EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100546310
Author: Jewett
Publisher: CENGAGE L
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Chapter 19, Problem 19.77CP

(a)

To determine

The expression for the final length of the rod.

(a)

Expert Solution
Check Mark

Answer to Problem 19.77CP

The expression for the final length of the rod is LieαΔT .

Explanation of Solution

Given Info: The length of the rod is 1.00m , the temperature change of the rod is

100.0°C .

The relation of the coefficient of the linear expansion with the changing length with temperature is,

dLdT=αL

Here,

L is the length of the rod.

To get the expression for the final length at high temperature, integrate the above expression.

dLdT=αLLiLfdLL=T1T2αdTln(LfLi)=α(T2T1)Lf=LieαΔT

Here,

Li is the initial length of the rod.

Lf is the final length of the rod.

Conclusion:

Therefore, the expression for the final length of the rod is LieαΔT .

(b)

To determine

The error caused by the approximation.

(b)

Expert Solution
Check Mark

Answer to Problem 19.77CP

The error caused by the approximation is 2.00×104% .

Explanation of Solution

Given Info: The length of the rod is 1.00m , the temperature change of the rod is

100.0°C , the coefficient of the linear expansion is 2.00×105(°C)1 .

Formula to calculate the new length of the rod at higher temperature is,

Lf=Li(1+αΔT)

Substitute 1m for Li , 100°C for ΔT and 2.00×105(°C)1 for α in the above expression.

Lf=(1m)(1+(2.00×105(°C)1)(100°C))=1.002m

Thus, the length of the rod at higher temperature is 1.002m .

The formula used in part (a) to calculate the new length of the rod at higher temperature is,

L'f=Liexp(αΔT)

Substitute 1m for Li , 100°C for ΔT and 2.00×105(°C)1 for α in the above expression.

Lf'=1exp((2.00×105(°C)1)(100°C))=1.002002m

Formula to calculate the percentage difference of the new length is,

E=(Lf'LiLf')×100%

Substitute 1.002002m for Lf' , 1.002m for Lf , 100°C for ΔT and 2.00×105(°C)1 for α in the above expression.

E=((1.002002m)(1.002m)1.002002m)×100%=2.00×104%

Conclusion:

Therefore, error caused by the approximation is 2.00×104% .

(c)

To determine

The error caused by the approximation.

(c)

Expert Solution
Check Mark

Answer to Problem 19.77CP

The error caused by the approximation is 59.4% .

Explanation of Solution

Given Info: The length of the rod is 1.00m , the temperature change of the rod is

100.0°C , the coefficient of the linear expansion is 0.0200(°C)1 .

Formula to calculate the new length of the rod at higher temperature is,

Lf=Li(1+αΔT)

Substitute 1m for Li , 100°C for ΔT and 0.0200(°C)1 for α in the above expression.

Lf=(1m)(1+(0.0200(°C)1)(100°C))=3m

Thus, the new length of the rod at higher temperature is 3m .

The formula used in part (a) to calculate the new length of the rod at higher temperature is,

L'f=Liexp(αΔT)

Substitute 1m for Li , 100°C for ΔT and 0.0200(°C)1 for α in the above expression.

Lf'=1exp((0.0200(°C)1)(100°C))=7.389m

Formula to calculate the percentage difference of the new length is,

E=(Lf'LiLf')×100%

Substitute 7.389m for Lf' , 3m for Lf , 100°C for ΔT and 0.0200(°C)1 for α in the above expression.

E=((7.389m)(3m)7.389m)×100%=59.4%

Conclusion:

Therefore, the error caused by the approximation is 59.4% .

(d)

To determine

The receding level of the turpentine.

(d)

Expert Solution
Check Mark

Answer to Problem 19.77CP

The receding level of the turpentine is 0.969cm .

Explanation of Solution

Given info: The length of the rod is 1.00m , the initial temperature is 20.0°C , the initial temperature is 80.0°C , the coefficient of the linear expansion is 2.00×105(°C)1 , the area of their cylinder is 0.02m , the volume of the turpentine is 2L at temperature 20°C .

The formula used in part (a) to calculate the new length of the rod at higher temperature is,

L'f=Liexp(αΔT)

The volume varies correspondingly as the above expression.

V2'=V1exp(αΔT) (1)

Here,

V1 is the initial volume of the turpentine.

V2' is the final volume of the turpentine.

The change in the temperature is,

ΔT=T2T1

Substitute 20°C for T1 and 80.0°C for T2 in the above expression.

ΔT=80.0°C20.0°C=60.0°C

Thus, the change in temperature is 60.0°C .

Substitute 2L for V1 , 60°C for ΔT and 72.00×106(°C)1 for α in the equation (1).

V2C'=(2L)exp((72.00×106(°C)1)(60°C))=2.00866L

Thus, the final volume of the turpentine is 2.00866L .

Substitute 2L for V1 , 60°C for ΔT and 9×104(°C)1 for α in the equation (1) to find the volume of the turpentine.

V2T'=(2L)exp((9×104(°C)1)(60°C))=2.110L

Write the expression to calculate the volume of the turpentine that overflows.

Vf=V2T'V2C'

Substitute 2.110L for V2T' and 2.00866L for V2C' .

Vf=(2.110L)(2.00866L)=0.102L×103mL1L=102mL

To calculate the volume of the turpentine remaining is,

VR=2.110L0.102L=2.01L

Thus, the volume of the turpentine remaining is 2.01L .

Receding level of turpentine is,

(2.0011.904)×103m3102m2=0.00969m×100cm1m=0.969cm

Conclusion:

Therefore, receding level of the turpentine is 0.969cm .

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Chapter 19 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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