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A copper bar is \(80 \mathrm{~cm}\) long at \(15^{\circ} \mathrm{C}\). What is the increase in length when it is heated to \(35^{\circ} \mathrm{C}\) ? The linear expansion coefficient for copper is \(1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\). $$ \Delta L=\alpha L_{0} \Delta T=\left(1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.80 \mathrm{~m})\left[(35-15){ }^{\circ} \mathrm{C}\right]=2.7 \times 10^{-4} \mathrm{~m} $$

Short Answer

Expert verified
The copper bar's length increases by 0.272 mm when heated from 15°C to 35°C.

Step by step solution

01

Determine Initial Parameters

Identify the given values: original length of the copper bar, temperature change, and the linear expansion coefficient. - Initial length, \( L_0 = 80 \) cm = 0.80 m (converted to meters).- Initial temperature, \( T_1 = 15^{\circ} \text{C} \).- Final temperature, \( T_2 = 35^{\circ} \text{C} \).- Linear expansion coefficient for copper, \( \alpha = 1.7 \times 10^{-5} \, ^{\circ}\text{C}^{-1} \).
02

Calculate Temperature Change

Find the change in temperature, \( \Delta T \), by subtracting the initial temperature from the final temperature. \[ \Delta T = T_2 - T_1 \]\[ \Delta T = 35^{\circ}\text{C} - 15^{\circ}\text{C} = 20^{\circ}\text{C} \].
03

Apply Linear Expansion Formula

Use the formula for linear expansion to find the increase in length, \( \Delta L \):\[ \Delta L = \alpha L_0 \Delta T \]\[ \Delta L = (1.7 \times 10^{-5} \, ^{\circ}\text{C}^{-1})(0.80 \, \text{m})(20^{\circ}\text{C}) \]
04

Perform the Calculation

Calculate the product:\[ \Delta L = (1.7 \times 10^{-5})(0.80)(20) \]\[ \Delta L = 2.72 \times 10^{-4} \, \text{m} \].
05

Conclusion

Interpret the result: The copper bar increases in length by \(0.000272\) meters or \(0.272\) mm when the temperature rises from \(15^{\circ}C\) to \(35^{\circ}C\).

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Key Concepts

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

Linear Expansion Coefficient
In thermal physics, the linear expansion coefficient, often symbolized as \( \alpha \), plays a crucial role in understanding how materials change size with temperature. This coefficient quantifies how a material's length changes per degree change in temperature. It's defined specifically for linear expansions, such as in long objects like rods or bars.

The formula for calculating linear expansion can be expressed as:
  • \( \Delta L = \alpha L_0 \Delta T \)
Here, \( \Delta L \) represents the change in length, \( \alpha \) is the linear expansion coefficient, \( L_0 \) is the initial length, and \( \Delta T \) is the change in temperature. Using this formula requires accurate initial measurements and a known coefficient, ensuring reliable predictions of length changes.

To put it into context, the copper's linear expansion coefficient is \( 1.7 \times 10^{-5} \, ^\circ\text{C}^{-1} \), indicating a very small change per degree Celsius increase — a characteristic shared with many metals that inherit this specific coefficient.
Temperature Change
Temperature change, noted as \( \Delta T \), is the difference between two temperature readings and is a crucial component in measuring thermal expansion. Understanding the temperature change helps in predicting how much a material, like copper, will expand or contract when subjected to different thermal environments.

The calculation for \( \Delta T \) is straightforward and is performed by subtracting the initial temperature from the final temperature:
  • \( \Delta T = T_2 - T_1 \)
This simple subtraction yields the thermal difference that drives the expansion or contraction process in materials.

In the given exercise, the copper bar experienced a temperature shift from \( 15^{\circ} \text{C} \) to \( 35^{\circ} \text{C} \), resulting in a \( 20^{\circ} \text{C} \) rise. This specific change sets the stage for calculating how much the copper's length increases using the linear expansion formula.
Copper Properties
Copper is a widely used metal known for its excellent thermal and electrical conductivity. It has distinct properties that make it interesting when studying thermal expansion. Some of the important thermal properties of copper include:
  • High thermal conductivity: Copper efficiently transfers heat, making it valuable in heat exchangers and other thermal applications.
  • Linear expansion coefficient: The coefficient for copper is relatively small, indicating that while it does expand with heat, the changes are quite minimal compared to other materials like plastics or certain other metals.
Additionally, copper retains its structural integrity across a wide range of temperatures, maintaining its usefulness in various industrial processes.

When heated, as in the exercise scenario, the copper bar expanded due to the increase in kinetic energy of its molecules, subtly increasing its length. Understanding copper's properties, particularly its expansion characteristics, is crucial for applications that demand precision and reliability, such as plumbing or electrical systems where copper pipes and wires are used.

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Most popular questions from this chapter

A steel wire of \(2.0 \mathrm{~mm}^{2}\) cross section at \(30^{\circ} \mathrm{C}\) is held straight (but under no tension) by attaching its ends firmly to two points a distance \(1.50 \mathrm{~m}\) apart. (Of course this will have to be done out in space so the wire is weightless, but don't worry about that.) If the temperature now decreases to \(-10^{\circ} \mathrm{C}\), and if the two tie points remain fixed, what will be the tension in the wire? For steel, \(\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) and \(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) If it were free to do so, the wire would contract a distance \(\Delta L\) as it cooled, where $$ \Delta L=\alpha L_{0} \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(1.5 \mathrm{~m})\left(40^{\circ} \mathrm{C}\right)=6.6 \times 10^{-4} \mathrm{~m} $$ But the ends are fixed. As a result, forces at the ends must, in effect, stretch the wire this same length \(\Delta L .\) Therefore, from \(Y=(F / A)\left(\Delta L / L_{0}\right)\), and Tension \(=F=\frac{Y A \Delta L}{L_{0}}=\frac{\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.0 \times 10^{-6} \mathrm{~m}^{2}\right)\left(6.6 \times 10^{-4} \mathrm{~m}\right)}{1.50 \mathrm{~m}}=176 \mathrm{~N}=0.18 \mathrm{kN}\) Strictly, we should have substituted \(\left(1.5-6.6 \times 10^{-4}\right) \mathrm{m}\) for \(L\) in the expression for the tension. However, the error incurred in not doing so is negligible.

A glass vessel is filled with exactly 1 liter of turpentine at \(20^{\circ} \mathrm{C}\). What volume of the liquid will overflow if the temperature is raised to \(86^{\circ} \mathrm{C}\) ? The coefficient of linear expansion of the glass is \(9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\); the coefficient of volume expansion of turpentine is \(97 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\).

A steel tape measures the length of a copper rod as \(90.00 \mathrm{~cm}\) when both are at \(10^{\circ} \mathrm{C}\), the calibration temperature for the tape. What would the tape read for the length of the rod when both are at \(30^{\circ} \mathrm{C}\) ? \(\alpha_{\text {stel }}=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} ; \alpha_{\text {copper }}=1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) At \(30^{\circ} \mathrm{C}\), the copper rod will be of length $$ L_{0}\left(1+\alpha_{c} \Delta T\right) $$ while adjacent "centimeter" marks on the steel tape will be separated by a distance of $$ (1.000 \mathrm{~cm})\left(1+\alpha_{s} \Delta T\right) $$ Therefore, the number of "centimeters" read on the tape will be $$ \frac{L_{0}\left(1+\alpha_{c} \Delta T\right)}{(1 \mathrm{~cm})\left(1+\alpha_{s} \Delta T\right)}=\frac{(90.00 \mathrm{~cm})\left[1+\left(1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(20^{\circ} \mathrm{C}\right)\right]}{(1.000 \mathrm{~cm})\left[1+\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(20^{\circ} \mathrm{C}\right)\right]}=90.00 \frac{1+3.4 \times 10^{-4}}{1+2.2 \times 10^{-4}} $$ Using the approximation $$ \frac{1}{1+x} \approx 1-x $$ for \(x\) small compared to 1 , we have $$ \begin{aligned} 90.00 \frac{1+3.4 \times 10^{-4}}{1+2.2 \times 10^{-4}} & \approx 90.00\left(1+3.4 \times 10^{-4}\right)\left(1-2.2 \times 10^{-4}\right) \approx 9000\left(1+3.4 \times 10^{-4}-2.2 \times 10^{-4}\right) \\ &=90.00+0.0108 \end{aligned} $$ The tape will read \(90.01 \mathrm{~cm}\).

A cylinder of diameter \(1.00000 \mathrm{~cm}\) at \(30^{\circ} \mathrm{C}\) is to be slid into a hole in a steel plate. The hole has a diameter of \(0.99970 \mathrm{~cm}\) at \(30^{\circ} \mathrm{C}\). To what temperature must the plate be heated? For steel, \(\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) The plate will expand in the same way whether or not there is a hole in it. Hence, the hole expands in the same way a circle of steel filling it would expand. We want the diameter of the hole to change by $$ \Delta L=(1.00000-0.99970) \mathrm{cm}=0.00030 \mathrm{~cm} $$ Using \(\Delta L=\alpha L_{0} \Delta T\) $$ \Delta T=\frac{\Delta L}{\alpha L_{0}}=\frac{0.00030 \mathrm{~cm}}{\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.99970 \mathrm{~cm})}=27^{\circ} \mathrm{C} $$ The temperature of the plate must be \(30+27=57^{\circ} \mathrm{C}\)

At \(20.0{ }^{\circ} \mathrm{C}\) a steel ball \(\left(\alpha=1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\) has a diameter of \(0.9000 \mathrm{~cm}\), while the diameter of a hole in an aluminum plate \(\left(\alpha=2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\) is \(0.8990 \mathrm{~cm}\). At what temperature (the same for both) will the ball just pass through the hole? At a temperature \(\Delta T\) higher than \(200^{\circ} \mathrm{C}\), the diameters of the hole and of the ball should be equal: $$ \begin{array}{r} 0.9000 \mathrm{~cm}+(0.9000 \mathrm{~cm})\left(1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) \Delta T= \\ 0.8990 \mathrm{~cm}+(0.8990 \mathrm{~cm})\left(2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) \Delta T \end{array} $$ Solving for \(\Delta T\), we find \(\Delta T=101^{\circ} \mathrm{C}\). Because the original temperature was \(20.0^{\circ} \mathrm{C}\), the final temperature must be \(121^{\circ} \mathrm{C}\).

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