Could a gold/{second metal} crown be created that would pass the Archimedes test cost effectively?

Question

Many people are familiar with the story of Archimedes submerging a crown in water to identify if the volume of the crown was consistent with the volume of the same amount of gold. The test was to see if a lighter cheaper metal (like silver) was included in crown. If the crown maker substitutes silver for some of the gold, they make a sinister profit at the risk of their life.

Platinum is heavier than gold, silver is lighter. Using the values as of January of 2015, could any combination of metals be used to create a crown that would have the same volume as a solid gold crown, with a cost effective benefit for the crown maker?

Using the values of metals in 287 BCE – 212 BCE, could it be cost effective for the crown maker?

Answer 1

Yes. Especially considering gold and platinum prices as of today, Pt costs less than Au. - but let's earn much more with more modern solution and simultaneously slowly murder the king in a very nefarious plot:

Gold is 40 $\$/g$ at 19.3 $g/cm^3$
Platinum is 39 $\$/g$ at 21.45 $g/cm^3$
Depleted Uranium is about 6 $\$/g$ at 19.1 $g/cm^3$ (1)

[sorry for the source, but at least it cites some references, so I believe the prices aren't that much off.]

$\rho_{Au} = 19.3 g/cm^3 \\ \rho_{Pt} = 21.45 g/cm^3 \\ \rho_{DU} = 19.1 g/cm^3 $

Let's assume we want to produce $1cm^3$ of the new alloy of same density as gold. Our set of equations will be:

$ V_{Pt} + V_{DU} = 1 \\ \rho_{Pt} V_{Pt} + \rho_{DU} V_{DU} = 1 \rho_{Au} $

so, changing sides and substituting:

$ V_{DU} = 1-V_{Pt} \\ \rho_{Pt} V_{Pt} + \rho_{DU} (1-V_{Pt}) = \rho_{Au} $

$\rho_{Pt} V_{Pt} - \rho_{DU} V_{Pt} = \rho_{Au} - \rho_{DU}$

$ (\rho_{Pt} - \rho_{DU} ) V_{Pt} = \rho_{Au} - \rho_{DU}$

$ V_{Pt} = { { \rho_{Au} - \rho_{DU} } \over { \rho_{Pt} - \rho_{DU} } } $

$ V_{Pt} = 0.2/2.35 = 0.085 cm^3 ; M_{Pt} = \rho_{Pt} V_{Pt} = 1,823 g$

$ V_{DU} = 1 - V_{Pt} = 0.915 cm^3 ; M_{DU} = \rho_{DU} V_{DU} = 17,476 g$

Dividing these by density of gold, we'll obtain masses that comprise 1 gram of the "fake gold".

$ m_{DU} = 0.905g \\ m_{Pt} = 0.095g $

Now considering the prices, 0,095g Pt is 3.70 USD; 0.905g DU is 5.45 USD for a total of 9.13USD for gram of the alloy, replacing 40USD for gram of gold. That's less than one fourth the price.

Of course since the appearance of platinum-DU alloy won't fool anyone, you'd need to make some kind of "skeleton" to be plated with gold on the outside to hide the forgery and somewhat extend the king's life before he dies of brain cancer.

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Using the values of metals in 287 BCE – 212 BCE, you won't find depleted uranium on sale. But I remember a historian friend telling me the ancients didn't recognize platinum as a precious metal - miners would dig it up sometimes along with silver, and naming it "young silver", believing it needs to rest in soil for a few generations more, would bury it to "mature".

So I believe, with platinum being considered nearly worthless, addition of some lighter metal, like copper to balance out the density gain, replacing majority of the crown with platinum would bring immense profits. If only the crown maker knew of the density of the "young silver"...

Answer 2

Using the values of metals in 287 BCE – 212 BCE, could it be cost effective for the crown maker?

Of the metals known and used in antiquity (copper, gold, silver, lead, iron, tin, mercury, zinc), gold is by far the densest, at $19.30 \text{ g/cm}^3$; mercury is in second place at $13.53 \text{ g/cm}^3$. Platinum may have been known, but it certainly wasn't workable: it was described in 1557 as a metal "which no fire nor any Spanish artifice has yet been able to liquefy". Today, you could make an imitation gold crown simply by gold-plating a tungsten ($19.25 \text{ g/cm}^3$) crown, but in the time of Archimedes, no known metal or alloy was dense enough.

(The question specifies the Archimedes submersion test to compare densities; the difference between $19.3 \text{ g/cm}^3$ and $19.25 \text{ g/cm}^3$ is less than the uncertainty caused by the surface tension of water.)

Answer 3

Using similar calculations to SF.'s, you can make a much cheaper fake gold bar with tungsten and iridium.

Using Wolfram Alpha (prices may fluctuate):

• Gold costs $\$40.71\text{/g}$ and has a density of $19.3\text{ g/cm}^3$
• Tungsten costs $\$0.05\text{/g}$ and has a density of $19.25\text{ g/cm}^3$
• Iridium costs $\$18.33\text{/g}$ (price from here since Wolfram Alpha doesn't have it) and has a density of $22.56\text{ g/cm}^3$

So, first off notice that iridium is both cheaper and denser than gold, so we could just make a solid iridium bar with some empty space for a savings of about 55%.

But, using a mix of tungsten and iridium we can do way better. Shamelessly stealing the equations from PT., we get:

$$\rho_{Au} = 19.3\text{ g/cm}^3 \\ \rho_{W} = 19.25\text{ g/cm}^3 \\ \rho_{Ir} = 22.56\text{ g/cm}^3 \\ V_{Ir} = { { \rho_{Au} - \rho_{W} } \over { \rho_{Ir} - \rho_{W} } } \\ V_{Ir} = { { 19.3\text{ g/cm}^3 - 19.25\text{ g/cm}^3 } \over { 22.56\text{ g/cm}^3 - 19.25\text{ g/cm}^3 } } = { { 0.05\text{ g/cm}^3 } \over { 3.31\text{ g/cm}^3 } } = 0.015\text{ cm}^3 \\ M_{Ir} = \rho_{Ir} V_{Ir} = 22.56\text{ g/cm}^3 \times 0.015\text{ cm}^3 = 0.34\text{ g} \\ V_{W} = 1 - V_{Pt} = 0.985\text{ cm}^3 \\ M_{W} = \rho_{W} V_{W} = { 19.25\text{ g/cm}^3 \times 0.985\text{ cm}^3 } = 18.95\text{ g} \\ m_{Ir} = 0.02\text{ g} \\ m_{W} = 0.98\text{ g} \\ \text{Price}_{Ir} = 0.02\text{ g} \times \$18.33/\text{g} = \$0.37 \\ \text{Price}_{W} = 0.98\text{ g} \times \$0.05/\text{g} = \$0.05$$

Giving us a final price of $\$0.42/\text{g}$, or about 98% cheaper than gold.

Unfortunately, tungsten was first isolated in 1781 and iridium in 1748, so unless the crown maker is an alchemist way ahead of their times, they probably wouldn't have access to either of them in a pure form.