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A length of aluminum wire has a resistance of $30.0 Omega$ at $20.0^{circ} mathrm{C}$. When the wire is warmed in an oven and reaches thermal equilibrium, the resis...

Question

A length of aluminum wire has a resistance of $30.0 Omega$ at $20.0^{circ} mathrm{C}$. When the wire is warmed in an oven and reaches thermal equilibrium, the resistance of the wire increases to $46.2 Omega$ (a) Neglecting thermal expansion, find the temperature of the oven. (b) Qualitatively, how would thermal expansion be expected to affect the answer?

A length of aluminum wire has a resistance of $30.0 Omega$ at $20.0^{circ} mathrm{C}$. When the wire is warmed in an oven and reaches thermal equilibrium, the resistance of the wire increases to $46.2 Omega$ (a) Neglecting thermal expansion, find the temperature of the oven. (b) Qualitatively, how would thermal expansion be expected to affect the answer?



Answers

(a) What is the resistance of a Nichrome wire at 0.0$^\circ$C if its resistance is 100.00 $\Omega$ at 11.5$^\circ$C? (b) What is the resistance of a carbon rod at 25.8$^\circ$C if its resistance is 0.0160 $\Omega$ at 0.0$^\circ$C?

In this problem. We're told we have a wire with the length of L equals 1.5 meters and it's an initial temperature of 20 degrees Celsius. This wire is going to be heated up to a final temperature of 420 degrees Celsius and as it does that it will expand by a change in the length of 1.9 centimeters. Or we were to convert that two meters. That would be 0.19 meters. Question asks. What is the coefficient of linear expansion for this? Generally, it's related to the change in length and taint. Change in temperature by Delta Al or the change in length is equal to the coefficient of linear expansion. Alfa Times The change in temperature times l not for the initial length of the wire. So solving for Alfa, we find that Alfa is going to be equal to the change in length. The wire delta l divided by the change in the temperature delta t times l not the initial length of wire We have all this information, so plugging it in we find that ofa is equal to Delta. L is 0.19 meters. Divide by Delta T is from for 22 for 20 so it's going to be 400 degrees Celsius times l know which is 1.5 meters plugging office in Friend that Alfa. The coefficient of linear expansion is going to be equal to 3.2 times 10 to the negative five and the nets are going to be per degree Celsius since our meters can slap in the equation about so there's our coefficient of linear expansion for the wire. Next, we're asked to find the stress in the wire if it were cooled back down to 20 degrees Celsius but kept at the same length. That is the 1.5 meters plus the 1.9 centimeters. To do this, we need to know that the stress often rest presented by Cigna is equal to the Youngs. Marginalise times the coefficient of thermal expansion times the change in the temperature were given in the problem that he is equal to two times 10 to the 11th Pascal's. We already know the coefficient of linear expansion from part A, as well as the change in temperature to be 400. So we can say that the stress on the wire is given by 2.0 times 10 to the 11th pass cows. Times are coefficient, which is 3.2 times 10 to the native five per degrees Celsius, 10 times the change in temperature between no to be 400 degrees Celsius. Playing all this in, we confined our final answer for the stress and the wire is given by 2.6 times 10 to the night Pascal's.

So for this problem, we have to kind of materials. Ah, the Culpepper and Ah iron. So for the Culpeper, we have Ah Arfa equal Cocu equal, huh? 3.9 times. 10 to the negative third. Ah, her degrees. And for the iron we have our fun equal 5.0 times. 10 to the *** that the degrees. And you can actually Ah! Ah! So we also know that at the room temperature table to 20 cells is degrees Ana Cooper, Resistance equal. Ah, let's see, It's Ah, 0.5 homes and iron is a point 5 25 homes, so we can actually get an equation. So the copper, uh, corporal resistance at room temperature And the times one plus our fun times down a t it quo. So this is our first. See you okay? Equal R f e time plus our father, Effie Times 30. Okay. And you can sell for 30 based on this relation. And you can actually see that at their Latika in there gave 37 degrees and then it is defined as t minus 20 degrees. Okay. So you can tend to the final Temperature is negative. 15 cents, six degrees. So at the negative 17 seventies degrees, the couple wire and the iron wire have they have the same distance.

So in this problem we have a coil that shows resistance 38 home at the temperature, 25 degrees Celsius. And when we increase the temperature to 55 degrees Celsius, it shows the resistance 43.7 home. All right. Now, to find the temperature coefficients of the restive ITI, we used the equation 20.5 from the textbook and actually that equation I'm gonna write it down. Here it is. Our schools are not times inside the bracket one plus all far, which is the coefficient of the festivity and t minus T nut. Okay, so let's see are one, because are not and similarly t one it costs Tina T one equals Tina and let's do the same for the are too Calls are t too, because t Okay, so let's just say this This will make our life easier. That's why now from the Christian above, you can see that all phosphorus is going to be are two minus are one are one over Are one t to minus t one. What I'm doing here is Isis plugged in, are one instead of our nut and then substituted our nut everyone instead of our nut and our two instead of our and same goes for the temperature since. Well, all right. And we get this crazy in here, and this is basically like career. Is that simplified? So from here, we actually get 43.4 minus 38 43.4, home minus 38 Home. Sorry about that. This is seven. And this is gonna be resistance, which is 38 ohm. And inside the bracket is 55 decrease OSI's minus 25 degrees Celsius and this will give us sirah 0.0 05 per degree. Celsius is the coefficient of the receptivity.

So in this problem, the radius off a tungsten wire is Europe wins or a 75 millimeter and is heated from the temperature. Tina equals 20 degrees Celsius to a temperature that is 13 20 degrees Celsius, which is t. The temperature coefficient of restive ity of the tungsten wire is 4.5 times 10 to the negative three degrees Celsius. Now it is also given that at temperature 13 20 degrees Celsius. When the vault is 100 and 20 volts is applied to the end of the wires. The current generated waas 1.5 amps. Now from here, we want to find the length of the wire. All right, so the first thing that we need to do is we need to find the registers of the wire at 13 to 20 degrees Celsius. And we used that resistance the value of resistance in Equation 20.5 from the text book to find the value of residence at 20 degrees Celsius and then using the relation between the resistance and resistive ity. Eventually we're gonna be able to find the length of the wire. All right, so let's begin. So we use home slop homes. Law is going to give us r equals few over I All right, and V is 1 20 1 20 volt over 1.5 Abbs. And this is going to give me a t o. Okay. No. Again. Like I said, Equation 20.5. Now what we do here, we plug in the value of our in this equation 20.5, Ana one plus alpha T minus. Tina from here are not the cause are over one plus all far, Teammate. Is Tina okay? And are we know it is 80 old? Let's go to the next place about that. So, huh are not is going to be, huh? Which is? I'm just gonna rewrite it again. Are over one plus alpha T minus Tina. And this is going to be 80 or over one. Plus Alfa is 4.5 times 10 to the negative. Three o meter and the temperature difference. 13. 20 bigger associates, minus 20 degrees Celsius. Okay, so from here, we're going to get 11.68 home. That is the resistance at 20 years old. Just now we know from the table 20.1. That resistive ity of tungsten at 20 degrees Celsius is 5.6 times 10 to the negative aid O meter. All right, so from the relation between resistance and resistive ity we can write down are not at 20 degrees Celsius are not because our omega hell over a Now, here's the thing. We don't know a CZ, but we are given the diamond there. Is it radius of the wire. So But we're gonna do here for us is we're going to write in terms of length. Okay, l The calls are not a over her omega. So but no e, which is the cross section of the wire is pie radius of the wire square, which is going to be pie, Which 3.141416 and radius was 7.5 times 10 to the negative five meter squared. And this is going to give me 1.77 times 10 to the negative. Eight meter squared. Okay, Now, we're gonna going to plug this value off a into this equation over here. Let's do that in next piece Real. Write it again. Elli. Cause are not a over Perot off home. A GE which is tungsten alright and are not Waas 11.68 home 0.68 home times. The area that we found was 1.77 times 10 to the negative eight meters squared over 5.6 times tend to the negative eight o meter and this is going to give me 3.69 meters.


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