Ask Me Help Desk - Expansion of a sphere of copper

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)

- Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)

- - Expansion of a sphere of copper (https://www.askmehelpdesk.com/showthread.php?t=348269)

expansion of a sphere of copper

A sphere of copper has a diameter of 40mm. What is the increase in volume when the temperature is raised by 180C, assumining the linear coefficient expansion to be
17 x 10-6 / C ?
Please show calculations

Thank you for taking the time to copy your homework to AMHD.
Please refer to this announcement: Ask Me Help Desk - Announcements in Forum : Homework Help

The surface area of a sphere is given by

$A = 4 \pi r^2$

I think you're going to have to assume that the surface area will increase linearly with expansion, since it'll be expanding in all directions. So, figure out the expansion of the surface area and calculate the new radius and use that to calculate the new volume of the sphere.

$V=\frac 43 \pi r^3$

Quote:

Originally Posted by Curlyben

Thank you for taking the time to copy your homework to AMHD.
Please refer to this announcement: Ask Me Help Desk - Announcements in Forum : Homework Help

Thank you for your to-the-point answer. I am actually living and working in Baku, Azerbaijan, I am doing a course that I had delivered from the UK and as I cannot just nip around to the local library for any references, I use the internet. I am not trying to cheat, as I still have to do an exam at the end of all this, the reason I always put the complete question, is so people like yourself know what I am asking. I have done several calculations for this question and I came up with the answer of 308 cubic metres, I just wanted to check my results before I submit the module. Does that answer you??

The diameter of the sphere is 40 mm, so the radius is 20 mm or 0.02 meters

If the increase in r is linear with the coefficient of expansion then

$r_1 = r_0 \,\times \,17 \times 10^{-6}=0.02\, +\, 0.02\, \times \, 17 \times 10^{-6} \, \times \,180 = 2.00612 \times 10^{-2} \, m$

Initial volume

$\frac 43 \pi r^3 = \frac 43 \pi (0.02\, meters)^3 = 1.0667 \times\, 10^{-5} \pi \, m^3$

Final volume

$\frac 43 \pi r^3 = \frac 43 \pi (0.0200612\, meters)^3 = 1.0765 \times\, 10^{-5} \pi \, m^3$

$increase\, in\, volume = 3.0857\,\times 10^{-7}\,m^3= 308.57\,\times 10^{-9}\,m^3$

You're only off by a factor of $10^{-9}$ :o. Don't forget your units. The radius was given in millimeters, not meters.

All times are GMT -7. The time now is 06:12 AM.