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Chapter 17: Problem 10

(II) To what temperature would you have to heat a brass rod for it to be \(1.0 \%\) longer than it is at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
You need to heat the brass rod to approximately 551.3掳C.

Step by step solution

01

Understand the Problem

We want the brass rod to be 1% longer than its original length at 25掳C. We need to find the temperature required to achieve this change in length.
02

Set Up the Formula

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

Express 1% Increase in length

Given that the rod is 1% longer than its original length, \( \Delta L = 0.01 L_0 \). Substitute this into the equation: \( 0.01 L_0 = L_0 \alpha \Delta T \).
04

Simplify the Equation

Cancel \( L_0 \) from both sides: \( 0.01 = \alpha \Delta T \).
05

Solve for \( \Delta T \)

Rearrange to solve for \( \Delta T \): \( \Delta T = \frac{0.01}{\alpha} \). Use \( \alpha = 19 \times 10^{-6} \text{掳C}^{-1} \) for brass.
06

Calculate \( \Delta T \)

Substitute \( \alpha \) into the equation: \( \Delta T = \frac{0.01}{19 \times 10^{-6}} \approx 526.32 \text{掳C} \).
07

Find the Final Temperature

The initial temperature \( T_0 = 25\text{掳C} \). Add \( \Delta T \) to \( T_0 \): \( T = 25 + 526.32 \approx 551.32 \text{掳C} \).

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Key Concepts

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

Understanding the Linear Expansion Formula
The linear expansion formula is the key to solving problems involving the change in length of materials due to temperature changes. It is expressed as \( \Delta L = L_0 \alpha \Delta T \).
This equation allows us to calculate how much a material will expand or contract with a change in temperature.
  • \( \Delta L \) represents the change in length.
  • \( L_0 \) is the original length of the material before heating or cooling.
  • \( \alpha \) is the coefficient of linear expansion, specific to each material.
  • \( \Delta T \) is the difference in temperature the material undergoes.
By rearranging this formula, we can determine any of the variables if the others are known. This makes it a versatile tool in thermal physics.
Role of the Coefficient of Linear Expansion
The coefficient of linear expansion \( \alpha \) is a critical factor in determining how much a material expands or contracts due to temperature changes.
It is a material-specific constant that provides the rate of expansion per degree change in temperature.
For brass, \( \alpha \) is typically 19 \( \times 10^{-6} \text{掳C}^{-1} \), indicating that for each degree Celsius increase in temperature, brass expands by 19 millionths of its original length.
  • A higher \( \alpha \) means the material expands more with temperature increases.
  • Makes materials selection crucial in applications where thermal stability is necessary.
  • Assists in predicting thermal stress in engineering projects.
Understanding \( \alpha \) helps engineers and scientists design structures that accommodate temperature changes.
Importance of Temperature Change in Thermal Expansion
Temperature change \( \Delta T \) is a fundamental aspect when considering thermal expansion in materials.
In problems involving linear expansion, \( \Delta T \) signifies the difference in temperature that causes the material to expand or contract.
  • Calculated as \( T_{final} - T_{initial} \).
  • In explicit problems like this, solving for \( \Delta T \) allows us to predict the behavior of materials under specific conditions.
  • The larger the \( \Delta T \), the more significant the expansion or contraction.
In our exercise, after calculating \( \Delta T \), we find that the temperature needed to expand the brass rod by 1% is 526.32掳C over its original state.
Exploring Brass Properties in Thermal Scenarios
Brass is an alloy primarily made of copper and zinc, known for its excellent thermal conductivity and expansion properties.
Given its coefficient of linear expansion, brass is an ideal material for applications involving moderate thermal shifts.
  • Possesses durable properties while still being capable of significant expansion when needed.
  • Used in applications like fittings and musical instruments, where temperature-induced length changes are crucial.
  • Its behavior under thermal conditions makes it invaluable for certain engineering and construction purposes.
Understanding brass and its properties help technicians and engineers leverage its expansion behavior to achieve desired outcomes in different scenarios.

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Most popular questions from this chapter

(II) A thermocouple consists of a junction of two different types of materials that produces a voltage depending on its temperature. A thermocouple's voltages were recorded when at different temperatures as follows: \begin{tabular}{lrrrr} \hline Temperature \(\left({ }^{\circ} \mathrm{C}\right)\) & 50 & 100 & 200 & 300 \\ Voltage \((\mathrm{mV})\) & 1.41 & 2.96 & 5.90 & 8.92 \\ \hline \end{tabular} Use a spreadsheet to fit these data to a cubic equation and determine the temperature when the thermocouple produces \(3.21 \mathrm{mV}\). Get a second value of the temperature by fitting the data to a quadratic equation.

(II) The lowest pressure attainable using the best available vacuum techniques is about \(10^{-12} \mathrm{~N} / \mathrm{m}^{2}\). At such a pressure, how many molecules are there per \(\mathrm{cm}^{3}\) at \(0^{\circ} \mathrm{C}\) ?

(II) Calculate the density of nitrogen at STP using the ideal gas law.

(III) A sealed container containing 4.0 mol of gas is squeezed, changing its volume from \(0.020 \mathrm{~m}^{3}\) to \(0.018 \mathrm{~m}^{3}\). During this process, the temperature decreases by \(9.0 \mathrm{~K}\) while the pressure increases by \(450 \mathrm{~Pa}\). What was the original pressure and temperature of the gas in the container?

(II) A tank contains 30.0 \(\mathrm{kg}\) of \(\mathrm{O}_{2}\) gas at a gauge pressure of 8.20 atm. If the oxygen is replaced by helium at the same temperature, how many kilograms of the latter will be needed to produce a gauge pressure of 7.00 atm?
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