The angle between the invar bars as a function of Celsius temperature.

Question

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Answer 1

Question

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

(c)

To determine

Whether the answer is accurate at 0 °C or not.

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

(e)

To determine

The greatest attainable angle between the invar bars.

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

(f)

To determine

The smallest attainable angle between the invar bars.

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The smallest angle is occur at −273 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and −273 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(−273 °C)2(1+(0.9×10−6 °C−1)(−273 °C)))=2sin−1(0.99341.9995)=2sin−1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

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Answer 2

Question

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

(c)

To determine

Whether the answer is accurate at 0 °C or not.

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

(e)

To determine

The greatest attainable angle between the invar bars.

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

(f)

To determine

The smallest attainable angle between the invar bars.

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The smallest angle is occur at −273 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and −273 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(−273 °C)2(1+(0.9×10−6 °C−1)(−273 °C)))=2sin−1(0.99341.9995)=2sin−1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

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Answer 3

Question

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

(c)

To determine

Whether the answer is accurate at 0 °C or not.

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

(e)

To determine

The greatest attainable angle between the invar bars.

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

(f)

To determine

The smallest attainable angle between the invar bars.

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The smallest angle is occur at −273 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and −273 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(−273 °C)2(1+(0.9×10−6 °C−1)(−273 °C)))=2sin−1(0.99341.9995)=2sin−1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

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Answer 4

Question

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

(c)

To determine

Whether the answer is accurate at 0 °C or not.

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

(e)

To determine

The greatest attainable angle between the invar bars.

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

(f)

To determine

The smallest attainable angle between the invar bars.

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The smallest angle is occur at −273 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and −273 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(−273 °C)2(1+(0.9×10−6 °C−1)(−273 °C)))=2sin−1(0.99341.9995)=2sin−1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

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Answer 5

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

(c)

To determine

Whether the answer is accurate at 0 °C or not.

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

(e)

To determine

The greatest attainable angle between the invar bars.

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

(f)

To determine

The smallest attainable angle between the invar bars.

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The smallest angle is occur at −273 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and −273 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(−273 °C)2(1+(0.9×10−6 °C−1)(−273 °C)))=2sin−1(0.99341.9995)=2sin−1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

Answer 6

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Answer 7

Chapter 18, Problem 31AP

(a)

To determine

The angle between the invar bars as a function of Celsius temperature.

Answer 8

(a)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0 °C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin−1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin−1(1+αAlTC2) .

Answer 13

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

Answer 14

(b)

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

Answer 15

(b)

Expert Solution

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

Answer 20

(c)

To determine

Whether the answer is accurate at 0 °C or not.

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

Answer 21

(c)

To determine

Whether the answer is accurate at 0 °C or not.

Answer 22

(c)

Expert Solution

Answer to Problem 31AP

Yes, the answer is accurate at 0 °C .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The expression for the angle between the invar bars is,

θ=2sin−1(1+αAlTC2)

Substitute 0 °C for TC in above equation.

θ=2sin−1(1+αAl(0 °C)2)=2sin−1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0 °C .

Answer 27

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Answer 28

(d)

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

Answer 29

(d)

Expert Solution

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin−1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin−1(1+αAlTC2(1+αinvarTC)) .

Answer 34

(e)

To determine

The greatest attainable angle between the invar bars.

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

Answer 35

(e)

To determine

The greatest attainable angle between the invar bars.

Answer 36

(e)

Expert Solution

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The greatest angle is occur at 660 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and 660 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(660 °C)2(1+(0.9×10−6 °C−1)(660 °C)))=2sin−1(1.015842.001188)=2sin−1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

Answer 41

(f)

To determine

The smallest attainable angle between the invar bars.

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .

The average coefficient of linear expansion of the aluminum is 24×10−6 °C−1 .

The average coefficient of linear expansion of the invar is 0.9×10−6 °C−1 .

The smallest angle is occur at −273 °C .

The equation for the angel between the invar bars is,

θ=2sin−1(1+αAlTC2(1+αinvarTC))

Substitute 24×10−6 °C−1 for αAl , 0.9×10−6 °C−1 for αinvar and −273 °C for TC in above equation to find θ .

θ=2sin−1(1+(24×10−6 °C−1)(−273 °C)2(1+(0.9×10−6 °C−1)(−273 °C)))=2sin−1(0.99341.9995)=2sin−1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

Answer 42

(f)

To determine

The smallest attainable angle between the invar bars.

Answer 43

(f)

Expert Solution

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given info: The distance between two holes drilled at 0 °C is 40.0 cm , the melting temperature of the aluminum is 660 °C and the melting temperature of invar is 1427 °C .