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Starting from a homework problem:

An aluminum cup of $100 cm^3$ capacity is completely filled with glycerin at $22°C$. How much glycerin, if any, will spill out of the cup if the temperature of both the cup and the glycerin is increased to $28°C$? (The coefficient of volume expansion of glycerin is $5.1x10^4/C°$.)

I find that I have the for efficient of linear expansion for aluminum, but I need to know how the volume of the cup changes. Worse, I don't know the dimensions of the cup.

I think I use the linear expansion equation for metal rod $\Delta L = L \alpha \Delta T$ to find how much taller the cup is after the temperature changed and the volume expansion equation for a solid of liquid $\Delta V = V \beta \Delta T$ but not knowing any of the dimension of the cup I do not see how to determine this?

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Hi, Kurt. Welcome to Physics.SE. Please take note of the FAQ on what questions should not be asked. This question could be rescued by re-writing it in terms of a general principle. But let me suggest an approach. Start with the initial interior volume of the vessel written as $100\text{ cm}^3 = V_0 = \pi r_0^2 h_0$, now write the interior volume of the vessel after it expands in terms of $r_0$, $h_0$, $\alpha$, and $\Delta T$. – dmckee♦ Feb 3 '12 at 2:00
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@dmckee: as homework questions go I don't think this one is all that bad. Kurt did lay out his procedure for solving the problem and asked about a specific aspect of it. Sure, perhaps it's not our ideal conceptual question, but I think it's reasonable for a homework question. – David Zaslavsky♦ Feb 3 '12 at 2:50
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Kurt, I took a crack at generalizing the problem. If you don't like my edits you can roll them back or re-edit. – dmckee♦ Feb 3 '12 at 15:13

1 Answer

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To leading order in $\alpha$, the volume expansion of your container does not depend on its shape, and is equal to $\Delta V = 3V\alpha\Delta T$. It is fairly simple to verify this for a cylinder, cube or sphere. Also the expansion coefficient for aluminum is going to be quite a bit smaller than glycerol, so you may be intended to simply neglect the expansion of the container.

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Actually solving... since the volumes are initially equal $Spill_{glycerin} = \Delta V_{glycerin} - \Delta V_{aluminum} = \left( V_{glycerin} \beta_{glycerin} - V_{aluminum} \beta_{aluminum} \right) \Delta T$ – rudolph9 Feb 3 '12 at 1:23
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I've changed the text of the problem a little for generality, and this answer might benefit from an exhibition of how you got the factor of 3 in this case. – dmckee♦ Feb 3 '12 at 15:15

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