator kind of the opposite of an engine a refrigerator has the same basic three components as a heat engine does right it's got a hot reservoir and a cold reservoir which are in contact with one another by the machine the machine in this case is a refrigerator and not an engine but it's the same idea now we can diagram the heat flow in a refrigerator just like we diagram heat flow in an engine except that it's sort of the opposite instead of heat entering the cold reservoir heat is leaving the cold reservoir and entering the refrigerator instead of work leaving work is entering the refrigerator and that work allows the refrigerator then to pump heat into the hot reservoir where as an engine would absorb heat from a hot reservoir if we use energy conservation which just tells us that the amount of heat leaving the system has to equal the amount of heat entering the system right those have to equal each other we get this equation which is the exact equation for energy conservation that we would have for a heat engine for a heat engine we would say that heat from the hot reservoir was entering, heat was leaving into the coal reservoir and work was leaving the engine and you would see those two are coming out of the engine this one's coming into the engine and those have to be equal to one another so it's the same energy conservation equation for refrigerator as it is for an engine just like the second law of thermodynamics limits how good an engine can be the second law also limits how good in quotes right a refrigerator can be what this means for an engine is that no engine it's not that it's not practically possible it's that it's not physically possible no engine period could ever convert all heat into work. All heat into work diagrammatically if this was the hot reservoir this was the engine and this was the cold reservoir that would look like heat flowing into the engine and work leaving the engine with no heat flowing into the cold reservoir and this is not physically possible with refrigerators what the second law means is that no refrigerator can transport heat from a cold reservoir into a hot reservoir with out any work right that would look like this, this is the cold reservoir this is the hot reservoir this is the refrigerator.
That would look like this we could just pretend like the refrigerator wasn't even there it would just block it out and this diagram looks the exact same this is simply saying that heat flows on its own from a cold reservoir to a hot reservoir which we know is not possible. Heat always flows from hot to cold it never flows from cold to hot. That limitation right this limiting of how good an engine can be we called the efficiency of that engine right that's how much work it can produce for a given heat input for refrigerator we don't use efficiency we use something called a coefficient of performance which is very similar to efficiency but defined slightly differently all right now something that's very important to remember and understand about refrigerators is that refrigerators need to have their hot and cold reservoirs thermally isolated if they don't you defeat the purpose of the refrigerator now what refrigerator does right is it uses work to take heat from the cold reservoir into the hot reservoir but remember guys that heat wants to flow in the opposite direction. A refrigerator is pumping that heat against its preferred direction of flow if the hot and cold reservoirs were in thermal contact right if they were not thermally isolated then the heat could simply flow backwards in the direction that it wants to go from the hot to the cold reservoir so if you pumped 10 Joules of heat from the cold into the hot then those 10 Jewels would just leave back into the cold and there will be no point to the refrigerator running the whole purpose would be defeated so you cannot have them in thermal contact they need to be thermally isolated otherwise heat will just flow back against the pumping direction and it defeats the whole purpose of having the refrigerator there now. The coefficiency performance right which measures how good of a refrigerator your machine is defined slightly differently than that of the efficiency for an engine for an engine remember the efficiency was defined as the work output divided by the heat input just as a note guys this is not for fridges. Do not see this equation written here and try to apply it to anything to do with refrigerators it's not for them this is for engines but I'm here I'm using it here to compare and contrast between the efficiency of an engine and the coefficient performance of a refrigerator we have the heat from the hot reservoir here and we have the heat from the cold reservoir there but in both cases it means the same thing this is how much heat enters the machine in an engine heat enters from the hot reservoir but in a refrigerator heat enters from a cold reservoir. Now a perfect engine which can convert all of the heat input into work would have an efficiency of 100% so the closer to 100% an engine is the better the engine is. A perfect refrigerator on the other hand would have a coefficient performance of infinity and the reason is for any amount of work even a very very small amount of work you get a large heat transferred a lot of heat leaves the cold reservoir and enters the refrigerator even for very very small works right so if this number is small and this number is large. The overall number is large so it'll be near infinity now a horrible engine right an idealistically horrible the most horrible an engine could ever be is going to have an efficiency at 0 % right a really bad engine the worse it is the nearest to 0% its efficiency is going to be because that means basically it's converting none of the heat input into work likewise a horrible refrigerator would have a coefficient performance near 0 because of the opposite no matter how much heat you put in sorry, no matter how small the heat input into the refrigerator is it requires an incredibly large work. So if the work is very very large and the heat input is still very small then that number is near 0 so you have the same interpretation between a coefficient of performance and efficiency right the lower the efficiency is the worse the engine is the lower the coefficient performances the worse the refrigerator is but a coefficient of performance stops at 1 or 100% but the efficiency for an engine I don't know if I said that correctly the efficiency of an engine stops at 100% or one but the coefficient of performs for a refrigerator can go all the way up to infinity can go as high as it wants to
Now a refrigerator with a coefficient of performance of 4 is connected to a 25 watt power supply how much work does the refrigerator use in 2 seconds? How much heat does the refrigerator generates in 5 seconds? The keyword here for the second question is generate and I'll get to that in a second. So first how much work is it producing in 2 seconds? Well work is just going to be the power at which it consumes work right the rate at which it consumes work times the amount of time that it is consuming work it's consuming energy at 25 watts right its connected to a 25 watt power supply so that's how much work is being used by 25 Joules every second times 2 seconds tells us 50 Joules of energy is being used in 2 seconds. Now we want to know how much heat is generated in 5 seconds? so that's how much heat leaves the refrigerator don't forget our energy conservation equation that says the amount of heat going into the hot reservoir which is how much heat the refrigerator generates equals the amount of heat coming from the cold reservoir this is how much heat enters the refrigerator so they are opposite how much heat enters versus how much heat leaves plus the amount of work used well we can find how much heat is required sorry how much heat is absorbed into the refrigerator through the coefficient of performance and we know the work done right the coefficient of performance is by definition Q C over W so Q C is going to be W times the coefficient of performance which in 2 seconds is 50 Joules and the coefficient of performance is just 4 so this is 200 Joules in 2 seconds 200 Joules in 2 seconds so now that we know this we can find how much heat is produced how much heat is generated by this refrigerator. It's 250 Joules in 2 seconds right now what did the question ask for ? Did it ask for how much heat is produced by the engine in 2 seconds? No it asked how much is produced by the engine in 5 seconds.The rate at which it's producing heat though is 250 Joules every 2 seconds so this is 125 Joules of the heat generated every second so what that means is in 5 seconds all we have to do is multiply the amount of time times how quickly heat is being generated so in 5 seconds the amount of heat being generated is just 125 which is how many Joules f heat is being generated per second times 5 seconds which is 625 Joules being generated in 5 seconds and those are our two answers 50 Joules of work is being used by the refrigerator in 2 seconds, 625 Joules of work is being generated right the amount of heat leaving the refrigerator 625 Joules in 5 seconds. All right guys that wraps up this sort of introduction into refrigerators. Thanks for watching guys.
Example #1: Heat Diagram for a Refrigerator
Hey guys, lets do an example a refrigerator requires 100 Joules of work to operate if the refrigerator releases right releases being the operative word here 150 Joules of heat to the environment every cycle diagram heat flow for this refrigerator and what I mean by the heat flow is the heat flow from the cold reservoir through the refrigerator into the hot reservoir including how much work is required to operate what is the coefficient of performance so two things diagram how the heat flows through the refrigerator including the work and what is the coefficient of performance. So the heat flow is going to look like this here is our hot reservoir here's our cold reservoir, here's our refrigerator we have heat flowing from the refrigerator into the hot reservoir right. Heat flowing from our cold reservoir into our refrigerator and work being used by the refrigerator work is entering the refrigerator so it's a 100 Joules of work and 150 Joules entering the hot reservoir but we don't know how much heat is leaving the cold reservoir but this is the diagram of the heat the diagram of the heat and the work we can use energy conservation I'm going to drop the absolute values just remember that they always need to be positive because this is about quantity of heat this isn't about positive or negative sign which tells us where the energy is going. So that tells us that the amount of heat leaving the cold reservoir and the refrigerator is just QH minus W. We know that this is 150 and this is 100 so that is 50 Joules right
So we can say then this is 50 Joules and now energy is conserved right you could clearly see a 100 and 50 are entering 150 is leaving but what's the coefficient of performance well that's how much heat enters the refrigerator right which is the heat from the cold reservoir divided by the amount of work the refrigerator uses 50 Joules of heat enters the refrigerator from the cold reservoir and it requires a 100 Joules of heat to operate so this is 0.5 that is the coefficients of performance. Alright guys that wraps up this problem. Thanks for watching.
Concept #2: Maximum Theoretical Coefficient of Performance
Hey guys, in this video we're going to talk about the maximum possible theoretical coefficient of performance for a refrigerator. Alright let's get to it. Remember guys that the carnot engine had the maximum possible efficiency for an engine placed between any hot and any cold reservoir. Any other engine placed between the two would have a lower efficiency. So if we had some hot reservoir and some cold reservoir the carnot engine would output some work WC let's say. Any other engine placed between the same reservoirs so T hot's the same T cold's the same would have to produce some W less than W carnot. So any other engine that's not a carnot engine is going to produce less work for the same temperature gradients. That's the idea with the carnot engine. Now likewise there's a maximum theoretical efficiency for a refrigerator placed between a hot and a cold reservoir, the same idea you have some hot and some cold reservoir if you put any other refrigerator between the same hot and the same cold reservoir you would have a lesser coefficient of performance. The larger the coefficient of performance the better the performance of the refrigerator. So if the carnot refrigerator which is what we call it. We call this theoretical refrigerator a carnot refrigerator, if the carnot refrigerator produces the most amount of heat pumped into the hot reservoir between any two thermal, sorry, between any two thermal reservoirs then any other refrigerator that' not a carnot refrigerator is going to produce less heat pumped for the same temperature difference. The same idea we have some hot reservoir, we have some cold reservoir, we have a carnot refrigerator with some work being pumped in, there's some a carnot heat entering the refrigerator from the cold reservoir. If we put in any other engine, sorry, any other refrigerator that was not a carnot refrigerator then for the same work done it would absorb a heat less than that of the carnot refrigerators so it can pump less heat, it has worse performance, it has a lesser coefficient of performance. The coefficient of performance of that ideal refrigerator of that perfect refrigerator is the maximum coefficient performance or the carnot coefficient of performance and it's the temperature of the cold reservoir divided by the temperature of the hot minus the temperature of the cold reservoir. This is for some hot reservoir at TH and some cold reservoir at TC. Very simple, very straightforward.
Let's do an example. A refrigerator is proposed to operate between a cold reservoir of 300 Kelvin and a hot reservoir of 500 Kelvin. The refrigerator will draw 15 watts of power from a power supply, will expel 20 joules of heat into the environment every second and operates at 50 cycles per second. Do you believe that such a refrigerator could exist? Now typically we define our coefficient of performance in terms of the amount of heat entering the refrigerator which comes from the cold reservoir over the amount of work being used by the refrigerator but we're not given an amount of heat or an amount of work being used, we're given the rate at which heat is, sorry, the rate at which work is being used and the rate at which heat is being produced. So we can define rates, we can say that the power is just the work per unit time or that the work produced is going to be the power times however long you're measuring. If it's one second, you produce 15 joules of work or sorry you use 15 joules of work. If it's 2 seconds it's 30 joules of work. Likewise we can define the rate of conduction or of the conductance which we talked about when discussing thermal conduction as the heat specifically for the cold reservoir divided by time as well. So if we want to know how much heat is being absorbed by the cold reservoir we could say that it was that conductance times time. Nothing in this problem tells us how much heat is being produced sorry how much heat is being absorbed by a refrigerator from the cold reservoir though, but it does tell us the rate at which heat is being generated. In one second, 20 joules of heat is being produced by the refrigerator that's 20 joules of heat being dumped into the hot reservoir and according to our work equation, 15 joules of work is being used in one second.
So we can use our conservation of energy requirement which says QH is QC plus W or that QC the amount of heat absorbed by the refrigerator is QH minus W which is 20 minus 15 which is 5 joules in 1 second. So we know that the refrigerator is absorbing 5 joules every 1 second or 5 joules every second. So using this definition right here we can just say that the conductance from the cold reservoir into the refrigerator is, if you bring a 1 not a 1 it's a 5 right 5 joules in 1 second or 5 watts. Now the reason we want to do this is because while the coefficient of performance is typically defined in terms of how much heat is being absorbed by the refrigerator and how much work is being used, we can actually define the coefficient of performance in terms of how much power is being used by the refrigerator and the conductance from the cold reservoir into the refrigerator and those times are just going to cancel so we can just say that this is the conductance from the cold reservoir over the power used and we know both of those. The conductance is 5 and the power used is 15 so this is 0.33. Now the question is could this refrigerator exist? Well the limit, the physical limit is that no coefficient of performance for a refrigerator can ever be larger than the carnot refrigerators coefficient of performance. So for a carnot refrigerator, the coefficient of performance maximum is going to the temperature of the cold reservoir divided by the temperature of the hot minus the temperature of the cold. This is going to be 300 Kelvin divided by 500 minus 300 which is 1.5. Notice that the coefficient of performance proposed is absolutely less than the coefficient of performance maximum and what that means is this engine could absolutely exist. So long as any proposed engine does not violate the second law of thermodynamics meaning that the coefficient of performance of that engine is never larger than the maximum allowed coefficient of performance, the carnot coefficient of performance then that engine absolutely could exist. It's just a matter of getting the engineering down to allow that coefficient of performance. Alright guys that wraps up the discussion on the carnot refrigerator and the maximum theoretical coefficient of performance for refrigerators. Thanks for watching
Concept #3: Refrigerators and PV Diagrams
Hey guys, in this video we're going to talk specifically about refrigerators and PV diagrams. We saw that PV diagrams were incredibly useful in discussing heat engines, they're also incredibly useful when discussing refrigerators. Let's get to it.
Remember that a heat engine requires a clockwise cycle on a PV diagram. Clockwise means the work of the gas making up that engine is negative and that means that work is released and you want an engine to release work. You want to produce usable work that the machine can use. This is required like I said clockwise cycles are required to have energy released for exactly this reason that I just wrote over here. Refrigerators can also be described as cycles on a PV diagram and gas in a refrigerator should absolutely undergo a cyclic process, that way it starts and ends at the same state so you can keep doing it cycle after cycle after cycle. It's the same reason why you want engines to be cyclic processes. However a refrigerator requires work to be input, work to act on the gas that means that the cycle needs to be counterclockwise because counterclockwise means that the work of the gas is positive which means that work was acting on the gas, that the gas absorbed energy. So in order for you to have a refrigerator described on a PV diagram, the cycle has to be counterclockwise because that's the requirement for having work being absorbed by the gas. These are two cycles right here we see that this is clockwise which means that work is released because the work of the gas is negative, the gas loses energy, that energy is released which means that this is an engine. On the right we have a counterclockwise rotation. Counter means work is absorbed, absorbed is not spelled with an O just a pro tip there guys, I'm not the greatest speller in the world. Work is absorbed the gas gains energy the work on the gas is positive and this means that this is a refrigerator. So that's just what two PV diagrams would look like one for refrigerator, one for an engine. Counterclockwise, clockwise.
Let's do an example. The cycle undergoes, sorry, the cycle a gas undergoes in a refrigerator shown in the following PV diagram. How much work is required to operate the refrigerator? How much work does the refrigerator need in each cycle in order to go on? Any cyclic process, the amount of work by the cycle is just the area enclosed by the cycle. This area is simply base times height, it's just a rectangle. The base is the difference in these volumes which is 0.002 this is the correct unit meters cubed and the height is the difference in these pressures from 1 to 5 is 4 but don't forget you're going to need that 10 to the 5 in order to have the right units. So this is 0.002 times 4 times 10 to the 5 Pascals which plugging it into calculator is 800 joules. So the work absorbed is 800 joules and that's how much work each cycle of the refrigerator requires to operate. Besides this guys, besides a cycle in a refrigerator being counter whereas a cycle in an engine is clockwise, all the same rules apply in all the same equations that we learned in the thermal processes apply for cyclic processes. This is just another cyclic process like an engine is a cyclic process but it's moving in the opposite direction. That wraps up this discussion on refrigerators and specifically cycles of gases on PV diagrams to describe refrigerators. Thanks for watching guys.
Practice: A monoatomic ideal gas is refrigerated in the following cycle. What is the coefficient of performance of the refrigerator?
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