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Chapter 12: Problem 15

ssm When the temperature of a coin is raised by 75 \({C}^{\circ}\) , the coin's diameter increases by \(2.3 \times 10^{-5} {m} .\) If the original diameter of the coin is \(1.8 \times 10^{-2} {m},\) find the coefficient of linear expansion.

Short Answer

Expert verified
The coefficient of linear expansion is approximately \(1.70 \times 10^{-5}\text{ }\text{m/}\text{m掳C}\).

Step by step solution

01

Understand the Problem

We need to find the coefficient of linear expansion \(\alpha\) of a coin given that its diameter increases by \(2.3 \times 10^{-5}\) meters when the temperature is raised by 75掳C. The original diameter is \(1.8 \times 10^{-2}\) meters.
02

Formula for Linear Expansion

The formula for linear expansion is given by \(\Delta L = \alpha L_0 \Delta T\), where \(\Delta L\) is the change in length (diameter, in this case), \(L_0\) is the original length (original diameter), \(\Delta T\) is the change in temperature, and \(\alpha\) is the coefficient of linear expansion.
03

Assign Values to Variables

Assign the given values to the corresponding variables: \(\Delta L = 2.3 \times 10^{-5} \) m, \(L_0 = 1.8 \times 10^{-2}\) m, and \(\Delta T = 75\) 掳C.
04

Solve for \(\alpha\)

Rearrange the formula \(\Delta L = \alpha L_0 \Delta T\) to find \(\alpha\):\[\alpha = \frac{\Delta L}{L_0 \Delta T}\]Then, substitute the values into the formula:\[\alpha = \frac{2.3 \times 10^{-5}}{1.8 \times 10^{-2} \times 75}\]
05

Calculate the Coefficient

Calculate \(\alpha\) using the formula from Step 4.\[\alpha = \frac{2.3 \times 10^{-5}}{1.35}\approx 1.704 \times 10^{-5}\text{ }\text{m/}\text{m掳C}\].
06

Write the Final Answer

Thus, the coefficient of linear expansion is approximately \(1.70 \times 10^{-5}\text{ }\text{m/}\text{m掳C}\).

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Key Concepts

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

Linear Expansion Formula
Linear expansion refers to the change in a material's dimension due to a change in temperature. This phenomenon can be described using a specific formula. The formula \[\Delta L = \alpha L_0 \Delta T\]provides a simple way to calculate the amount by which a material expands.
The variables in the formula are:
  • \(\Delta L\): Change in length, such as lengthening or shortening of materials.
  • \(\alpha\): Coefficient of linear expansion, a constant specific to each material indicating how much it expands per degree change in temperature.
  • \(L_0\): Original length (or other dimension) of the material.
  • \(\Delta T\): Change in temperature.
This formula helps in predicting how materials will behave under thermal stress, which is fundamental in engineering and construction.
Temperature Change Effects
When temperature changes, it directly influences the size of the material due to atomic level shifts. As the temperature rises?
  • Atoms in the material gain energy and vibrate more, resulting in an expansion of the material.
  • This increase in temperature leads to a proportional increase in dimensions when other conditions remain constant.
  • Conversely, when the temperature decreases, materials may contract as atoms lose energy and vibrate less.
Understanding these effects is essential for designing structures and devices that experience temperature fluctuations. It ensures stability and integrity in numerous applications ranging from buildings to electronic circuits.
Material Properties
Every material has distinct properties which determine how it reacts to temperature changes. The coefficient of linear expansion \(\alpha\) is a critical property that characterizes this behavior.
  • Materials with a high coefficient of linear expansion tend to expand more for a given temperature change, while those with a low coefficient will expand less.
  • Metals typically have higher coefficients than non-metals, making them more responsive to temperature changes.
  • Engineers must consider these properties during the material selection process, particularly in environments subject to thermal cycling.
By understanding these properties, one can predict how materials will expand or contract, aiding in the development of more reliable and efficient designs.
Dimensional Change Calculation
Calculating the change in dimensions of a material due to temperature is vital in preventing structural failures. To perform this calculation, follow these steps:First, identify the original dimension (\(L_0\)), the temperature change (\(\Delta T\)), and the material's coefficient of linear expansion (\(\alpha\)).
Then, apply the linear expansion formula:\[\Delta L = \alpha L_0 \Delta T\]Substitute the values into the formula to find \(\Delta L\). This result will tell you how much the material's dimension will change with the given temperature shift.Using this calculation, engineers can ensure that the structures they work on remain safe and functional under varying temperatures. It allows for proactive measures, like choosing materials that meet environmental demands, to be applied in design and maintenance.

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Most popular questions from this chapter

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?

An ice chest at a beach party contains 12 cans of soda at \(5.0^{\circ} {C}\) . Each can of soda has a mass of 0.35 \({kg}\) and a specific heat capacity of 3800 \({J} /({kg} \cdot {C}^{\circ}) .\) Someone adds a 6.5 \({kg}\) watermelon at \(27^{\circ} {C}\) to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon.

mmh During an all-night cram session, a student heats up a one-half liter \(\left(0.50 \times 10^{-3} {m}^{3}\right)\) glass (Pyrex) beaker of cold coffee. Initially, the temperature is \(18^{\circ} {C}\) , and the beaker is filled to the brim. A short time later when the student returns, the temperature has risen to \(92^{\circ} {C}\) . The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?

A thermos contains 150 \({cm}^{3}\) of coffee at \(85^{\circ} {C}\) . To cool the coffee, you drop two \(11-{g}\) ice cubes into the thermos. The ice cubes are initially at \(0^{\circ} {C}\) and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water.

A spherical brass shell has an interior volume of \(1.60 \times 10^{-3} {m}^{3}\) Within this interior volume is a solid steel ball that has a volume of \(0.70 \times 10^{-3} {m}^{3}\) . The space between the steel ball and the inner surface of the brass shell is filled completely with mercury. A small hole is drilled through the brass, and the temperature of the arrangement is increased by 12 \({C}^{\circ}\). What is the volume of the mercury that spills out of the hole?
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