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PHSC101 Homework Heat Interactions Kol Dtlha ut Elkd ~thk plcu Uitio & 37 Lr quuler #iut nimic cluleelie follawing Mkaon G/R to Ice nd &ze Ivo |el, wudl tbo...

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PHSC101 Homework Heat Interactions Kol Dtlha ut Elkd ~thk plcu Uitio & 37 Lr quuler #iut nimic cluleelie follawing Mkaon G/R to Ice nd &ze Ivo |el, wudl tbo txtlo ot Wutot cools donn ilingtIIU &l thl AtunektaHe [micthcta: Wtor MattuThcnnn Uanctu_ITtxrmi Fueeyso IElak tale dlngcamn corecl' Juytlf: KereutKeAUtuxleeCOmcredDmutUu: AoUll Acnelelnm Dotine Mhee orute Comtact PIP |ntotucexin baauol Ulo block Mu| ehe table. #ua Knoue -hat cnly consictry I ~ole nny ablect tunur (iaIL @utu

PHSC101 Homework Heat Interactions Kol Dtlha ut Elkd ~thk plcu Uitio & 37 Lr quuler #iut nimic cluleelie follawing Mkaon G/R to Ice nd &ze Ivo |el, wudl tbo txtlo ot Wutot cools donn ilingtIIU &l thl Atunekta He [micthcta: Wtor Mattu Thcnnn Uanctu_ ITtxrmi Fueey so IElak tale dlngcamn corecl' Juytlf: Kereut KeA Utuxlee COmcred DmutUu: AoUll Acnelelnm Dotine Mhee orute Comtact PIP |ntotucexin baauol Ulo block Mu| ehe table. #ua Knoue -hat cnly consictry I ~ole nny ablect tunur (iaIL @utuide Juinenent Siioo Lim uloce nnd td Votu #itDCT{MLII Lnclr puIIuuudLn JoNOD thcy ~uM ehieaeloro havo hent EACLIOIA Uli E mninii; 'cnplla | G/R dlaram hlEuuntko FW cl: 4Iui Anununt OlTm- Uuck Knmu-Wrau Tat A Heneenice Gulenutlul Ealyexatteue



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State precisely the meaning of each of the following terms. You may need to review Chapter 1 to refresh your memory about terms introduced there. (a) heat; (b) temperature; (c) system; (d) surroundings; (e) thermodynamic state of svstem; (f) work.

Question 63 says a black has dimensions L too Well and three hour when and then we're just gonna change the orientation of his block in space. But we're always going between two temperature differences t one and T too. So the Delta T is the same anyways, it says when one of the EL by two l faces so l by two l So that would be this face, right? Eyes at t one and the other l by two. Will face is held AT T to the rate of heat conduction through them is Pete So that tells me P is equal to And we've got to be very careful how we do this right. We gotta pay attention to this, but it's got some constant. And then we multiply that by its area. That's cross sectional area is l by two l and then we multiply that by the delta T at's Delta T. And then we divide that by the length. Well, in this case, that's three l right? And so what I have is the power. When the temperature differences is held between these two faces is given by K Delta T and then one of these else cancels. The other one does not write. And so I have times to over three l. Now what it asks us in part is what is the rate of heat conduction in the block? If one of the EL by three l faces is held a temperature t one and the other is held by three. Ellis held a t two so same temperature difference. But now we're going to use the top and bottom faces. So this is part B. We're going to go across there, which means now our power is given by, um, this constant, which is so same Delta T, which is still the same. But now the area is given by L. A times three l and then the Delta t. Like I said, that's still the same. And now our length is actually a cross this block this way, which is of course, to l. And so I'm left with this powers equal to K times three halves l delta T times K. So how is power one related to power, too. Well, in order to make three halves into our 2/3 into three halves, I would have to multiply by nine over four, right? If I take 2/3 and divide it by three halves, that means multiplied by the reciprocal there. Sorry, the other way around, if I take three have been divided by 2/3. That's multiplied by the reciprocal. So three has times three halves is, of course, nine over four. So the power here the rate is nine over four times the power here, which makes sense because it's going through a much bigger area. So go faster and it's going through a shorter distance so it can go faster. Now how are three? It says the last face. It says, What do we do if the rate of heat conduction is the If the block is one of the two well by three l faces, and so now our area is going to be too well, times three out on and then over the length. But now, since we're going through this face, our link is only L. And this is of course, six L. So what do I have to multiply 2/3 by right 2/3 times what is going to get me six? And so again, I multiplied by that reciprocal, which is three have six times three over two is, of course, nine. So there's nine times the power here, so have nine times the power for Part B, and we have 9/4 times the power for party.

So here will be fine. Hate temperature system surroundings thermodynamic state of system and work. So now he'd hit is defined amount of energy transport from one place to another. It is because of result of temperature difference. No, and Bridget. It is a fundamental quantity. Used to measure the degree of hotness or coldness. Usually we use a scale to measure temperature. Uh Now come to we come to systems a set of substances that are being studied that is being focused that are we focusing on that whole thing is the system. Now the surroundings. Now, if we leave the system, everything else is surrounding for everything. External legal system is known as surrounding, not thermodynamic state of system. This is defined other set of values of properties of a thermodynamic system that are specified in order to define a system. So, uh some values of properties, properties are there uh that we have to specify in order to define a system. These individual parameters are also known as thermodynamic variables. Now coming to work what is simply defined as quantity of energy transport from one system to another. It's society. Unity. Also jewel. Thanks

Here. The information we have, we are given a mallet 0.35 kg. The initial temperature is 22 degrees Celsius, convert to kelvin. The melting point is negative. 20 self Equus Celsius also converted kelvin. And the final temperature is the temperature of the nitrogen, the 77 kelvin's. And we have the specific heat uh in liquid form, then specific hidden solid farm. And uh specific latent heat of fusion. Yes, so that's l so that the total hit required in three stages. So the first stage is the heat required to go some 95-50 trick open still in the liquid state. Then we have the heat of fusion to melt to convert the liquid solid. Then we have the heat required to go from 2 53 2 to 77 kilometers. Okay, so we used the three questions here. The first question here in the first parenthesis is the heat required for the first stage and the second is for the heat of fusion. And the third is that he recorded the solid state. Okay, so we plug in our numbers, you can factor out the mass, keep it outside and put everything else in the parenthesis. Do the calculation and we get to 8733350 jewels. And that's 2.9 10, 10 of the far five jewels. That's option

So we are cold. A piece of hot metal, A supposed to be rapidly cooled or quenched to 25 degrees C. And that requires that we take, um um 1000 killer jewels out of the metal. Which means that that energia energy heat must go into the working fluid s. So that would be our heat into the fluid. Um, the cold space that absorbs the the energy could be one of the possible. Okay, so, one, it could be a submerging the metal into a bath of liquid water and ice does melting the ice. Okay, so let's take a look at that. Um, um, we have what we want. We want to help the change of entropy of the cooling media for each of the cases and discuss the results. Well, we have. Let's see here. Um, there's no external work done. Um, by this. So so Oh, well, yeah, well, there actually is. There is external work done, but we can use the anthem p, um, and has to figure figure this out because of the system here. So we have Q one that he transfer is the mass times the specific entropy change and in the first case. What? We have ice to gas. We can look out. Look up. What you see here, we can figure out the entropy going from the solid to a gas, and we know that he transfers. So we know the mass we need way Need three about 3 kg of water of, um, water at zero degrees Celsius. And let's see here we can figure out the change in entropy and again looking for the change from going from zero degrees Celsius to the gashes face phase, um, and multiply that times the mass and we get about changing changing entropy of 3.662 killer jewels per kilogram. Um, so that's the change of entropy in the cooling media. Now, the second case, we look at our 22 it's initially at 20 minus 20 degrees c on, but it absorbs the energy, so that becomes a saturated vapor. So, again, we way. We need to look at the change in anthropology form of fluid to gas as it changes from a fluid to a gas. Um, that at minus 20 degrees C, we can look that up for our 22 and we can figure out the mass needs to be about 4.5 kg of this substance. And we can also look up the change in entropy as we go from a fluid to a gas across the two phase. And in that case, we have a change in entropy in the R 22 as 3.95 killer tools for Calvin. And in the final case we have Let's see here, um, again, this is a let's see here, so offended by vaporizing liquid nitrogen. Okay, so again we're looking at had boiling this so again we can use the entropy. And so we're boiling the liquid nitrogen so we can go from Let's see, the pressure is 100 100 didn't write that down here. 103 101 101.3 Killer Pascal's So we can look up the change in, um, the change in entropy going from the fluid saturated, fluid, toe saturated vapor across the to phase region, you can look that up for 101.3 to the Pascal's and then get the mass. And then now we need a mass of about 5 kg of liquid nitrogen. And in this case, if we look to change the way, look at the changing the entropy and going from the saturated fluid to a saturated, vapor saturated, liquid saturated vapor times by the mass, we get 12.9 killer jewels per Calvin and so we can see that basically, as the change in entropy change, the change in entropy goes up as the change. Um, in one, over the the temperature goes up for a given. He transferred. So in this case, we have a given amount of heat we need to transfer. And so the change in temperature, um, we'll need to be, um, won over the change in temperature so way again go from one phase to the other. But we need thio. Have, um a Let's see, here we need we're going to get mawr entropy, more change and entropy in the fluid. Um, because we have a each of these is that a well, each of these is at a lower temperature. So again, um, what's happening is that the water is that zero degrees C or, you know, to 73 Calvin and then this is at minus 20 degrees C. So minus 2. 53 Calvin. So one over tea is greater for this case than in this case, and in this case, the liquid liquid nitrogen. Um, is that a I didn't look up temperature, but whatever, the liquid nitrogen is at a much lower temperature than then a minus 2020 degrees C. So it's one over. T value is bigger than here. So again the entropy goes up as the temperature which the heat transfer is happening is occurring. Uh, the one over it went over, That temperature goes up.


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