WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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Today's lesson is on thermodynamics.
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Our objectives are going to be to understand that energy is transferred spontaneously from a higher temperature system to a lower temperature system, to explain the first law of thermodynamics in terms of conservation of energy involving the internal energy of a system, and to represent transfers of energy through work and heat by using PV diagrams.
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Let us begin by talking about the zeroth law of thermodynamics.
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The zeroth law of thermodynamics, they added after some other laws of thermodynamics because they needed it to help make all of their proofs work out.
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It saids if Object (A) is in thermal equilibrium with Object (B), and Object (B) is in thermal equilibrium with Object (C), then Object (A) must be in thermal equilibrium with Object (C).
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Sounds kind of obvious, but just so we have everything in there, that is the zeroth law of thermodynamics.
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The first law of thermodynamics is a little bit more practical for our purposes.
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It says that the change in the internal energy of a closed system is equal to the heat added to the system plus the work done on the system.
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ΔU, change in internal energy is heat added to plus work done on, and those are for the positive values.
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This is really just a restatement of the law of conservation of energy applied in the thermal sense.
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The sign conventions are extremely important.
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Positive heat is heat added to the system; positive work is work done on the system.
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If heat is taken from the system, it is negative and if work is done by the system, the work is negative.
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All right. Let us talk about work done on a gas.
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Typically we will use the first law of thermodynamics to analyze the behavior of ideal gases.
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It may be useful to explore our understandings of the work done on a gas a little bit though.
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If you recall, work is force times the displacement -- and we are going to assume that we have it in the same direction so that we do not have to worry about sines/cosines.
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That is a reasonable assumption as we are talking about thermodynamics, which implies then -- well if we know pressure is force over area, then force must be pressure times area.
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I could rewrite this as work is equal to pressure times area times Δr, but we are going to take another step here.
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Change in volume is equal to A(Δr) and because we have the convention, that work done on the gas is positive, corresponding to a decrease in volume, we will put a negative sign there, so our sign conventions work out.
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Then we could say that work is equal to -P × Δv.
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All right if work is force multiplied by displacement, then work is pressure times area times displacement and negative -- just there for the sign convention -- replace A × Δr with Δv and we get that work is minus P(Δv).
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That is going to be extremely helpful as we start analyzing these gas systems.
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Let us take an example.
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Five thousands joules of heat are added to a closed system which then does 3,000 J of work.
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What is the net change in the internal energy of the system?
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Well, Δu is (Q) + (W) -- 5,000 J of heat are added to, added to, so that must be positive, so 5,000 J is positive, which then does 3,000 J of work.
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If the system is doing the work, that is negative, so -3,000 -- our total change in net internal energy, must be 2,000 J.
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Or a second example -- a gas is expanded at atmospheric pressure, 101,325 Pa.
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The volume of the gas was 5 × 10^6m³.
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The volume of the gas is now 5 × 10^-3m³.
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How much work was done in the process?
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Well, work equals -P(Δv), so that's (-P) and Δ anything is always the final value minus the initial.
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So that is V-final - V-initial; P is 101,325 Pa; V-final is 5 × 10^-3m³... 0291 ...V-initial is 5 × 10^-6m³, which implies then that the work is -506 J.
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All right. Let us talk about another useful tool for analyzing gas systems.
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It is called the pressure volume diagram or PV diagram.
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We put pressure on the y-axis, volume on the x-axis and we are going to keep the amount of gas constant, so when we talk about PV = NRT, our ideal gas law, pressures on the graph, volumes on the graph, the amount of gas is constant, so that stays constant, (R) is already a gas constant...
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...we can solve for (T) using the ideal gas law, so a PV diagram shows us pressure, volume, and indirectly temperature, so we can find (T) once we know these other quantities.
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If we transition from state (A) to state (B) on a PV diagram, the volume is increasing, so our pressure is decreasing.
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The work done then is going to be the area under the curve from (A) to (B).
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That area here is going to be our work.
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As the volume expands, the gas is doing work, so (W) would be negative and as the volume compresses, the work is being done on the gas, (W) is positive.
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Also important to note here is that as you move up into the right on the graph, you move to higher temperatures.
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Let us take a look at some analysis using a PV diagram.
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Using the PV diagram below, find the amount of work required to transition from state (A) to state (B) and then the amount of work required to go from state (B) to state (C).
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Well let us start out with the work going from (A) to (B).
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The work in going from state (A) to state (B) is the area under the graph and as we go from (A) to (B), that is just a straight line, there is no area -- no work done.
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How about the work done as we go from (B) to (C)?
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Well that is -P × ΔV or minus 50,000 Pa -- V, is ΔV is final minus initial, so that is going to be 4 m³- 2 m³ or -100,000 J.
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Notice that the gas was expanding, the gas was doing work.
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Work is positive if the work is done on the gas since the gas is doing work it makes sense that we get a negative value for the work done in going from (B) to (C).
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There are several different types of PV processes that we ought to point out, special PV processes.
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They have some goofy names and they are kind of vocabulary words, so you really just have to memorize these.
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Adiabatic -- This is when heat (Q) is not transferred into or out of the system; the heat remains constant.
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That is adiabatic and a PV graph for an adiabatic process looks like this here in the light blue -- adiabatic.
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Isobaric -- pressure (P) remains constant and in an isobaric process, since (P) remains constant, you have a horizontal line.
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Isochoric means volume remains constant so that means you have a vertical line and you stay at the same (V).
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An isothermal means temperature (T) remains constant and you get an isotherm that looks like this -- isothermal lines on a PV diagram, we call isotherms.
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And we will dive into these in a little bit more detail right away.
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Adiabatic process -- heat is not transferred into or out of the system -- Q = 0 -- therefore by the first law of thermodynamics, if ΔU is equal to Q + W, and we know that Q = 0 in an adiabatic process, then the change in internal energy of the gas is the work done on the gas, ΔU = W.
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Pretty straightforward and the processes have that sort of shape.
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An isobaric process -- pressure remains constant.
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Isobaric -- constant pressure -- the PV diagram shows a horizontal line and if PV = NRT, (P) is constant and then (R) are constant, we can rearrange this to say that V/T = NR/P.
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If all of that is constant, that means that V/T, that ratio remains constant for any gas processes.
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That happens in an isobaric or constant pressure process.
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In an isochoric process, the volume remains constant.
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In an isochoric, we have constant volume or a vertical line and the work done on the gas is 0, because remember work done on a gas is the area under the graph and in a vertical line, you do not have any area under it and if PV = NRT and volume remains constant, well constant P/T = NR/V.
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All of those are constant, so the ratio of P/T remains constant for all of your processes.
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In an isothermal processes where the temperature remains constant, the lines on the PV diagram for these are called isotherms; there is an isothermal process.
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If PV remains constant, the internal energy of the gas must remain constant.
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Let us look at an example for an adiabatic expansion.
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An ideal gas undergoes an adiabatic expansion -- adiabatic -- Q = 0 -- no transfer -- doing 2,000 J of work.
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How much does the gases internal energy change?
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Well, Δu = Q + W, but since it is adiabatic, we know that Q = 0, so Δu = W, which must be - 2,000 J.
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The biggest trick here is remembering the definitions of these terms.
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Example 5: Removing some heat -- Heat is removed from an ideal gas as its pressure drops from 2,000 Pa to 100,000 Pa.
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The gas then expands from a volume of 0.05 m³ to 0.1 m³ as shown in the PV diagram below.
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If curve (AC) represents an isotherm, find the work done by the gas and the heat added to the gas.
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Well, right away the work in going from (A) to (B) is 0, because there is no area under that graph and the work going from (B) to (C) is just -P(Δv)...
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... or -100,000 Pa × V-final - V-initial or 0.1 - 0.05, which is -5,000 J.
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That is the work done by the gas, that is why it is negative.
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Now we are on an isotherm going from (A) to (C), so (U) must be constant; our internal energy has to stay the same.
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Δu = 0, which equals Q + W, therefore, Q = -W = 5,000 J.
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You must have added 5,000 J to the gas.
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Our key answers -- find the work done by the gas -- the work done by the gas was 5,000 J and the heat added to the gas -- we added 5,000 J.
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The gas did 5,000 J of work and we added 5,000 J to it.
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Let us take a look at the PV diagram below and answer these questions.
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During which process is the most work done by the gas?
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Well, work done by the gas, that is a negative work or an expanding gas.
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We see that -- that is the area under the graph going to the right here from (A) to (B), so that must be (A) to (B) here.
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Going from (B) to (C) is no work or no area and from (C) to (A), we are compressing the gas, so work is being done on it.
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Again, during which process is the most work done on the gas?
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That must be going from (C) to (A).
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We have the most area going from (C) to (A) and we are compressing the gas, so work is being done to the gas.
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In which state is it the highest temperature?
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Remember temperature gets bigger as you go up into the right, so that must be state (C).
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On to the second law of thermodynamics.
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Heat flows naturally from a warmer object to a colder object and cannot flow from a colder object to a warmer object without doing work on the system.
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Heat energy also cannot be completely transformed into mechanical work or another way to say that is nothing is 100% efficient.
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Now all natural systems tend toward a higher level of disorder or entropy.
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The only way to decrease the entropy of a system is to do work on it.
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An entropy is kind of a state of disorder.
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For example, if I had a really cool Lego castle here right now and I dropped it, it is going to become more messy.
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In the natural state of the world, I am never going to have a bunch of Lego's in all different pieces dropping and then when I look down and go to pick it up, the castle is already built.
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Things do not get more ordered unless you do work on it.
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That is the second law of thermodynamics.
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Now, another way to look at this is in terms of heat engines.
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Heat engines convert heat into mechanical work.
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And the efficiency of a heat engine is the ratio of the energy you get out in the form of work to the energy you put in, so typically how these work... you have a high temp reservoir, a place where you create a lot of heat.
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You use that to do some sort of work.
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If you have heat energy at the high temp reservoir, some of it becomes productive output and some of it goes into the low temp reservoir, where it is not very useful.
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The work that you get out is equal to what you put in minus what is left over -- what goes to that low temp reservoir, and the efficiency of your system is going to be what you wanted to work out divided by what you put in.
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And we will put the absolute value signs around that, just so you do not have to deal with negatives.
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But W = Qh - Qc/Qh, so you could rewrite that if you wanted as 1 - Qc/Qh.
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A couple of key things, but the efficiency is one of the key formulas from this slide.
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Power in heat engines -- Power is the rate at which work is done, work over time.
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We talked about that back in mechanics.
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From a heat engine perspective, though, we can take this a little bit further.
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If efficiency is work over the high temp heat, then we could rewrite that as work is equal to the efficiency times (Qh) or dividing both sides by time -- W/t is efficiency × Qh/t.
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Work over time is power, so since P = W/t, then power on the left hand side becomes efficiency × Qh/t, but let us go another step.
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We just found that efficiency could also be written as 1 - Qc/Qh, therefore, P = 1 - Qc/Qh × Qh/t.
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Well with a little bit more rearrangement and a little more Algebra, P = Qh/t - -- well the Qh's will cancel -- Qc/t.
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A couple of other ways to help you calculate the power from heat engines.
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All right. Heat engines and PV diagrams -- On a PV diagram, a heat engine is a closed cycle.
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For clockwise processes, these are heat engines.
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If you go in the other direction, counter-clockwise processes -- those are refrigerators.
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Now let us talk a little bit about the Carnot engine.
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The Carnot engine is not something that you just go out and buy.
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It is a theoretical model, a theoretical idea of an engine that has the maximum possible efficiency.
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It uses only isothermal and adiabatic processes and Carnot's theorem states that no engine operating between two heat reservoirs can be more efficient than the Carnot engine operating between those same two reservoirs.
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So the Carnot engine is kind of the theoretical model of the maximum efficiency you could get from an engine and the efficiency of the Carnot engine is equal to the temperature of the hot reservoir minus the temperature of the cold reservoir, divided by the temperature of the hot reservoir.
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When you actually utilize this to do calculations, keep a note that the temperature must be in standard SI units or Kelvins.
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Let us take another look at a Carnot engine problem.
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A 35% efficient Carnot engine absorbs 1,000 J of heat per cycle from a high temp reservoir held at 600 K.
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Find the heat expelled per cycle as well as the temperature of the cold reservoir.
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Well, if our efficiency is 35% or 0.35, we also know that our Qh is 1,000 J per cycle and that the temperature of our high temp reservoir is 600 K.
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We could start with efficiency as our high temperature when its our cold temperature divided by our hot temperature for the engine, therefore, efficiency equals 1 - cold temperature/hot temperature or cold temperature/hot temperature is 1 - efficiency.
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Therefore, to find the cold temperature, (TC) is going to be equal to the hot temperature times 1 - the efficiency or 600 K × 1 - 0.35 = 0.65 × 600 or 390 K.
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Now we have the heat expelled per cycle as well as the temperature of the cold reservoir, so if we want E = W/Qh and we want to find what that W is, that is going to be E × Qh or our efficiency 0.35 × the heat on the hot side (1,000 J) or 350 J.
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So then W = Qh - Qc.
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Therefore, Qc = Qh - W or 1,000 - 350 = 650 J.
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Let us look at a maximum efficiency problem.
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Determine the maximum efficiency of a heat engine with a high temperature reservoir of 1200 K and a low temperature reservoir of 400 K.
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Now, this is not really asking for a Carnot efficiency because the most efficiency you can have is the Carnot engine.
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The Carnot efficiency is Th - Tc/Th or 1200 K - 400 K/1200 K = 0.667 or about 66.7%.
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One last problem here -- Which of the following terms best describes a PV process in which the volume of the gas remains constant?
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Constant volume -- So I check on vocabulary words from those PV processes -- Adiabatic, no that is constant (Q); isobaric -- that is constant pressure; isochoric -- that is constant volume, and isothermal of course is constant temperature.
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Our correct answer there must be C.
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Hopefully that will give you a good start in thermodynamics.
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I appreciate your time and thanks for coming to visit us at Educator.com.
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Make it a great day everyone!