Problem 31 - 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 (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 \(20 \times 10^{11} \mathrm{Pa}\) .

Step-by-Step Solution

Answer

a) \(\alpha = 3.17 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1}\); b) Stress = \(2.536 \times 10^{10} \text{ Pa}\).

Step 1: Understand the Problem for Average Coefficient of Linear Expansion

In part (a) of the exercise, we need to find the average coefficient of linear expansion (\(\alpha\)) for a wire that changes length as the temperature changes from initial \(T_i= 20.0^{\circ} \mathrm{C}\) to final \(T_f = 420.0^{\circ} \mathrm{C}\). The initial length of the wire, \(L_i\), is \(1.50 \, \text{m}\) or \(150 \, \text{cm}\), and the change in length \(\Delta L\) is \(1.90 \, \text{cm}\).

Step 2: Apply Formula for Linear Expansion

The formula to find the average coefficient of linear expansion is:\[ \alpha = \frac{\Delta L}{L_i \times \Delta T} \]where \(\Delta T = T_f - T_i\) is the change in temperature. Substitute the values: \(\Delta L = 1.90 \, \text{cm}\), \(L_i = 150 \, \text{cm}\), \(\Delta T = 420.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 400^{\circ} \mathrm{C}\).

Step 3: Calculate Average Coefficient of Linear Expansion

Substitute the given values into the formula:\[ \alpha = \frac{1.90 \text{ cm}}{150 \text{ cm} \times 400^{\circ} \mathrm{C}} \]\[ \alpha = \frac{1.90 \text{ cm}}{60000 \text{ cm}\cdot^{\circ} \mathrm{C}} \]\[ \alpha = 3.17 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1} \]

Step 4: Understand the Problem for Stress in the Wire

In part (b), the wire does not contract when cooled from \(420.0^{\circ} \mathrm{C}\) to \(20.0^{\circ} \mathrm{C}\). The stress \(\sigma\) in the wire can be calculated using the formula: \(\sigma = Y \times \alpha \times \Delta T\), where \(Y\) is Young's modulus.

Step 5: Calculate the Stress in the Wire

Substitute the values we have into the formula for stress:\[ \sigma = 20 \times 10^{11} \text{ Pa} \times 3.17 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1} \times 400^{\circ} \mathrm{C} \]\[ \sigma = 20 \times 10^{11} \text{ Pa} \times 1.268 \times 10^{-2} \]\[ \sigma = 2.536 \times 10^{10} \text{ Pa} \]

Step 6: Finalize the Answers

a) The average coefficient of linear expansion for the wire is \(\alpha = 3.17 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1}\).b) The stress in the wire when cooled without contraction is \(2.536 \times 10^{10} \text{ Pa}\).

Key Concepts

Coefficient of Linear Expansion

When materials are heated, they often expand, and this expansion can be quantified using the coefficient of linear expansion (\( \alpha \) ). This coefficient tells us how much a material's length will increase per degree change in temperature.
In the given exercise, when the wire is heated from \( 20.0^{\circ} \mathrm{C} \) to \( 420.0^{\circ} \mathrm{C} \) , its length increases by 1.90 cm. The initial length of the wire is 1.50 m (or 150 cm), and the temperature change (\( \Delta T \)) is 400°C.

• Formula: \[ \alpha = \frac{\Delta L}{L_i \times \Delta T} \]
• Substitute the given values: \( \alpha = \frac{1.90\, \text{cm}}{150\, \text{cm} \times 400^{\circ} \mathrm{C}} \).
• Calculate: \( \alpha = 3.17 \times 10^{-5}\, ^{\circ} \mathrm{C}^{-1} \).

This shows that for each degree Celsius of temperature increase, the wire extends by 3.17 millionths of its length.

Stress and Strain

Stress and strain are fundamental concepts when discussing materials under forces. Stress (\( \sigma \)) is defined as force per unit area applied on an object being stretched or compressed.
In this exercise, when the heated wire is cooled to \( 20.0^{\circ} \mathrm{C} \) without the chance to contract, stress is induced in the wire.

• Formula for stress: \[ \sigma = Y \times \alpha \times \Delta T \]
• Since the wire does not contract, the thermal expansion build-up leads to stress.
• Substitute given values: \( \sigma = 20 \times 10^{11} \text{ Pa} \times 3.17 \times 10^{-5}\, ^{\circ} \mathrm{C}^{-1} \times 400^{\circ} \mathrm{C} \).
• Calculation results in: \( \sigma = 2.536 \times 10^{10} \text{ Pa} \).

This stress implies that if the wire were initially free to contract, it would have exerted a force resisted by its surroundings.

Young's Modulus

Young's Modulus (\( Y \)) is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. It's often called the "stiffness" of a material.
In our scenario, the wire has a given Young's Modulus of \( 20 \times 10^{11} \text{ Pa} \). This high value suggests a stiff, less pliable wire.

• Young's Modulus Formula: \[ Y = \frac{\sigma}{\epsilon} \], where \( \epsilon \) is the strain.
• Even though we do not calculate strain directly in this exercise, Young’s Modulus helps us understand how stress correlates to potential elongation (strain) over a material's length.
• A higher modulus indicates that a material stretches less when a force is applied.

Young's Modulus is crucial for assessing materials in construction or engineering due to its relation to both tensile strength and resistance to length changes.