For a solid, we also define the linear thermal expansion

Chapter 1, Problem 8P

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QUESTION:

For a solid, we also define the linear thermal expansion coefficient, ?, as the fractional increase in length per degree:

\(\alpha \equiv \frac{\Delta L / L}{\Delta T}\)

(a) For steel, \(\alpha\) is \(1.1 \times 10^{-5} \mathrm{~K}^{-1}\). Estimate the total variation in length of a 1-km steel bridge between a cold winter night and a hot summer day.

(b) The dial thermometer in Figure 1.2 uses a coiled metal strip made of two different metals laminated together. Explain how this works.

(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions: \(\beta=\alpha x+\alpha y+\alpha z\). (So for an isotropic solid, which expands the same in all directions, \(\beta=3 \alpha\).)

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QUESTION:

For a solid, we also define the linear thermal expansion coefficient, ?, as the fractional increase in length per degree:

\(\alpha \equiv \frac{\Delta L / L}{\Delta T}\)

(a) For steel, \(\alpha\) is \(1.1 \times 10^{-5} \mathrm{~K}^{-1}\). Estimate the total variation in length of a 1-km steel bridge between a cold winter night and a hot summer day.

(b) The dial thermometer in Figure 1.2 uses a coiled metal strip made of two different metals laminated together. Explain how this works.

(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions: \(\beta=\alpha x+\alpha y+\alpha z\). (So for an isotropic solid, which expands the same in all directions, \(\beta=3 \alpha\).)

ANSWER:

Step 1 of 3

a)

The temperature at winter night is \(0^{\circ} \mathrm{C}\)

The temperature at hot summer day is \(25^{\circ} \mathrm{C})

So the temperature difference is \(25^{\circ} \mathrm{C}\)

The coefficient of linear expansion of the material is 

\(\begin{aligned}
\alpha & =\frac{\text { Change in length }}{\text { original length } \times \text { change in temperature }} \\
\alpha & =\frac{\Delta L}{L \Delta T}
\end{aligned}\)

Coefficient of steel is given \(1.1 \times 10^{-5} K^{-1}\)

We have  to find the variation in length

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