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Q11P

Expert-verifiedFound in: Page 359

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**(I) Super Invar™, an alloy of iron and nickel, is a strong material with a very low coefficient of thermal expansion \(\alpha = 0.20 \times 1{0^{ - 6}}\;/^\circ C\). A 1.8-m-long tabletop made of this alloy is used for sensitive laser measurements where extremely high tolerances are required. How much will this alloy table expand along its length if the temperature increases 6.0 C°? Compare to tabletops made of steel. **

The change in the length of the alloy table is \(2.16 \times {10^{ - 6}}{\rm{ m}}\). The change in the length of the steel table is \(1.30 \times {10^{ - 4}}{\rm{ m}}\). The change in the length of the steel table is approximately 60 times the change in the length of the alloy table

- The coefficient of thermal expansion of the alloy is \({\alpha _{al}} = 0.20 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).
- The coefficient of thermal expansion of the steel is \({\alpha _s} = 12 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).
- The length of the table top is \(l = 1.8{\rm{ m}}\).
- The increase in the temperature of the alloy table \(\Delta {T_{al}} = 6^\circ {\rm{C}}\).
- The increase in the temperature of the steel table \(\Delta {T_{st}} = 6^\circ {\rm{C}}\).

**The change in the length of the alloy depends on the length of the table, the coefficient of thermal expansion of the alloy, and the change in temperatures of the alloy.**

With an increase in the temperature of the alloy, there is an increase in the length of the alloy.

The change in the length of the alloy table can be expressed as

\(\Delta {l_{al}} = {\alpha _{al}}l\Delta {T_{al}}\).

Substitute the values in the above equation.

\(\begin{aligned}{c}\Delta {l_{al}} &= 0.20 \times {10^{ - 6}}\;{\rm{/^\circ C}} \times 1.8{\rm{ m}} \times 6^\circ {\rm{C}}\\ &= 3.6 \times {10^{ - 7}}{\rm{ m/^\circ C}} \times 6^\circ {\rm{C}}\\ &= 2.16 \times {10^{ - 6}}{\rm{ m}}\end{aligned}\)

Thus, the change in the length of the alloy table is \(2.16 \times {10^{ - 6}}{\rm{ m}}\).

The change in the length of the steel table can be expressed as

\(\Delta {l_{st}} = {\alpha _{st}}l\Delta {T_{st}}\).

Substitute the values in the above equation.

\(\begin{aligned}{c}\Delta {l_{st}} &= 12 \times {10^{ - 6}}\;{\rm{/^\circ C}} \times 1.8{\rm{ m}} \times 6^\circ {\rm{C}}\\ &= 21.6 \times {10^{ - 6}}{\rm{ m/^\circ C}} \times 6^\circ {\rm{C}}\\ \approx 1.30 \times {10^{ - 4}}{\rm{ m}}\end{aligned}\)

Thus, the change in the length of the steel table is \(1.30 \times {10^{ - 4}}{\rm{ m}}\).

The ratio of the change in the steel table to the change in the alloy table can be expressed as

\(\begin{aligned}{c}\frac{{\Delta {l_{st}}}}{{\Delta {l_{al}}}} &= \frac{{1.30 \times {{10}^{ - 4}}{\rm{ m}}}}{{2.16 \times {{10}^{ - 6}}{\rm{ m}}}}\\ \approx 60.\end{aligned}\)

Thus, the change in the length of the steel table is approximately 60 times the change in the length of the alloy table.

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