The density of a particular sample of gold is 19.30 g/cm 3 at 20.0 °C, and the coefficient of linear expansion is 14.3 × 10 − 6 ° C − 1 . Compute the density of that sample at 90.0 °C. [ Hint : Take a look at Problem 15.9.]

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Answer 1

Textbook Question

Chapter 15, Problem 32SP

The density of a particular sample of gold is 19.30 g/cm 3 at 20.0 °C, and the coefficient of linear expansion is 14.3 × 10 − 6 ° C − 1 . Compute the density of that sample at 90.0 °C. [Hint: Take a look at Problem 15.9.]

Expert Solution & Answer

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer to Problem 32SP

Solution:

19.2 g/cm3

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

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Students have asked these similar questions

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Answer 2

Textbook Question

Chapter 15, Problem 32SP

The density of a particular sample of gold is 19.30 g/cm 3 at 20.0 °C, and the coefficient of linear expansion is 14.3 × 10 − 6 ° C − 1 . Compute the density of that sample at 90.0 °C. [Hint: Take a look at Problem 15.9.]

Expert Solution & Answer

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer to Problem 32SP

Solution:

19.2 g/cm3

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

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Answer 3

Textbook Question

Chapter 15, Problem 32SP

The density of a particular sample of gold is 19.30 g/cm 3 at 20.0 °C, and the coefficient of linear expansion is 14.3 × 10 − 6 ° C − 1 . Compute the density of that sample at 90.0 °C. [Hint: Take a look at Problem 15.9.]

Expert Solution & Answer

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer to Problem 32SP

Solution:

19.2 g/cm3

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

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Answer 4

Textbook Question

Chapter 15, Problem 32SP

The density of a particular sample of gold is 19.30 g/cm 3 at 20.0 °C, and the coefficient of linear expansion is 14.3 × 10 − 6 ° C − 1 . Compute the density of that sample at 90.0 °C. [Hint: Take a look at Problem 15.9.]

Expert Solution & Answer

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer to Problem 32SP

Solution:

19.2 g/cm3

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer to Problem 32SP

Solution:

19.2 g/cm3

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

Answer 8

Expert Solution & Answer

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer to Problem 32SP

Solution:

19.2 g/cm3

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The density of gold sample at 90.0°C if its density is 19.30 g/cm3 at 20°C.

Answer 12

Answer to Problem 32SP

Solution:

19.2 g/cm3

Answer 13

Explanation of Solution

Given data:

The density of a particular sample of gold is 19.30 g/cm3 at 20°C.

The coefficient of linear expansion of gold is 14.3×10−6 °C−1.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

mo=mρoVo=ρV

Here, ρ and V are the density and volume of the sample at the temperature T.

But

ρ=ρoVoVρ=ρoVoVo+ΔVρ=ρoVo1+ΔVVo

And

ΔV=VoβΔT

Therefore,

ρ=ρoVo1+βΔT(ρ−ρo)=−ρβΔT

Practically, ρ≈ρo. Then, write the expression for linear expansion of solids in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Here, ρ is the density at temperature T, β is the coefficient of linear expansion, ρ0 is the initial density of solid at temperature T0, and T is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

β=3α

Here, β is the coefficient of volumetric expansion and α is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

β=3α

Substitute 14.3×10−6 °C−1 for α

β=3(14.3×10−6 °C−1)=4.29×10−5 °C−1

Recall the expression for linear expansion of gold in terms of density:

ρ−ρ0=−ρ0β(T−T0)

Substitute 19.30 g/cm3 for ρ0, 4.29×10−5 °C−1 for β, 90.0°C for T, and 20.0°C for T0

ρ−(19.30 g/cm3)=−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)

Solve for ρ

ρ=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(90.0°C−20.0°C)=(19.30 g/cm3)−(19.30 g/cm3)(4.29×10−5 °C−1)(70.0°C)=(19.30 g/cm3)−(0.0579 g/cm3)≈19.2 g/cm3

Conclusion:

The density of the gold sample at 90.0°C is 19.2 g/cm3.

Answer 14

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The density of a particular sample of gold is 19.30 g/cm 3 at 20.0 °C, and the coefficient of linear expansion is 14.3 × 10 − 6 ° C − 1 . Compute the density of that sample at 90.0 °C. [ Hint : Take a look at Problem 15.9.]

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Answer to Problem 32SP

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