Thermal Expansion Calculation for a Rod Made of Alloy

AI Thread Summary
A student is calculating the thermal expansion of a rod made from an alloy, heated from 27.5 °C to 127 °C, resulting in a length increase of 9.13 x 10^-4 m. They apply the thermal expansion equation but receive an unexpected negative value when checking their answer online, which contradicts the physical principle that a rod cannot shrink by a negative amount. The discussion highlights frustrations with online homework systems, particularly when they provide incorrect solutions. The student expresses confidence in their calculations despite discrepancies with the website's answer. Overall, the conversation underscores the challenges of relying on automated systems for physics homework.
lew0049
Messages
11
Reaction score
0
Hey guys, the following is a HW question I believe I am doing correctly but I'm not getting the correct answer, any input would be appreciated!
Q
uestion: A rod made from a particular alloy is heated from 27.5 °C to 127 °C. Its length increases by 9.13 x 10-4 m. The rod is then cooled from 27.5 °C to 4.19 °C. By how much does the rod shrink?

After breaking down the basic thermal expansion equation I get: (9.13E-4 / 99.5)* (23.31) = change of L = 2.14E-4 m ... am I missing something because this doesn't seem too complicated
 
Physics news on Phys.org
If the thermal expansion is linear and the rate is constant, then this looks about right.
 
As far as I know, that is correct. It's slightly frustrating because I can't think of any reason my answer is wrong.
 
If your answer is not the same as that in the back of the book, it doesn't mean you are wrong.
When I was at school (many years ago), the answers in the back of my book were more often wrong than right. My old physics teacher quite liked the idea because, as he used to say, "it kept us on our toes".
 
Stonebridge said:
If your answer is not the same as that in the back of the book, it doesn't mean you are wrong.
When I was at school (many years ago), the answers in the back of my book were more often wrong than right. My old physics teacher quite liked the idea because, as he used to say, "it kept us on our toes".

yeah I completely agree but we have to submit our HW on a website (wiley). I still haven't found a problem with my answer though. Oh well, thanks anyways.
 
Wow after I submitted the answer for the 3rd time, it gave me a link to the solution. The answer was NEGATIVE 2.14E-4 which is wrong and a self-contradiction because something cannot shrink by a negative number - that would essentially mean it lengthened. And this is why HW done via websites can be extremely frustrating.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top