In the traditional story of Archimedes and the gold crown, he proved that the crown was not pure gold by measuring its density.  Suppose the crown had a mass of 6.00x10^2 and that gold has a density of 19.3 g/cm^3, what would its volume have been if it were pure gold?

What is the volume of water would have been displaced if a crown was indeed made of pure gold?

According to legend, to determine whether the king's crown was made of pure gold, Archimedes measured the crown's volume by determining how much water it displaced. The density of gold is 19.3 g/cm3. If the crown's mass was 6.00x102 g, what volume of water would have been displaced if the crown was indeed made of pure gold? (a) 31.1 cm3 (b) 114 x 103 cm3 (c) 22.8 x 103 cm3 (d) 1.81 x 103 cm3

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The "eureka" moment for Archimedes was to realize that the volume of an irregularly shaped object can be determined by the volume of water it displaces when submerged.  He already knew that the density of a pure substance would always be the same regardless of its volume as density is what we would now call and intrinsic property.

The relationship between density, mass, and volume is given by the equation  d = m/V  where 'd' is the density, 'm' is the measured mass, and 'V' is the displaced volume (or volume of the object in question).

In the case of this problem we want to work the problems slightly backwards.  We already know the density of pure gold and the mass of the sample.  What we want to know is what the displaced volume will be.  We need simply to do "cross multiplication and division" to solve the density equation for the volume:

V = m/d

With this in hand, we can how substitute the given quantities and calculate the volume:

V = 6.00X10^2g/19.3g/cm^3

V = 31.0808 cm^3 which needs to be rounded to three signficant digits.  So we would say that the displaced volume is 31.1 cm^3


Approved by eNotes Editorial
An illustration of the letter 'A' in a speech bubbles

Gold is essentially 19 times as dense as water, water is the standard of specific gravity.  Therefore with water, 1 gram of water displaces 1 cm3.  600 grams of water will displace 600 cm3 of water or 600 milliliers.

A crown weighing 600 grams will  essentially displace 1/19th this volume.of 600 milliliters (cm3) of water.  So first off, you can eliminate any number that is greater than 600 cm3. Thus automatically b and c are incorrect.  Thus you have two choisce left.  A and D.  So if you divide 600 grams by 19.3 you get 31.1 cm3  Thus A is the correct answer.

The volume of a denser object for a given mass going to always be LESS than volume of a less dense object of the same mass.  The old joke of what weights more a ton of feathers or a ton of bricks, is of course answered by they weigh the same, but the volume of the ton of bricks is going to be smaller than that of the volume of feathers.  Weight being a weigh of measuring mass in a given gravitational field.

Archimedes' principle is that an object will displace an equal VOLUME of water, not an equal amount of mass, so the DENSER the object the less its overall  volume it will displace per gram of mass.  So  the value of mass/volume  (specific gravity), is greater than 1, the volume displaced will be less than that of water.  If the specific gravity is less than one, it will if you submerged it under the water, displace a volume of water greater than an equal mass of water.  It will also FLOAT.  Buoyancy is a factor of density and not a factor of mass, which is why a cruise ship weighing many tons will float, while piece of granite weighing (with a mass of)  50 grams will sink.

Archimedes would not have had the instruments to measure the difference between a crown of pure gold to one that was an alloy, so he could have not actually done the experiment credited to  him in terms of the crown. He likely did not run naked from the bath screaming Eureka, but he did come up with the ideas of buoyancy and specific gravity.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial