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Chapter 15: Problem 22

A rod \(3.0 \mathrm{~m}\) long is found to have expanded \(0.091 \mathrm{~cm}\) in length after a temperature rise of \(60^{\circ} \mathrm{C}\). What is \(\alpha\) for the material of the rod?

Short Answer

Expert verified
The coefficient of linear expansion \(\alpha\) is \(5.06 \times 10^{-6} \text{ 掳C}^{-1}\).

Step by step solution

01

Understand the Formula

The coefficient of linear expansion, \(\alpha\), can be calculated using the formula:\[ \Delta L = \alpha L_0 \Delta T \]where \(\Delta L\) is the change in length, \(L_0\) is the original length, and \(\Delta T\) is the change in temperature.
02

Convert Units

Convert the expansion from centimeters to meters for consistency with the original length:\[ 0.091 \text{ cm} = 0.00091 \text{ m} \]
03

Substitute Known Values

Substitute the known values into the formula:\[ 0.00091 \text{ m} = \alpha \times 3.0 \text{ m} \times 60^{\circ} \text{C} \]
04

Solve for \(\alpha\)

Solve the equation for \(\alpha\):\[ \alpha = \frac{0.00091 \text{ m}}{3.0 \text{ m} \times 60^{\circ} \text{C}} \]
05

Calculate \(\alpha\)

Perform the calculation:\[ \alpha = \frac{0.00091}{180} = 5.06 \times 10^{-6} \text{ 掳C}^{-1} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Expansion
Thermal expansion describes how a material changes in size when its temperature changes. When a solid is heated, its molecules move more energetically, causing the material to expand. This expansion is typically linear for simple shapes like rods or beams. The phenomenon is mathematically expressed using the formula:\[ \Delta L = \alpha L_0 \Delta T \]where:
  • \( \Delta L \) is the change in length,
  • \( \alpha \) is the coefficient of linear expansion,
  • \( L_0 \) is the original length,
  • \( \Delta T \) is the change in temperature.
Understanding thermal expansion is essential in designing structures and devices that may experience temperature changes. For example, engineers must consider this property when building bridges and railways to prevent damage or deformation.
Temperature Change
Temperature change is a critical factor in thermal expansion. It refers to the difference between two temperature readings, typically an increase or decrease. In the context of the rod exercise, a temperature change of \(60^{\circ} \mathrm{C}\) was observed. This kind of temperature variation can cause material size changes, as they respond to heating or cooling, making knowledge of this change vital to calculating expansion effects. Whether you're calculating how much a rod lengthens, or how much a building might shift due to temperature fluctuations, capturing accurate temperature variations is essential. It determines how dramatically a material will expand or contract, especially given the unique properties of different materials, which react variously depending on their coefficients of linear expansion.
Material Properties
Material properties, specifically the coefficient of linear expansion \( \alpha \), play a crucial role in understanding how materials behave under temperature changes. This coefficient is a unique property for each type of material, quantifying the change in length per degree change in temperature.Different materials have different coefficients:
  • Metals generally have higher coefficients than non-metallic materials, leading to more expansion with the same temperature rise.
  • For instance, an aluminum rod will expand more than a glass rod under the same conditions.
These properties help in predicting how much a material will expand or contract. In engineering applications, this knowledge ensures that components fit together and remain functional under varying temperatures, preventing failures and improving the reliability of structures.

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Most popular questions from this chapter

A cylinder of diameter \(1.00000 \mathrm{~cm}\) at \(30^{\circ} \mathrm{C}\) is to be slid into a hole in a steel plate. The hole has a diameter of \(0.99970 \mathrm{~cm}\) at 30 \({ }^{\circ} \mathrm{C}\). To what temperature must the plate be heated? For steel, \(\alpha=\) \(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) The plate will expand in the same way whether or not there is a hole in it. Hence, the hole expands in the same way a circle of steel filling it would expand. We want the diameter of the hole to change by $$ \Delta L=(1.00000-0.99970) \mathrm{cm}=0.00030 \mathrm{~cm} $$ Using \(\Delta L=\alpha L \Delta T\), $$ \Delta T=\frac{\Delta L}{\alpha L_{0}}=\frac{0.00030 \mathrm{~cm}}{\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.99970 \mathrm{~cm})}=27^{\circ} \mathrm{C} $$ The temperature of the plate must be \(30+27=57^{\circ} \mathrm{C}\).

Lead melts at \(621^{\circ} \mathrm{F}\). What temperature is that in kelvins?

When a building is constructed at \(-10{ }^{\circ} \mathrm{C}\), a steel beam (crosssectional area \(45 \mathrm{~cm}^{2}\) ) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force on the beam when the temperature is \(25^{\circ} \mathrm{C}\) ? For this kind of steel, \(\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) and \(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). Proceed much as in Problem 15.11: $$ \begin{array}{c} \frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{-5} \mathrm{C}^{-1}\right)\left(35^{\circ} \mathrm{C}\right)=3.85 \times 10^{-4} \\\ F=Y A \frac{\Delta L}{L_{0}}=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(45 \times 10^{-4} \mathrm{~m}^{2}\right)\left(3.85 \times 10^{-4}\right)=3.5 \times 10^{5} \mathrm{~N} \end{array} $$

A glass vessel is filled with exactly 1 liter of turpentine at \(20{ }^{\circ} \mathrm{C}\). What volume of the liquid will overflow if the temperature is raised to \(86^{\circ} \mathrm{C}\) ? The coefficient of linear expansion of that glass is \(9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\); the coefficient of volume expansion of turpentine is \(97 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\).

Dry ice freezes at a temperature of \(-109.3^{\circ} \mathrm{F}\). What is that in Celsius?
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