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$\bullet$ As a budding mechanical engineer, you are called upon todesign a Carnot engine that has 2.00 moles of He gas (seeTable 15.4 ) as its working substance and that operates from a high-temperature reservoir at $500^{\circ} \mathrm{C}$ . The engine is to lift a15.0 kg weight 2.00 $\mathrm{m}$ per cycle, using 500 $\mathrm{J}$ of heat input. Thegas in the engine chamber can have a minimum volume of 5.00 $\mathrm{L}$ during the cycle. (a) Draw a $p V$ diagram for this cycle. In yourdiagram, show where heat enters and leaves the gas. (b) Whatmust be the temperature of the cold reservoir? (c) What is the thermal efficiency of the engine? (d) How much heat energy doesthis engine waste per cycle? (e) What is the maximum pressurethat the gas chamber will have to withstand?

(a) Draw a Carnot cycle(b) $T_{C}=45.0^{\circ} \mathrm{C}$(c) eCarnot $=59.0 \%$(d) $\left|Q_{C}\right|=206 \mathrm{J}$(e) The maximum pressure is at point a $=2570 \times 10^{3} \mathrm{Pa}$

Physics 101 Mechanics

Chapter 16

The Second Law of Thermodynamics

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

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University of Washington

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In physics, the second law…

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here, we're going to look at a particular Carnot engine that is using helium, an ideal gas as the working substance. So, a reminder of carnot engine, um I have a PV diagram for one below, but it consists of two idiomatic segments along which there is no heat exchange with the environment. I've showed showing those in blue and to ice a thermal segments along which there is no change in internal energy if you have an ideal gas, because the internal energy of an ideal gas depends solely on the temperature. So uh there are two temperatures in such a system A. T. High and A. T. Low. The T. High is associated with points A. And B. Uh and the work gets done because of the absorption of heat along with the expansion of the gas. And then there's a tea cold for the lower is a thermal segment um along which exhaust heat is expelled back into the environment and the area enclosed by that for segment loop, is the work done. So we have a lot of things specified about this engine. What I see is we have the work done per cycle specified as the force needed to lift a particular weight by a certain distance. So we can quickly calculate the work done per cycle jules. We also have the amount of heat that this engine is supposedly absorbing um As it does, it's useful work. And so we can determine the efficiency of this engine. There is a general expression for efficiency as work, output per heat input and then times 100%. I usually don't or you too much about that. Um And that is 0.588 I'll leave the 100% off to the side. But what we know about a carnot engine is it has the best efficiency possible. And that efficiency is related to the two temperatures, the hot and the cold in a very simple manner based on definition of entropy and since we know the efficiency and we know the high temperature, we can't use that. Yeah, there's 100%. Again, sorry we can use this carnot efficiency in order to solve for the lower temperature and um I won't go through the outbreak details. You can see why I like to keep writing the 100% off to the side. Um solving for this, we get the t. Low equals 318 kelvin. Yeah, that's definitely lower than 773. Yeah. Um We can also find the lower bound to the uh expelled heat through using conservation of energy. Um That is what we have is a no internal energy change throughout the entire cycle. So delta U. Is equal to zero per cycle and that is the Q. Net minus the work done. So the work done is the difference between the heat absorbed and the heat expelled. I'll put an absolute value around that because usually you consider that to be negative. Um But we can then solve for the expelled waste heat. That may be an important consideration for cooling the engine. Yes. So it doesn't overheat. So that's 500 jewels minus the heat expelled into the cold reservoir. And so we can solve for Q. Expelled. And we can see that that's a certain percent of the total energy was absorbed. It's a 0.412 So we could have found that also by subtracting the 0.588 from one and multiplying by 500 jewels. But I kind of like using the first law of thermo explicitly. Just kind of a good practice. Okay. And the last thing we may want to know is what is the maximum pressure that the system will need to handle? Um and where that maximum pressure is, I've labeled the points on my cycle and that's that point A which is at the high temperature and it's at the low volume, so that's at point A and of course we'll need to use our uh huh. Good old ideal gas law. Nice thing about using ideal gases is they have a simple equation. So we know just about everything in there. There are two moles of gas, of course, we know the gas constant are we know the temperature at that point is 773 kelvin and we know the volume is five times 10 to the minus three cubic meters. And those are on S. I. Units. So that will give us units of pascal. So um a couple of times 10 times one atmosphere. We won't worry too much about it, except it is a fairly high pressure to deal with

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