Problem 23 - University Physics with Modern Physics

Question

(a) A wire that is 1.50 \(\mathrm{m}\) long at \(20.0^{\circ} \mathrm{C}\) is found to increase in length by 1.90 \(\mathrm{cm}\) when warmed to \(420.0^{\circ} \mathrm{C}\) . Compute its average coefficient of linear expansion for this temperature range. (b) The wire is stretched just taut (zero tension) at \(420.0^{\circ} \mathrm{C}\). Find the stress in the wire if it is cooled to \(20.0^{\circ} \mathrm{C}\) without being allowed to contract. Young's modulus for the wire is \(2.0 \times 10^{11} \mathrm{Pa}\)

Step-by-Step Solution

Answer

(a) \( \alpha = 3.17 \times 10^{-5} \underline{\phantom{xxx}} ^{\circ}\text{C}^{-1} \). (b) Stress \( = 2.54 \times 10^{8} \text{ Pa} \).

Step 1: Understanding Coefficient of Linear Expansion

The average coefficient of linear expansion, \( \alpha \), can be calculated using the formula: \( \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \), where \( \Delta L \) is the change in length of the wire, \( L_0 \) is the original length, and \( \Delta T \) is the change in temperature.

Step 2: Calculate Change in Length

The change in length of the wire is given by \( \Delta L = 1.90 \text{ cm} = 0.0190 \text{ m} \).

Step 3: Calculate Change in Temperature

Calculate the change in temperature \( \Delta T \) by subtracting the initial temperature from the final temperature: \( \Delta T = 420.0^{\circ} \text{C} - 20.0^{\circ} \text{C} = 400.0^{\circ} \text{C} \).

Step 4: Calculate Average Coefficient of Linear Expansion

Using the formula from Step 1 and the values from Steps 2 and 3, calculate \( \alpha \):\[\alpha = \frac{0.0190 \text{ m}}{1.50 \text{ m} \times 400.0^{\circ} \text{C}} = 3.17 \times 10^{-5} \underline{\phantom{xxx}} ^{\circ}\text{C}^{-1}\]

Step 5: Understanding Stress in the Wire

When the wire is cooled without contraction, it experiences thermal stress. The stress \( \sigma \) can be calculated using the formula: \( \sigma = Y \cdot \alpha \cdot \Delta T \), where \( Y \) is Young's modulus.

Step 6: Calculate Stress in the Wire

Substituting the known values into the equation for stress:\[\sigma = 2.0 \times 10^{11} \text{ Pa} \times 3.17 \times 10^{-5} \underline{\phantom{xxx}} ^{\circ}\text{C}^{-1} \times 400.0^{\circ} \text{C} = 2.54 \times 10^{8} \text{ Pa}\]

Key Concepts

Thermal Expansion

When materials are heated, they often expand. This is known as thermal expansion. It's an important concept in understanding how materials change with temperature. Thermal expansion happens because the atoms or molecules in a substance move more when they are heated, making the substance take up more space.
There are several types of thermal expansion, but for solids, we often focus on linear expansion. Linear expansion describes how the length of an object changes due to temperature changes. The formula for linear expansion is:

• equation: \[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \] where:
• \( \alpha \) is the coefficient of linear expansion,
• \( L_0 \) is the original length,
• \( \Delta T \) is the change in temperature,
• and \( \Delta L \) is the change in length.

As you can see, knowing the coefficient of linear expansion is key when working with thermal expansion problems.

Young's Modulus

Young's modulus is a measure of a material's ability to withstand changes in length when under lengthwise tension or compression. It is often represented by the symbol \( Y \). A higher Young's modulus indicates that a material is stiffer, meaning it does not stretch or compress easily.
Understanding Young's modulus is essential when calculating how much stress a material can endure. It helps predict how materials will react when subjected to forces, making it crucial in engineering and construction.
The formula involving Young's modulus in relation to stress and strain is:

• equation: \[ \sigma = Y \cdot \epsilon \] where:
• \( \sigma \) is the stress,
• \( Y \) is Young's modulus,
• \( \epsilon \) is the strain (change in length/original length).

Young's modulus is used in various formulas, including calculations involving thermal stress.

Thermal Stress

Thermal stress occurs when a material is restrained and cannot expand or contract freely with temperature changes. This stress can cause material to bend or break.
In the scenario given, when the wire is cooled back to its original temperature, it experiences thermal stress because it's not allowed to contract to its original length. This restriction creates stress within the material.
The thermal stress formula is:

• equation: \[ sigma = Y \cdot \alpha \cdot \Delta T \]
where:
• \( Y \) is Young's modulus,
• \( \alpha \) is the coefficient of linear expansion,
• \( \Delta T \) is the temperature change.

Recognizing when thermal stress occurs is important in design to prevent structural failures.

Temperature Change

Temperature change, denoted as \( \Delta T \), plays a critical role in understanding how materials behave under thermal effects. It is simply the difference between the final temperature and the initial temperature of an object.
This change can cause various effects:

• Expansion or contraction of materials
• Changes in resistance and conductivity
• Influences on chemical reactions

In calculations involving thermal expansion or stress, the temperature change is vital. For instance, in the given problem, the temperature changes from \( 20.0^{\circ}C \) to \( 420.0^{\circ}C \), resulting in \( \Delta T = 400.0^{\circ}C \).
Having precise temperature change values ensures accurate calculations for thermal responses of materials.