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Chapter 17: Problem 31

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

Short Answer

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

Step by step solution

01

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 \(1.90 \, \text{cm}\).
02

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: \(\Delta L = 1.90 \, \text{cm}\), \(L_i = 150 \, \text{cm}\), \(\Delta T = 420.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 400^{\circ} \mathrm{C}\).
03

Calculate Average Coefficient of Linear Expansion

Substitute the given values into the formula:\[ \alpha = \frac{1.90 \text{ cm}}{150 \text{ cm} \times 400^{\circ} \mathrm{C}} \]\[ \alpha = \frac{1.90 \text{ cm}}{60000 \text{ cm}\cdot^{\circ} \mathrm{C}} \]\[ \alpha = 3.17 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1} \]
04

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.
05

Calculate the Stress in the Wire

Substitute the values we have into the formula for stress:\[ \sigma = 20 \times 10^{11} \text{ Pa} \times 3.17 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1} \times 400^{\circ} \mathrm{C} \]\[ \sigma = 20 \times 10^{11} \text{ Pa} \times 1.268 \times 10^{-2} \]\[ \sigma = 2.536 \times 10^{10} \text{ Pa} \]
06

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 \(2.536 \times 10^{10} \text{ Pa}\).

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Key Concepts

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

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掳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鈥檚 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.

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Most popular questions from this chapter

Bulk Stress Due to a Temperature Increase. (a) Prove that, if an object under pressure has its temperature raised but is not allowed to expand, the increase in pressure is $$\Delta p=B \beta \Delta T$$ where the bulk modulus \(B\) and the average coefficient of volume expansion \(\beta\) are both assumed positive and constant. (b) What pressure is necessary to prevent a steel block from expanding when its temperature is increased from \(20.0^{\circ} \mathrm{C}\) to \(35.0^{\circ} \mathrm{C} ?\)

A crate of fruit with mass 35.0 \(\mathrm{kg}\) and specific heat 3650 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) slides down a ramp inclined at \(36.9^{\circ} \mathrm{C}\) below the horizontal. The ramp is 8.00 \(\mathrm{m}\) long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

A wood ceiling with thermal resistance \(R_{1}\) is covered with a layer of insulation with thermal resistance \(R_{2} .\) Prove that the effective thermal resistance of the combination is \(R=R_{1}+R_{2}\) .

Before going in for his annual physical, a \(70.0-\mathrm{kg}\) man whose body temperature is \(37.0^{\circ} \mathrm{C}\) consumes an entire \(0.355-\mathrm{L}\) can of a soft drink (mostly water) at \(12.0^{\circ} \mathrm{C}\) . (a) What will his body temperature be after equilibrium is attained? Ignore any heating bythe man's metabolism. The specific heat of the man's body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . (b) Is the change in his body temperature great enough to be measured by a medical themometer?

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.
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