/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Multiple-Concept Example 4 revie... [FREE SOLUTION] | 91Ó°ÊÓ

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Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} {C}\) . When the temperature drops to \(-14^{\circ} {C}\) , the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} {N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross- sectional area of \(1.6 \times 10^{-5} {m}^{2},\) and it is made from a material whose coefficient of linear expansion is $$2.5 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}$$ . What is Young's modulus for the material from which the ruler is made?

Short Answer

Expert verified
Young's modulus is \( 7.69 \times 10^{10} \, \text{N/m}^2 \).

Step by step solution

01

Understand the Problem

The problem involves thermal expansion, force, and Young's modulus. When the temperature changes, the ruler expands or contracts. Applying a force to stretch it back to its original length allows us to calculate Young's modulus.
02

Use Thermal Expansion Formula

Thermal expansion causes the length change \( \Delta L \) due to temperature change \( \Delta T = -14^{\circ}C - 25^{\circ}C = -39^{\circ}C \). The formula for linear expansion is:\[\Delta L = \alpha L_0 \Delta T\]Here, \( \alpha = 2.5 \times 10^{-5} \,(\mathrm{C}^{\circ})^{-1}\) is the coefficient of linear expansion.
03

Relate Force to Strain and Stress

Force stretches the ruler back to its original length, inducing stress \( \sigma \) given by:\[\sigma = \frac{F}{A}\]where \( F = 1.2 \times 10^{3} \, N \) and \( A = 1.6 \times 10^{-5} \, m^{2} \). Calculate \( \sigma \):\[\sigma = \frac{1.2 \times 10^{3}}{1.6 \times 10^{-5}} = 7.5 \times 10^{7} \, \text{N/m}^2\]
04

Relate Strain to Thermal Expansion

The strain \( \epsilon \) that restores the ruler is related to the change in ruler length \( \Delta L / L_0 \), which comes from thermal expansion:\[\epsilon = \alpha \Delta T = 2.5 \times 10^{-5} \times -39 = -9.75 \times 10^{-4}\]
05

Calculate Young's Modulus

Young's modulus \( E \) is the ratio of stress to strain:\[E = \frac{\sigma}{\epsilon}\]Substitute the values:\[E = \frac{7.5 \times 10^{7}}{-9.75 \times 10^{-4}} = -7.69 \times 10^{10} \, \text{N/m}^2\]Since modulus is absolute, we take the positive value, thus:\[E = 7.69 \times 10^{10} \, \text{N/m}^2\]
06

Verify Correctness

Ensure the calculation steps accurately reflect the physical concepts of thermal expansion and material mechanics, confirming the compatibility of units and formulas used.

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Key Concepts

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

Thermal Expansion
Thermal expansion is the phenomenon where materials change their shape, area, or volume with a change in temperature. This is crucial in many daily applications and understanding it helps in designing structures that can withstand temperature variations.

The basic idea is simple: when materials are heated, their particles move faster and tend to move apart. Conversely, when cooled, particles move slower and get closer. This process affects how structures, like a ruler in this problem, function across different temperatures.
  • For linear thermal expansion, the formula is: \[\Delta L = \alpha L_0 \Delta T\]
  • \(\Delta L\) is the change in length due to a temperature change \(\Delta T\),
  • \(\alpha\) is the coefficient of linear expansion, and
  • \(L_0\) is the original length of the material.
This is why, in the exercise, the ruler needs to be adjusted when the temperature drops. It shrinks and creates errors in measurement.
Stress and Strain
Stress and strain are central concepts in understanding how materials deform under force. They describe how pressure and deformation relate to the structural properties of a material.

Stress (\(\sigma\)) quantifies the internal forces a material experiences when an external force is applied. It's calculated by dividing the force (\(F\)) by the area (\(A\)) over which the force is applied:\[\sigma = \frac{F}{A}\]

Strain (\(\epsilon\)), on the other hand, describes the deformation experienced by a material. It is dimensionless and is calculated by the change in length divided by the original length. For thermal expansion:\[\epsilon = \frac{\Delta L}{L_0} = \alpha \Delta T\]
  • Stress has the unit of pressure, \(\text{N/m}^2\),
  • Strain has no units, it’s merely a ratio,
  • These two concepts help in determining how much a material will stretch or compress under certain conditions.
In the exercise, stress from the applied force compensates for the strain caused by thermal contraction, helping us find Young's modulus.
Coefficient of Linear Expansion
The coefficient of linear expansion (\(\alpha\)) describes how much a material's length changes per degree change in temperature. It is a material-specific property and varies from one material to another.

This coefficient is a constant used to predict the degree to which a material will expand or contract with temperature variations. For instance, materials with a high \(\alpha\) will expand or contract significantly for small temperature changes.

In practical terms:
  • Measured in \(\text{C}^{-1}\),
  • Dictates the rate of linear expansion or contraction for the material,
  • Helps in calculating the thermal stress and resulting strain in conjunction with Young's modulus.
In our exercise, the coefficient \(2.5 \times 10^{-5} \text{C}^{-1}\) tells us exactly how much the ruler length changes with a decrease in temperature. This factor directly affects the calculations for stress and strain, ensuring accurate compensation for thermal effects.

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Most popular questions from this chapter

(a) Objects A and B have the same mass of 3.0 kg. They melt when \(3.0 \times 10^{4} {J}\) of heat is added to object \({A}\) and when \(9.0 \times 10^{4} {J}\) is added to object B. Determine the latent heat of fusion for the substance from which each object is made. (b) Find the heat required to melt object A when its mass is 6.0 kg.

ssm At a fabrication plant, a hot metal forging has a mass of 75 \({kg}\) and a specific heat capacity of 430 \({J} / {kg} \cdot {C}^{\circ}\) . To harden it, the forging is immersed in 710 \({kg}\) of oil that has a temperature of \(32^{\circ} {C}\) and a specific heat capacity of 2700 \({J} / {kg} \cdot {C}^{\circ}\) ). The final temperature of the oil and forging at thermal equilibrium is \(47^{\circ} {C}\) . Assuming that heat flows only between the forging and the oil, determine the initial temperature of the forging.

When it rains, water vapor in the air condenses into liquid water, and energy is released. (a) How much energy is released when 0.0254 m (one inch) of rain falls over an area of \(2.59 \times 10^{6} {m}^{2}\) (one square mile)? (b) If the average energy needed to heat one home for a year is \(1.50 \times 10^{11} {J}\) , how many homes could be heated for a year with the energy determined in part (a)?

An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can's initial volume at \(5^{\circ} {C}\) is \(3.5 \times 10^{-4} {m}^{3} .\) The coefficient of volume expansion for aluminum is \(69 \times 10^{-6}({C}^{9})^{-1}\) When the can and the liquid are heated to \(78^{\circ} {C}, 3.6 \times 10^{-6} {m}^{3}\) of liquid spills over. What is the coefficient of volume expansion of the liquid?

The heating element of a water heater in an apartment building has a maximum power output of 28 kW. Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of \(14 \times 10^{-5} {m}^{3} / {s} .\) If the water going into the heater has a temperature of \(11^{\circ} {C},\) what is the maximum possible temperature of the hot water that each showering resident receives?

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