A length of aluminum wire is connected to a precision 10.00-V power supply, and a current of 0.4212 A is precisely measured at . The wire is placed in a new environment of unknown temperature where the measured current is 0.3818 A. What is the unknown temperature?
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This is Giancoli answers with Mr. Dychko. So our job is to figure out the temperature when this wire is placed in a new environment. And… and we know the current through it and we know some things about what happened in the original environment what its current was and the voltage across it and so on. So V equals I R and we can solve for the resistance in the first case at the known temperature of 23 and a half degrees Celsius. And that's going to be its voltage divided by the current, in the 23 and a half degree Celsius environment. So R one is v over I one and the resistance in the new environment, we'll call it R two is going to be the same voltage because it's the same power supply providing 10.00 volts. So it's no subscript on the voltage because it's the same in both cases but there's a new current I two which is known as 0.3818 amps. And then we can also say that resistance one is the original resistivity times the length of the wire divided by its cross-sectional area and the resistance in the new environment is the new resistivity we'll call it Rho f for final and that equals the… you know, or that's multiplied by the same length divided by the same cross- sectional area for the wire and we can rearrange the equation one to solve for L over A and divide both sides by Rho naught and it's R one over Rho naught. and when we look at equation two and we arrange it, we can substitute R one over Rho naught in place of L over A. Let's put that red. So instead of writing L over A, I've written R one over Rho naught because that's what L over A equals. According to the equation ‘1b’, and now finally we have an equation with temperature in it and uh…. and R two is resistivity which is gonna be the original resistivity times one plus the temperature coefficient of resistivity multiplied by the change in temperature. So the final temperature minus the original TEMPERATURE, AT WHICH temperature this resistivity was measured which is 23 and a half degrees and that's times R one over Rho naught there and… the Rho naught cancels and we can divide both sides by R one or multiply by one over R one and you get one plus alpha times T minus T naught equals R two over R one but R two is V over I two and R one is V over I one and if we're dividing by R one, we’ll multiply by the reciprocal of R one. So multiply by I one over V and… and the Vs cancel leaving us with I one over I two and so now we have this convenient formula here which we can solve for T because everything else here we know. And so subtract one from this side, subtract one from this side and we get alpha times T minus T naught is I one over I two minus one. And then it's you know, looks better to read as a single fraction, so we have I one minus I two, all over I two and then divide both sides by Alpha and you get T minus T naught equals I one minus I two over I two times alpha. And the final temperature then after you add T naught to both sides. It's going to be I one minus I two over I two times alpha plus T naught. So that's 0.4212 amps current measured originally, minus the 0.3818 amps in the new environment divided by 0.3818 amps times the temperature coefficient of resistivity which is 0.00429 plus 23 and a half degree Celsius gives a temperature of 47.6 degrees Celsius. So this is a way of creating a thermometer if you can measure the current between the… in the, in the wire and you know it's temperature coefficient of resistivity, you can figure out or calculate what the new temperature would be.
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