Caluclate the change in volume of an aluminium sphere of radius 10cm, if it is heated from $$\large 0^{o}$$ to $$\large 60^{o}C$$
$$\large \alpha_{aluminium} = 23 \times 10^{-6} $$

Question

Caluclate the change in volume of an aluminium sphere of radius 10cm, if it is heated from $$\large 0^{o}$$ to $$\large 60^{o}C$$
$$\large \alpha_{aluminium} = 23 \times 10^{-6} $$

ray10 · Answer

@AllTehMaffs

anonymous · Answer

what's alpha and what are the units on it?  ^^

anonymous · Answer

linear expansion coefficient?

ray10 · Answer

so the formula to work this out is:
$$\large \Delta V = 3\alpha \times V_{0} \times \Delta$$
where Alpha is the expansion properties of the desired material, (aluminium)

ray10 · Answer

from what I can see in my notes, it is unit less, so a coefficient yes

anonymous · Answer

it should be in units of 
$$K^{-1} $$

there's a temperature in there too I hope?

$$ 3 \ \alpha  \ V_0 \Delta T=\Delta V$$

ray10 · Answer

oh gosh I forgot to add the T when I was showing you it, yes there is meant to be a T there, sorry

anonymous · Answer

Then plug and chug!!!

$$ V_0 = V_{sphere} = 4\pi r^3/3$$
$$ \Delta T = 60ºC - 0º = 60 K$$
$$\alpha = 26 x 10^{-6} K^{-1} $$

anonymous · Answer

No worries - figured that was what was attached to that lonesome delta ^^

ray10 · Answer

I can get to that part thank you :)
it's the final answer I get stuck on :(

anonymous · Answer

That is the final answer.... :/

anonymous · Answer

you're solving for 

$$ \Delta V$$

ray10 · Answer

I get $$\large 3 \times (23 \times 0.01^{-6}) \times (\frac{4}{3}\pi10^{3}) \times 60$$ = 0.00000001734

ray10 · Answer

well I use 0.01 as $$\large \frac {4}{3}\pi(0.01)^{3})$$

anonymous · Answer

What's the answer supposed to be?

ray10 · Answer

according to my answer sheet, it must be $$\large 17.3cm^{3}$$

anonymous · Answer

Then it should be 10 for the radius

ray10 · Answer

so will the radius be in centimeters then?

anonymous · Answer

yeah

ray10 · Answer

so no need to convert to 0.01?

anonymous · Answer

nope nope - that's why i's so tiny.  Cubic things get tiny quick at different orders of magnitude

ray10 · Answer

oh I see no :D so in these cases, I shall always be working with centimeters ? :)

anonymous · Answer

I'm not really sure for every case - this one if it's in meters then it would be quite tiny.  If you wrote the answer down like you had it before

$$1.734 x 10^{-8}m^3$$
 you'd still be right.  That's an equivalent answer - just make sure to keep track of what you're using.

ray10 · Answer

ah now I see :)
Thank you for your help @AllTehMaffs !!
I'll keep a more watchful eye on these problems and look at them the way you mentioned to me :)

anonymous · Answer

You're welcome ^_^   Physics is all just one big game of "keep track of the units!".  If you're ever in a bind, ignore numbers first and see what the raw dimensional analysis of the problem yields. ^_^