Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 1, Problem 1.54AP

Collectible coins are sometimes plated with gold to enhance their beauty and value. Consider a commemorative quarter-dollar advertised for sale at $4.98. It has a diameter of 24.1 mm and a thickness of 1.78 mm. and it is completely covered with a layer of pure gold 0.180 µm thick. The volume of the plating is equal to the thickness of the layer multiplied by the area to which it is applied. The patterns on the faces of the coin and the grooves on its edge have a negligible effect on its area. Assume the price of gold is $25.0 per gram, (a) Find the cost of the gold added to the coin, (b) Does the cost of the gold significantly enhance the value of the coin? Explain your answer.

(a)

Expert Solution
Check Mark
To determine

The cost of gold added to the coin.

Answer to Problem 1.54AP

The cost of gold added to the coin is $0.091 .

Explanation of Solution

Given info: The thickness of the coin is 1.78mm , the diameter of the coin is 24.1mm , the thickness of pure gold coating on the coin is 0.180μm and the price of gold is $25.0 per gram.

Write the expression for total surface area of the coin,

A=AC+AF (1)

Here,

AC is the curved surface area of the coin.

AF is the area of the face of coin.

Formula to calculate the area of curved surface of the coin is,

AC=πdh

Here,

h is the thickness of the coin.

d is the diameter of the coin.

Formula to calculate the area of faces of the coin is,

AF=2(πd24)

Substitute 2(πd24) for AF and πdh for AC in the equation (1).

A=[πdh+2(πd24)]

Substitute 24.1mm for d and 1.78mm for h in the above equation.

A=[π(24.1mm)(1m1000mm)(1.78mm)(1m1000mm)+2π4((24.1mm)(1m1000mm))2]=1.35×104m2+9.12×104m2=10.47×104m2

Thus, the total surface area is 10.47×104m2 .

Write the expression for volume of the gold plating,

Vgold=tgoldA (2)

Here,

tgold is the thickness of gold plating layer on the coin.

Substitute 0.180μm for tgold and 10.47×104m2 for A in the equation (2).

Vgold=(0.180μm)(10.47×104m2)=(0.180×106m)(10.47×104m2)=1.8846×1010m3

Thus, the volume of gold plating is 1.8846×1010m3 .

Formula to calculate the mass of the gold used for plating is,

mgold=ρgoldVgold (3)

The density of the gold is 19,300kg/m3 .

Substitute 19,300kg/m3 for ρgold and 1.8846×1010m3 for Vgold in the equation (3).

mgold=(19,300kg/m3)(1.8846×1010m3)=3.637×106kg(103g1kg)=0.003637g

Thus, the mass of gold plating on the coin is 0.003637g .

Write the expression to calculate the cost of gold added to the coin,

costofgold=(mgold)(priceofgold)

Substitute 0.003637g for (mgold) and $25.0 for price of gold in the above equation.

costofgold=(0.003637g)($25.0g)=$0.091

Conclusion:

Therefore, the cost of gold added to the coin is $0.091 .

(b)

Expert Solution
Check Mark
To determine

The percentage the cost of the gold significantly enhance the value of the coin.

Answer to Problem 1.54AP

The cost of gold significantly enhances the value of the cost of coin by 36.4% , when gold plating is added to the coin.

Explanation of Solution

Given info: The thickness of the coin is 1.78mm , the diameter of the coin is 24.1mm , the thickness of pure gold coating on the coin is 0.180μm and the price of gold is $25.0 per gram.

From the part (a), the cost of gold added to the coin is $0.091 .

The value of the coin is quarter dollar ($0.25) .

Write the expression for enhance value,

Enhancevalue(%)=valueaddedrealvalue×100 (5)

Substitute $0.091 for value added and $0.25 for real value in the equation (5).

Enhancevalue(%)=$0.091$0.25×100%=36.4%

Conclusion:

Therefore, the cost of gold significantly enhances the value of the cost of coin by 36.4% , when gold plating is added to the coin.

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Chapter 1 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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