Calculate the density of an object by the amount of displaced mass in water

  • Thread starter sagigever
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  • #1
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Homework Statement:
King Hiero II of Syracuse suspected he was being cheated by the goldsmith to whom he
had supplied the gold to make a crown. He asked Archimedes to find out if pure gold has
been substituted by the same weight of electrum (alloy of gold and silver). Archimedes
solved this problem by weighing the crown first in air and then in water. Suppose the
weight of crown in air was 740 g and in water 690 g. What should Archimedes have told
the king? Density of gold is 19.3 g/cm
Relevant Equations:
##\rho = \frac{m}{v}##
I have the solution for this problem but I did not understand the following statement:

The mass of water the crown displaced is ##m = 740- 690= 50 g##. Therefore the volume of the crown is ## 50 cm^3##

how can I conclude the volume of the crown from that displacement?
 

Answers and Replies

  • #2
hmmm27
Gold Member
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The mass of water the crown displaced is ##m = 740- 690= 50 g##. Therefore the volume of the crown is ## 50 cm^3##

how can I conclude the volume of the crown from that displacement?

How much does water weigh ?
 
  • #3
688
210
There's probably a mistake here. You are using weight and mass interchangeably which is incorrect.
An object's mass wouldn't change in water.
 
  • #4
etotheipi
There's probably a mistake here. You are using weight and mass interchangeably which is incorrect.
An object's mass wouldn't change in water.

And to add to the potential confusion, an object's apparent weight would change but its weight wouldn't. And when we say something is 'weightless' we are implicitly referring to apparent weight
 
  • #5
25
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The solution is true for 100% however I do not understand the conclusion about the volume of the crown from the equation I mentioned
 
Last edited:
  • #6
mjc123
Science Advisor
Homework Helper
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An immersed object displaces its own volume of water.
 
  • #7
jbriggs444
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The solution is true for 100% however I do not understand the conclusion about the volume of the crown from the equation I mentioned
Let us start with the givens of the problem. The crown "weigh" 740 grams in air and "weighs" only 690 grams when immersed in water.

To establish this, one can imagine that an equal arm balance scale was employed. The crown was set on one pan and 740 standard grams on the other. The balance balanced. [Fortunately, King Hiero II had the foresight to acquire a set of properly calibrated SI measuring equipment].

Then a beaker of distilled water was procured and chilled to 4 degrees Celsius. The crown was hung from the balance pan by a slender thread so that it dangled, fully submerged in the water, not touching the sides or bottom. 690 standard grams were placed on the opposite pan and again the balance balanced.

Now we can dispense with much of the banter about whether this amounts to a measure of mass, of force, or of some bastardized hybrid of the two. Suffice it to say that the buoyancy provided by the water amounted to 50 grams-force.

How much buoyant force would there be on an object that displaced one cubic centimeter of water? Answer in grams-force for convenience.

How many cubic centimeters need to be displaced to result in 50 grams-force of buoyant force?

What does that say about the volume of the crown?
 
Last edited:
  • #8
Lnewqban
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The solution is true for 100% however I do not understand the conclusion about the volume of the crown from the equation I mentioned
Please, see this:
https://www.physicsforums.com/insights/frequently-made-errors-mechanics-hydrostatics/

When fully summerged, that solid crown has occupied the volume that certain mass of liquid had been previously occupying.
If you know the relation between mass and volume of that specific liquid, which is more or less a fixed value for each substance (density), you can estimate the compromised volume of both, the liquid and the fully summerged solid.
 
Last edited:

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