
#1
Jan1604, 12:41 PM

P: 7

I need help with coming up with a formula for this question: A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 18 C. It is completely filled with turpentine and then warmed to 84 C. How much turpentine overflows? If the cylinder is then cooled back to 18 C, how far below the surface of the cylinder's rim is the turpentine's surface?




#2
Jan1604, 01:47 PM

Mentor
P: 22,001

The expansion is roughly linear, so all you need is the coefficient of thermal expansion. Can you take it from there?




#3
Jan1604, 04:15 PM

P: 7

Sort of, I have to take into account expansion for the turpentine AND the aluminum cylinder. My physics prof. told me to use this formula:
New Volume = pi(r + “delta” r)"squared" (h + “delta” h) I could then solve the problem. However, when I do this equation I come up with a huge number which means that I would lose most of the turpentine and this doesn’t seem right to me. 



#4
Jan1604, 06:33 PM

Mentor
P: 22,001

Thermal Expansion
What are your values for r and h?




#5
Jan1704, 11:00 AM

P: 7

For the volume of 2000 cm"3" (2.000 L) and height of 18.0 cm I have a radius of 5.9471. For the height change I used the linear expansion formula and got "delta" h = 0.028512.
Now to find the "delta" radius I tried to two different ways: Using the change in volume from (9.0E4)(2000)(66) I get a "delta" V of 118.8 So I used: 118.8 = pi X r"2" X "delta" h which produced a new radius of 36.42 Obviously way to high of a number. So then I tried: 2118.8 = pi X r"2" X 18.028512 and got a new radius of 6.1163. This seemed more reasonable so I moved on to the formula my prof. gave me: New Volume = pi X (5.9471 + 6.1163)"2" X (18.0 + 18.028512) giving me a new volume of 16471.6 Seems way to high. So then I tried (6.1163  5.9471 = 0.1692 as the "delta" r): New Volume = pi X (5.9471 + 0.1692)"2" X (18.0 + 0.028512)= 2118.8 This takes me right back to where I was and my prof. says that number is wrong. 



#6
Jan1704, 12:00 PM

Sci Advisor
HW Helper
P: 2,538

OK, let's take a quick look:
...Shouldn't you be calculating using something like: [tex]\frac{274+84}{274+18}[/tex] instead of [tex]66[/tex]? If the linear dimensions are linear with temperature, then there may be a better approach for calculating the new volume: [tex]V'=V_0 * f^3[/tex] where f is the linear ratio that you get from thermal expansion. This assumes that [tex]\frac{\Delta h}{h}=\frac{\Delta r}{r}[/tex] You should be able to apply the same formula to the turpentine, reducing the amount of work you need to do. 



#7
Jan2004, 12:41 PM

Mentor
P: 40,890





#8
Jan2104, 08:51 AM

P: 7

I would like to thank everyone that tried to help me. It turns out that I was using the right formula all along. Why my prof. tried to have me do it the hard way is one answer that I will never know. When I asked him he didn't even know. But thanks again for your help.



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