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A typical coal-fired power plant generates 1000 $\mathrm{MW}$ ofusable power at an overall thermal efficiency of 40$\%$ . (a) Whatis the rate of heat input to the plant? (b) The plant burmsanthracite coal, which has a heat of combustion of $2.65 \times 10^{7} \mathrm{J} / \mathrm{kg} .$ How much coal does the plant use per day, ifit operates continuously? (c) At what rate is heat ejected intothe cool reservoir, which is the nearby river? (d) The river's temperature is $18.0^{\circ} \mathrm{C}$ before it reaches the power plant and$18.5^{\circ} \mathrm{C}$ after it has received the plant's waste heat. Calculatethe river's flow rate, in cubic meters per second. (e) By howmuch does the river's entropy increase each second?

(a) $\left(\frac{Q_{H}}{t}\right)=2500 \mathrm{MW}$(b) $m=82.0 \times 10^{5} \mathrm{kg}$ of coal(c) $\left(\frac{Q_{C}}{t}\right)=1500 \mathrm{MW}$(d) Flow rate $=716 \mathrm{m}^{3} / \mathrm{s}$(e) $\Delta S=52.0 \times 10^{5} \mathrm{J} / \mathrm{K}$

Physics 101 Mechanics

Chapter 16

The Second Law of Thermodynamics

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

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in part A. We're going to use the fact that efficiency is able to work over Cuba. Sewing for Q H. We get work over efficiency and then now transforming this into power. We get th is equal to PW over deficiency, and we know what P. W is. And we know what the efficiency is so we can calculate he ate. And so P. W. Is 1,000 megawatts and efficiency is 40%. And so this gives 2.5 times 10 to the third. Make it once and this is the answer to party. In part be, we know that the he input and one day is going to be this time is the time of one day. Now there's 8.64 times 10 to the four seconds a day because they're gonna multiply this time by this power to get the toll heat going in. But this is making a lot, so I'm going to convert it into Watts. And what that does is it turns this three into a nine. Since that mega won is a lot of watts. In fact, it's 1,000,000 once what finds out gives us 2.16 times 10 to the 14 Jules and to figure out the mass of the coal used. What weaken use is the jewels for kilogram that's given in the problem. And so if we take this amount here and divide by 3.65 times 10 to the seven Jules per kilogram, this is given the problem, and it converts the energy into a mass that we need to use. You'll see the jewels would cancel, the coherence will flip up and we get 8.15 times 10 to the six kilograms. Thats how much coal we need to supply to get this amount of energy, and that's the amount of energy that we're going to get in one day. So it's the Parsi to figure out the power going into the cold reservoir. We need to use this equation. Actually, it's supposed to be in a tch. Q H is equal to W plus QC, and so, you see, is equal to Kyu age minus of you and now just taking the dirt of respects time or converting it into power like we did in party we get PC is equal to P. H minus PW and then now we can plug in for P H and P. W. PH, we found was 2.5 times 10 to the third megawatts. It's three in the exposure and POW is 1,000. Maybe it wants. And so this ends up being 1.5 times 10 to the third. Mayor wants This is the amount of power going into the cold reservoir. And so is the answer to Marcy Snow in the party. We just found what PC was, and this is the amount of power going into the river. And so, just to reiterate that he employed into the river is 1.50 times 10 to the nine jewels for second and recall Jules per second isn't want. So this year is why we're in here. And that's the heat and put into the river. And so now we're going to use this equation. Q is equal to Elm Sea Delta. T C is the specific e It was the mass, and lt is a change in temperature are going to use. Lt is equal to 0.5 degrees Celsius. This has given the problem and we're going to solve this for elm. When we do that, we get you over. See Delta T we know CIA's we know adult in tears and so we can plug in what this is for one second if we take one second here than this just turns into 151.50 times 10 to 9 jewels. That's the he entering the river in one second. And this will be equal to our cue in this equation here. And so playing this this and see for water in which is a known constant 4,190 Jules per kilogram. Kelvin, we can solve this and we get seven point 16 times 10 to the fifth kilograms. This is the kilograms of water that is heated by this amount. In one second, we can convert. Convert this into a volume by using the density of water supply musicals of mass over the density, and the density of water is well known. And so it's just a number. You go well, if you want. The volume is of being 716 meters tube and so this is the mass of the water being heated by this much in one second This is the volume of the water being heated by this much in one second. And so the river floor. It must be seven sixteen 16 meters Q and A one second. So says the Florey. So now we have to figure out the entropy change. And since the temperature of the river is nearly constant, we can imply the so called ice, a thermal expanse of ice, a thermal approximation. And this is discussed in 16 8 of the books of Section 16 8 And this is Thie Isil thermal X approximation for finding entropy. And so in one second, 7.16 times 10 to the fifth kilograms of water goes from 2 91.1 killing 2 to 91.6 kelvin. All right. It changes by half a degree, and the amount of heat added is 1.50 times 10 to the nine jewels. Heat added to get this change in temperature to happen. And so the interview change is just equal to 1.50 times 10 to the nine. Where is the heat added over the 2 91.6 and the reason we used to 91.6 is the Isil thermal approximation. If you want more information, c section 16.8 In the textbook, when you do the sound, you get 5.1 times 10 to the six jewels for killing and that.

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