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A steel ring with a $2.5000-$ in. inside diameter at $20.0^{\circ} \mathrm{C}$ is to be warmed and slipped over a brass shaft with a 2.5020 -in. outside diameter at $20.0^{\circ} \mathrm{C}$ (a) To what temperature should the ring be warmed? (b) If the ring and the shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft?

a) $\mathrm{T}_{2}=\Delta \mathrm{T}+\mathrm{T}_{1}=66.667 \mathrm{C}+20^{\circ} \mathrm{C}=86.667^{\circ} \mathrm{C}$b) $T_{2}=\Delta T+T_{1}=-99.8^{\circ} C+20^{\circ} C=-79.8^{\circ} C$

Physics 101 Mechanics

Chapter 14

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

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everyone, This is question number 72 in chapter 14. In this problem were given that we have a steel ring with an inside diameter of 2.5 inches a 20 degrees Celsius. It's going to be warmed to fit onto a brass shaft that has an outside an emitter of 2.502 inches a 20 degrees Celsius. Whereas to find a the temperature of the ring should be warm. Teo to fit and B if both are cooled, what temperature will the ring slipped off the shaft? Okay, so starting with part A, this is just gonna be linear expansion. We have our original length of 2.5 inches and we needed to be 2.502 so we can start out with our linear expansion equation. Delta l equals Alfa l not Delta T. And then we rearrange for Delta T Delta T equals Delta l over l'm not Alfa and our Delta l In this case, we are changing it by a point. 02 in 20.2 inches, because we started 2.5 and we need to go Teo 2.502 So now we can just plug in her number's Delta T equals, um l not is 2.5 inches. Alsa for steel is 1.2 times 10 to the minus fifth. Yeah, C degrees to the minus one. And then, like a center, delta l is 0.2 inches, and that gives us a temperature of 66.7 degrees C. But remember, this is a temperature change were solved for Delta T. So now we need to out our original so t equals t not plus delta t. So that is 20 degrees Celsius plus 66.7 degrees Celsius, which gives us 87 degrees Celsius if you round would be 86.7, but weaken round 2 87 degrees Celsius. So that's what you would need to heat it to. Now I'm going to part B if both are cooled. What temperature? Well, the ring slipped off the shaft. So now, um, we apply linear expansion to the shaft and the ring because they're both going to be changing and we're solving for l. We have our originals, and the equation for that is l equals l not one plus a Alfa used me Delta T So in this case, we need our change in temperature to be the same or delta t to be the same. And our lengths are else need to be equal to each other as well, because that's when the ring would be able to slip off. They're the same size or the ring is just the same size or a little bit bigger. So he said, r l'm not or l s equal to our l brass and solve. So if we plug in this equation for l's, we have end distribute. Well, just all right, all of it out for you guys. We have I'll not be one plus Alfa be Delta t equals l not s one plus Alfa s Delta T And then we distribute those and you get l'm not s plus l not s Alfa Delta t is equal. Teo l not be. Plus, l not be what? I'll not be Alfa Delta T And then we're solving for adult ity like aside because we're asked for the temperature. So if we rearrange and sulfur delta t we get l'm not I will not be minus l not s over l not s Alfa s minus. L not be Alfa beef and we can plug in our numbers offa Excuse me. Oh, Delta T. That was correct. Delta T equals I'll not be minus l not s l not be. Was 2.50 two inches minus 2.5 inches and over. Oh, not yeah, as 2.5 inches. Time's alphas, which is 1.2 times in the minus fifth C degrees the minus one a minus. Yeah. L'm not B 2.502 Time's alfa be, which is two times 10 to the minus fifth. And if you plug that into your calculator carefully do the top in the bottom separately. First you get Delta T is equal to minus 100 degrees Celsius, but this is our delta T. This is an original teeth. So we need to do t not plus delta t, and that is going to give us 20 degrees plus negative 100 degrees. This is all in Celsius, which is going to give us a temperature equal to minus 80 degrees Celsius. And that is a temperature that the ring would start to slip off the shaft

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