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The outer diameter of a glass jar and the inner diameter of its iron lid are both 725 \(\mathrm{mm}\) at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) What will be the size of the difference in these diameters if the lid is briefly held under hot water until its temperature rises to \(50.0^{\circ} \mathrm{C}\) without changing the temperature of the glass?

Short Answer

Expert verified
The size difference in diameters will be approximately 0.239 mm.

Step by step solution

01

Identify relevant material properties

We need the coefficient of linear expansion for iron, which is approximately \(11 imes 10^{-6} \, ^{\circ}\mathrm{C}^{-1}\). This coefficient will be used to determine how much the iron lid expands when heated.
02

Apply the formula for linear expansion

The formula for linear expansion is given by \( \Delta L = \alpha \cdot L \cdot \Delta T \), where \( \Delta L \) is the change in length, \( \alpha \) is the coefficient of linear expansion, \( L \) is the original length, and \( \Delta T \) is the temperature change.
03

Calculate the change in diameter of the iron lid

\( \Delta T = 50.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 30.0^{\circ} \mathrm{C} \). The original diameter \( L = 725 \mathrm{mm} \). Applying the formula: \( \Delta L = 11 \times 10^{-6} \cdot 725 \cdot 30 \approx 0.239 \mathrm{mm} \).
04

Determine the size difference without temperature change of the glass

Since the glass remains at \(20.0^{\circ} \mathrm{C}\), its diameter does not change. Therefore, the change in the size difference between the lid and the jar is solely due to the expansion of the iron lid, which is \(0.239 \mathrm{mm}\).

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Key Concepts

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

Understanding Linear Expansion
Linear expansion is a phenomenon where a material changes its length when its temperature changes. This is an important concept in physics, especially when dealing with materials that expand or contract due to temperature fluctuations.
Linear expansion is calculated using a simple formula: \[ \Delta L = \alpha \times L \times \Delta T \] where:- \( \Delta L \) is the change in length.- \( \alpha \) is the coefficient of linear expansion.- \( L \) is the original length.- \( \Delta T \) is the temperature change.
The change in length \( \Delta L \) depends directly on how much the temperature changes, and it's linear, meaning that if you double the temperature change, you double the expansion. The original length also plays a role; longer objects will have a larger overall expansion.
Coefficient of Linear Expansion Explained
The coefficient of linear expansion, \( \alpha \), is a material-specific value indicating how much a material expands per degree change in temperature. It is expressed in units of inverse degrees Celsius \( (/^{\circ} \mathrm{C}^{-1}) \).- Different materials have different coefficients. For example, metals typically have higher coefficients than materials like glass.- In the example problem, the iron's coefficient of linear expansion is given as approximately \( 11 \times 10^{-6} \, /^{\circ}\mathrm{C}^{-1} \).
This means that for every degree Celsius increase in temperature, the iron expands by \( 11 \times 10^{-6} \) times its original length. Knowing the value of \( \alpha \) helps predict how much a material will expand or contract when its temperature changes. This can be crucial for applications where precise fit and measurements are necessary.
Temperature Change and Its Effects
Temperature change, denoted as \( \Delta T \), is one of the critical factors influencing the degree of linear expansion. It is the difference between the final and initial temperatures of the material.
In the example, the temperature change for the iron lid is calculated by subtracting the initial room temperature from the higher temperature after heating: \[ \Delta T = 50.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 30.0^{\circ} \mathrm{C} \]This 30°C increase leads to expansion in the iron lid.
Understanding temperature change is vital in solving problems related to expansion, as a greater temperature difference results in more significant expansion. In practical terms, engineers and designers must always consider potential temperature fluctuations in materials to ensure optimal performance and safety of structures.

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Most popular questions from this chapter

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