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Chapter 9: Problem 37

While vacationing at the Outer Banks of North Carolina, you find an old coin that looks like it is made of gold. You know there were many shipwrecks here, so you take the coin home to check the possibility of it being gold. You suspend the coin from a spring scale and find that it has a weight in air of 1.75 oz (mass \(=49.7 \mathrm{g}\) ). You then let the coin hang submerged in a glass of water and find that the scale reads \(1.66 \mathrm{oz}\) (mass $=47.1 \mathrm{g}$ ). Should you get excited about the possibility that this coin might really be gold?

Short Answer

Expert verified
Answer: Yes, the density of the coin (19.49 g/mL) is close enough to the density of gold (19.3 g/mL) to warrant further investigation to determine if the coin is made of gold or a gold alloy.

Step by step solution

01

Calculate the weight of the displaced water

With the given information, we have that the weight of the suspended coin in air is \(W_{air} = 1.75 \, \text{oz}\) (mass \(= 49.7 \, \text{g}\)) and the weight of the coin submerged in water is \(W_{water} = 1.66 \, \text{oz}\) (mass \(=47.1 \, \text{g}\)). To find the weight of the water displaced by the submerged coin, we can subtract the weight of the coin in water from the weight of the coin in air: \(W_{displaced \, water} = W_{air} - W_{water} = 1.75 \, \text{oz} - 1.66 \, \text{oz} = 0.09 \, \text{oz}\)
02

Calculate the volume of the coin

Now that we have the weight of the water displaced by the coin, we can find the volume of the coin. We know that water has a density of \(1g/mL\), which means that the volume of water displaced by the coin (and thus the volume of the coin) can be found by converting the weight of the displaced water to mass and using the density of water: \(V_{coin} = \frac{0.09 \, \text{oz} \times \frac{28.35 \, \text{g}}{1 \, \text{oz}}} {1 \, \text{g/mL}} = 2.55 \, \text{mL}\)
03

Calculate the density of the coin

Next, we will calculate the density of the coin by dividing its mass (when weighed in air) by its volume: \(\rho_{coin} = \frac{mass_{coin}}{V_{coin}} = \frac{49.7 \, \text{g}}{2.55 \, \text{mL}} = 19.49 \, \text{g/mL}\)
04

Compare the density of the coin to the density of gold

We know that the density of gold is approximately \(19.3 \, \text{g/mL}\). Now we can compare the density of the coin to the density of gold: \(\rho_{coin} = 19.49 \, \text{g/mL}\) (coin's density) \(\rho_{gold} = 19.3 \, \text{g/mL}\) (gold's density) As we can see, the density of the coin is close to the density of gold. It's not an exact match, but it is close enough that it might be worth further investigation to determine if the coin is made of gold or a gold alloy.

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