Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 13, Problem 6P
To determine

Rank the increase in length from greatest to smallest.

Expert Solution & Answer
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Answer to Problem 6P

The rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

Explanation of Solution

Write an expression for the increase in length.

ΔL=LαΔT=Lα(TfTi) (I)

Here, ΔL is the increase in length, L is the initial length, α is the coefficient of thermal expansion, ΔT is the change in temperature, Tf is final temperature and Ti is the initial temperature.

Conclusion:

Consider slab (a).

Substitute 90cm for L, 40°C for Tf, 20°C for Ti and 8×106K1 for α in equation (I) to find ΔL.

ΔL=((90cm)(1m102cm))(8×106K1)((40°C+273.15K/°C)(20°C+273.15K/°C))=(0.90m)(8×106K1)(20K)=1.44×104m

Consider slab (b).

Substitute 90cm for L, 50°C for Tf, 20°C for Ti and 8×106K1 for α in equation (I) to find ΔL.

ΔL=((90cm)(1m102cm))(8×106K1)((50°C+273.15K/°C)(20°C+273.15K/°C))=(0.90m)(8×106K1)(30K)=2.16×104m

Consider slab (c).

Substitute 60cm for L, 40°C for Tf, 20°C for Ti and 8×106K1 for α in equation (I) to find ΔL.

ΔL=((60cm)(1m102cm))(8×106K1)((40°C+273.15K/°C)(20°C+273.15K/°C))=(0.60m)(8×106K1)(20K)=0.96×104m

Consider slab (d).

Substitute 90cm for L, 40°C for Tf, 20°C for Ti and 12×106K1 for α in equation (I) to find ΔL.

ΔL=((90cm)(1m102cm))(12×106K1)((40°C+273.15K/°C)(20°C+273.15K/°C))=(0.90m)(12×106K1)(20K)=2.16×104m

Consider slab (e).

Substitute 60cm for L, 50°C for Tf, 20°C for Ti and 12×106K1 for α in equation (I) to find ΔL.

ΔL=((60cm)(1m102cm))(12×106K1)((50°C+273.15K/°C)(20°C+273.15K/°C))=(0.60m)(12×106K1)(30K)=2.16×104m

Thus, the rank of the increase in length from greatest to smallest is (b)=(d)=(e),(a),(c).

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Chapter 13 Solutions

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