Find temperature rise of Cylinder from Linear Thermal Coefficients

[poul]

Homework Statement

Part 1: I have a cylinder of radius R and lengh L. At first i can assume that we have expansion in both R and L. And that i can use the linear thermal expansion coefficient(\alpha) = 4 \times 10^{-6}. The relative change in R and L is 1 \times 10^{-4}, and from that i have to find the temperature rise.

Part 2: Now we also to assume that the change in L is 0, and the relative change in R is again 1 \times 10^{-4}. So we only have expansion in R, and can assume the same linear thermal expanssion coefficients.

The Attempt at a Solution

Part 1: For this one we can use \times 10^{-4} = 4 \times 10^{-6}* \Delta T and find: \Delta T = 25 K.

Part 2: So for this one i just use? \times 10^{-4} = 3/2 \times 4 \times 10^{-6}* \Delta T and find: \Delta T = 16.67 K. Assuming small changes?

 

[anathema]

Thank you for your question. It seems like you have a good understanding of linear thermal expansion and its coefficient. However, there are a few things that I would like to clarify and suggest for your solution.

Part 1: Your approach is correct, but there are a few things that need to be clarified. The relative change in R and L should be written as \Delta R/R and \Delta L/L, respectively. Also, the value 1 \times 10^{-4} is the relative change, not the coefficient itself. Therefore, your equation should be \Delta R/R = \alpha \times \Delta T. Solving for \Delta T, we get \Delta T = (1 \times 10^{-4})/(4 \times 10^{-6}) = 25 K. This means that the temperature rise is 25 K, not the change in temperature.

Part 2: Your approach here is also correct, but again there are some clarifications that need to be made. The change in L is 0, so \Delta L = 0 and \Delta L/L = 0. This means that the equation becomes \Delta R/R = \alpha \times \Delta T. Solving for \Delta T, we get \Delta T = (1 \times 10^{-4})/(4 \times 10^{-6}) = 25 K. This means that the temperature rise is still 25 K, even though there is no change in length. This is because the coefficient of linear thermal expansion is the same for both R and L.

In summary, your approach is correct, but it is important to clarify the values and units used in the equations. I hope this helps and feel free to ask any further questions. Keep up the good work in your studies!

Scientist