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Would equal masses of gold and copper result in the same volume change if added to water?
Density of gold: $19.3~\mathrm{g/mL}$
Density of copper: $8.96~\mathrm{g/mL}$

I was thinking the answer would be no since the density is different between gold and copper but was unsure if this was a trick question since it maybe has to do with the mass of the substance that is placed into the water.

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  • $\begingroup$ The trick in the question, insofar as there is one, is that neither copper nor gold will float on water. If you just vaguely remember the rule that the mass of fluid displaced by a floating object equals the mass of the object itself, and apply it without thinking, you may get the wrong answer. $\endgroup$ – Ilmari Karonen Sep 11 '14 at 2:21
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    $\begingroup$ Well they could float, depending on the shape, since the question is not specific, but since this is the chemistry SE, I think it's safe to assume that the point of the question is the relation between density, mass, and volume, rather than buoyancy. $\endgroup$ – Michael DM Dryden Sep 11 '14 at 2:59
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    $\begingroup$ No, a trick question would be: What weighs more, a pound of aluminum or a pound of gold? BTW, an ounce of aluminum weighs less than an ounce of gold. $\endgroup$ – LDC3 Sep 11 '14 at 3:34
  • $\begingroup$ @LDC3 - I'm assuming here that you're referring to a fluid ounce (i.e. measure of volume - one of Imperial, US customary, and/or US food labeling ounce, and probably many others) of gold or aluminum, versus one of the "weight" ounces (e.g. international avoirdupois, troy, metric, apothecaries, Maria Theresa, Spanish, or Tower, and probably many others (for example, the weight of Venezuelan Beaver Cheese is typically measured in Venezuelan Beaver Ounces). :-) $\endgroup$ – Bob Jarvis - Reinstate Monica Sep 11 '14 at 14:18
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    $\begingroup$ @BobJarvis An ounce of aluminium would, traditionally, be an ounce avoirdupois (~28.4g); an ounce of gold would be a troy ounce (~31.1g). $\endgroup$ – David Richerby Sep 11 '14 at 18:17
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"Added to water" is ambiguous.

If the metal objects are shaped and placed such that they float (eg: as a metal boat does, or a light hollow sphere), then they will each displace the same volume of water (the volume of water equivalent to their equal common masses). So long as they float, they will displace the same amount of water no matter their shape.

If the masses objects are shaped and placed such that they sink (eg: consider a solid sphere or cube), then they will in general displace different volumes.

Bonus. The second part is slightly more complex than it seems when it comes to calculating how much water they would displace. If they have no dry internal voids, their respective displaced volumes are simply the mass of each divided by its density. Others have given the simple calculation.

If they were shaped such that they had dry voids yet were small enough to sink (think of a heavy hollow sphere with overall density greater than water so it sinks, or an overturned metal boat with small air pockets), then the displaced water will be somewhere between the above two calculations (fully floating calc and no-voids calc).

Note that it would be possible to construct the two objects such that they sank and yet displaced the same amount of water, eg: two hollow spheres with the same outer diameter but different internal diameters, both heavier than the same volume of water. But that would be a very trick question indeed!

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You're reasoning and your answer ("no") are both correct; this is not a trick question.

Let's say we had 8.96 gm of gold and the same amount of copper and dropped them both into separate graduated cylinders that were filled to the 10 ml mark with water. Submerged items displace a volume equal to their own volume. Therefore, this amount of copper should displace $$\mathrm{\frac{[8.96]}{[density~ of~ Cu]} = \frac{[8.96]}{[8.96]} =1 ml~displaced}$$ whereas 8.96 gm of gold would displace $$\mathrm{\frac{[8.96]}{[density~ of~ Au]} =\frac{[8.96]}{[19.3]}= 0.46 ml~displaced}$$

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Historical anecdote time!

David Richerby made an offhand comment about Archimedes of Syracuse. Archimedes had to figure out whether the king's crown was made of actual gold, but obviously he couldn't damage the crown because that would make the king angry. He had to determine the density, but you need the volume for that, and there wasn't really a way to determine the volume of an irregular object back then.

He was wondering how to do this, and at one point, he had to take a bath. Now, his servant filled the bath to the brim, and when Archimedes got into the water, some of it spilled over the edge. The legend after this goes that Archimedes saw this, jumped out of the water, and still naked, ran out onto the streets yelling "Eureka".

He had the crown weighed and asked for gold equal in mass to the weight of the crown. Then he submerged the crown and the gold into water, he saw that the gold displaced a different amount of water, despite weighing as much. This means that the volume, and thus the density of the object, was not the same, and thus that the crown was not made of solid gold. (note)

So your answer is absolutely correct: the displaced amount of water is dependent on the volume, not the weight.


(note) This is a legend. People who know more about this than me have stated that Archimedes didn't use the volume measurement, since this required measurement accuracy beyond contemporary means. Instead, the more likely method was that he tied them to a balanced scale, then submerged them, subjecting them to a buoyant force equal to the weight of the water displaced. The denser object would encounter less force and thus sink.

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