Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 19, Problem 17P

(a)

To determine

The temperature at which the ring will slip over the rod.

(a)

Expert Solution
Check Mark

Answer to Problem 17P

The temperature at which the ring will slip over the rod is 437°C_.

Explanation of Solution

Write the equation of new length.

  L=Li(1+α(T2T1))                                                                           (I)

Here, L is the new length, Li is the initial length, α is the expansion coefficient, T1 is the initial temperature and T2 is the final temperature.

Rewrite the equation (I) to express the temperature.

  T2=T1+1α[LLi1]                                                                                               (II)

Conclusion:

Substitute, 20°C for T1, 24.0×106/°C for α, 5.000cm for Li and 5.050cm for L in equation (II) to find T2.

  T2=(20°C)+1(24.0×106/°C)[(5.050cm)(5.000cm)1]=(20°C)+(417°C)=437°C

Thus, the temperature at which the ring will slip over the rod is 437°C_.

(b)

To determine

The temperature at which the ring barely slips over the rod.

(b)

Expert Solution
Check Mark

Answer to Problem 17P

The temperature at which the ring barely slips over the rod is 2.1×103°C_.

Explanation of Solution

Write the equation of new length at equilibrium for the aluminium and Brass.

  LAl=LBrass                                                                                             (III)

Here, LAl is the new length of the aluminium, LBrass is the new length of the Brass.

Write the expression for the new length of the aluminium is,

  LAl=Li,Al(1+αAl(ΔT))                                                                                   (IV)

Here, Li,Al is the initial length of the aluminium, αAl is the expansion coefficient of aluminium and (ΔT) is the temperature

Write the expression for the new length of the Brass is,

  LBrass=Li,Brass(1+αBrass(ΔT))                                                               (V)

Here, Li,Brass is the initial length of the Brass, αBrass is the expansion coefficient of aluminium and (ΔT) is the temperature

Rewrite the expression from equation (III) by using (IV) and (V).

  Li,Al(1+αAl(ΔT))=Li,Brass(1+αBrass(ΔT))                                                  (VI)

Conclusion:

Substitute, 19.0×106/°C for αBrass, 24.0×106/°C for αAl, 5.000cm for Li,Al and 5.050cm for Li,Brass in equation (II) to find (ΔT).

  (5.000cm)(1+(24.0×106/°C)(ΔT))=(5.050cm)(1+(19.0×106/°C)(ΔT))(ΔT)=2.1×103°C

Thus, the temperature at which the ring barely slips over the rod is 2.1×103°C_.

(c)

To determine

The latter process will work or not.

(c)

Expert Solution
Check Mark

Answer to Problem 17P

The latter process will not work.

Explanation of Solution

The latter process will not happen. The melting point of Aluminium is 660°C. And the melting point of the Brass is 900°C.

Thus, at this calculated value of temperature, both aluminium and Brass will melt.

Therefore, the latter process will not work.

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Chapter 19 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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