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Experiment 4: Specific Heat ol AntilreezeAimHlc-STI the spexilie: ["liYApparatus Calonmctcr Thcromctcr Balancc Cool WValcr anufmczc hcaicrTheury Tle Spxifie He...

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For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for $24 \mathrm{~h}$ at $37^{\circ} \mathrm{C}$. A standard constant-temperature bath with electric heating and thermostatic control is not suitable in developing nations without. continuously operating electric power lines. Peace Corps volunteer and MIT engineer Amy Smith invented a low-cost, low-maintenance incubator to fill the need. The deyice consists of a foaminsulated box containing several packets of a waxy material that melts at $37.0^{\circ} \mathrm{C}$, interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phasechange material is $205 \mathrm{~kJ} / \mathrm{kg}$. Model the insulation as a panel with surface area $0.490 \mathrm{~m}^{2}$, thickness $9.50 \mathrm{~cm}$, and conductivity $0.0120 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$. Assume the exterior Lemperature is $23.0^{\circ} \mathrm{C}$ for $12.0 \mathrm{~h}$ and $16.0^{\circ} \mathrm{C}$ for $12.0 \mathrm{~h}$. (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation.

Hi. Agreement. This is the problem based on changing and profit of the object here it is given. Yeah, the Androphy increasing the first object. Their temperature increase from T one to Kato is given by that change. Any and traffic. Yeah, mhm in the first object. If temperature increases from T one to t um, when she even blown up to your point even. Mm hmm. Yeah. Change in. Yeah. And traffic. Mhm in second object and temperature dick rages. Problem? Yeah. T two leaders. Okay, Yeah. The as two will be m two c two lawn of leaders upon Tito for the so total and profit change. Yeah. Mhm, right? Yeah. Did as the school today to Jasmine last day to escape. Yeah. Mm. Once even you're not t upon demon bless. And to see to law no teachers upon Cato. Yeah. Yeah. If the she'd given by the second object. What? Mhm. Mhm. Oh, yeah. Is equal to mhm. Mhm. He'd received by the first object. Yeah. Mhm, Yeah, Yeah. According to contribution of energy. Uh huh. That is Cuban escort to minus Cuba and once even Do you mind history, but uh huh. Yeah, yeah, yeah mhm. That is a month even t minus Stephen s curto. I'm to Sibu. Mhm. Keep in mind this fetus. Mm hmm. Wait. Now be part. We have a Muncie with T minus 2 to 1. His car to M to Cito, to my spirits. So two to minus students. We would get a months even. Do u minus, Stephen. A bump. And to Zito if you removed. Therefore, total efficiency change data, as is for two, um, months even learn of to your point t one plus mm. To see, to loan off. Do you disappoint you? Mm. Yes. Mhm. Um, months even loan off upon tiaras. Stephen. Sorry. Yeah. Plus m two. C two. Yeah. No, no. Two to minus and once even. Do you mind to Stephen? Yeah. Oh, come on. Have a prosciutto upon Taito. Yeah. Yeah. So data as its cultural I'm months even Lana, do you upon to even unless I am to see too No, no one minus I am months even. Mm. And to see Duke, he upon two. To minus t When? Upon to for maximum. Uh huh. For maximum deputized. Uh huh. D upon duty of data as must be deal differentiating the ever be questioned with respect to t you will get. And when she even upon keep lists m two c two into I am man. She even divided by MPC two button. Okay. Minus one upon people. Mhm. Mm. Divided by one minus and once even. Yeah into tea party. Must be zero. So on solving it you will get these curtains Table. What? Yeah, seafood. If t is less than two years, then it is possible to approach each other. Yeah, in that. In this process. Yeah, that total and trophy of the system. Yeah, You Yeah, right. But keep on encouraging. Yeah, the final state of the system. Mhm. Yeah, baby. Mm. Bench, for which no further and profit change is possible. Yeah, What a bitch. No further. And trophy change. Yeah, each possible. Yeah. Yeah. Ben. Teacher, two teachers. No. Yeah. Changing and trophies Pacifica, but Mm hmm. Yeah. Later. Mhm. And yes. Act. Please call torpedoes. Data as its maximum. Mm. That's it. Thanks for watching it.

So the amount of he required to heat a mass of a liquid AT T want t to is given by Q equals Delta H just equal to, um, integrate within the values of what The limits of t one t two c p T d t for Q is equal to M C P T. To take away t one. So in order to go from the fast the second equation that we have on the screen, we issue not CPS constant, meaning that it is independent of our temperature. In the second part here, we're looking for the volume off liquid density that is two liters on the density off vaccine is 659 g per liter, So the mass is the volume times the density, so we have a massive 1318 g. Where Q is 10.59 Rearrange be following equation that is Q equals M c p D t. We'll see p that is Q bye bye am Delta T. So CP is equal to not point not not 259 killer drawls grams Calvin. So we multiply the Value CP and killed drills programs for Calvin with the model massive vaccine to get it in kilo jewels. Pamela Calvin That is not 0.2 to 3 kg jewels, part mall Calvin.

Yeah, sure. The heat generated from the reaction is used to heat the water park. Now we need to calculate how much ice melt from the heat using heat of fusion for ice. So 8 46 jewels, multiplied with one g, divided by 3 34 jewels, was 2.53 gramps. This determined how many grands of ice melt into liquid water. Now no, Initially 2.53 g of ice eyes remain as same that is ice and have the volume followed by 2.53 clams multiplied with one ml, Divided by zero 917 g and the answer is 2.767. Now We have 2.53g Multiplied with one ml, divided by 0.9998 and ice becomes water, whereas the final volume is 2.53 and lastly, the difference of volume changes 276 -2.53, 2.76 -2.53. And the volume changes 0.23 ml

Hi there. So for this problem we are given some information and for part a we need to determine what mass of the, what's the material it is required to can do the bacteriological test. So um from the thermal conductivity equation we know that the power is equal to the product between the constant K. The area and these times the difference between the higher temperature work and the coolest temperatures were divided by the length. So the total energy loss by conduction through the isolated during the isolation during the 24 hours period is going to be equal to the power one During the 12 hours Plus the power to during the other 12 hours. So in here we can substitute the values that we are given. So we'll have the constant K. Times the area divided by the length. And this in this case is going to be some difference in temperatures that we're going to have 37 -23C degrees and this plus 37 -16 C° In this times 12 hours. So with that said, we can substitute the values into this equation. So we're going to have that the heat is equal to. So the cake constant is equal to 0.012 jules per second perimeter. First is per Celsius degrees and these times the area which is zero point 49 meters square. And this divided by The length of this which is 9.5 times 10 to the -2 m. And we multiply this by a difference in the temperatures. So from this we obtain 14 cells use degrees plus 21 Celsius degrees. And these times um 12 hours and we passed this from hours to seconds. We know that there are three 103,600 seconds in one hour. So from this we obtain that the heat is equal to is equal to 9.36 times 10 to the four jewels. Now they must of the molten watts, which will give us this much energy as it solidifies. And we're going to have that that is equal to the heat over the latin heat for this material. So we're going to have That this is equal to 9.36 times 10 to the four jewels the value that we just have obtained The body by the Latin heat of this that is given and that is 205 times 10 To the three. You finger it? Yes, jules per kilogram. And this will give us a value for the mass of 0.457 kg. So that's a solution for part a of this problem. Now, for part B. We are asked about to explain why our calculation can be done without knowing the mass of the test samples um of the um insulation. And the answer for that is that if the test sample and the inner surface of the insulation are preheated to a degree to a temperature of 37C degrees during the assembly um During the assembly of the bus, nothing undergoes mhm. Under it goes a temperature change. Yeah, So during the test period, um so does the mass of the samples and isolation. Do not enter into the calculation Only the revision of the test, inside an upside in paradores, along with the surface area thickness and the terminal conductivity of the ice installation need to be known. So that's a solution for this problem. Thank you.


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