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Chapter 17: Problem 54

When a 50.0 -m-long metal pipe is heated from \(10.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\), it lengthens by \(2.85 \mathrm{~cm}\). a) Determine the linear expansion coefficient. b) What type of metal is the pipe made of?

Short Answer

Expert verified
Answer: The linear expansion coefficient of the metal pipe is \(1.9\times 10^{-5}\,\text{C}^{-1}\), and it is most likely made of aluminum.

Step by step solution

01

Identify the given values

We are given the following values: Initial length of the pipe, \(L_{1} = 50.0\,\text{m}\) Final length of the pipe, \(L_{2}=50.0\,\text{m} + 2.85\,\text{cm}\) Initial temperature, \(T_{1} = 10.0^{\circ}\mathrm{C}\) Final temperature, \(T_{2} = 40.0^{\circ}\mathrm{C}\) Convert the final length of the pipe to meters for consistency. \(L_{2}=50.0\,\text{m} + 0.0285\,\text{m}=50.0285\,\text{m}\)
02

Calculate the change in length and temperature

We need to calculate the change in length and temperature. Use the formula: \(\Delta L = L_{2} - L_{1}\) \(\Delta T = T_{2} - T_{1}\) \(\Delta L = 50.0285\,\text{m} - 50.0\,\text{m} = 0.0285\,\text{m}\) \(\Delta T = 40.0^{\circ}\mathrm{C} - 10.0^{\circ}\mathrm{C} = 30.0^{\circ}\mathrm{C}\)
03

Determine the linear expansion coefficient

The formula for linear expansion is given by: \(\alpha = \frac{\Delta L}{L_{1}\Delta T}\) Plug in the values to calculate the linear expansion coefficient: \(\alpha = \frac{0.0285\,\text{m}}{(50.0\,\text{m})(30.0^{\circ}\mathrm{C})}\) \(\alpha = 1.9\times 10^{-5}\,\text{C}^{-1}\) a) The linear expansion coefficient is \(1.9\times 10^{-5}\,\text{C}^{-1}\).
04

Identify the type of metal

By comparing the linear expansion coefficient found above with known expansion coefficients of common metals, we can find that the value is close to the expansion coefficient of aluminum, which has a linear expansion coefficient of approximately \(2.3\times 10^{-5}\,\text{C}^{-1}\). b) The type of metal is most likely aluminum.

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Most popular questions from this chapter

The background temperature of the universe is a) \(6000 \mathrm{~K}\). b) \(288 \mathrm{~K}\). c) \(3 \mathrm{~K}\). d) \(2.73 \mathrm{~K}\). e) \(0 \mathrm{~K}\).

A copper cube of side length \(40 . \mathrm{cm}\) is heated from \(20 .{ }^{\circ} \mathrm{C}\) to \(120{ }^{\circ} \mathrm{C}\). What is the change in the volume of the cube? The linear expansion coefficient of copper is \(17 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

A plastic-epoxy sheet has uniform holes of radius \(1.99 \mathrm{~cm}\). The holes are intended to allow solid ball bear- ings with an outer radius of \(2.00 \mathrm{~cm}\) to just go through. Over what temperature rise must the plastic-epoxy sheet be heated so that the ball bearings will go through the holes? The linear expansion coefficient of plastic-epoxy is about \(1.3 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

One thermometer is calibrated in degrees Celsius, and another in degrees Fahrenheit. At what temperature is the reading on the thermometer calibrated in degrees Celsius three times the reading on the other thermometer?

You are designing a precision mercury thermometer based on the thermal expansion of mercury \(\left(\beta=1.8 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\right)\) which causes the mercury to expand up a thin capillary as the temperature increases. The equation for the change in volume of the mercury as a function of temperature is \(\Delta V=\beta V_{0} \Delta T\) where \(V_{0}\) is the initial volume of the mercury and \(\Delta V\) is the change in volume due to a change in temperature, \(\Delta T .\) In response to a temperature change of \(1.0^{\circ} \mathrm{C}\), the column of mercury in your precision thermometer should move a distance \(D=1.0 \mathrm{~cm}\) up a cylindrical capillary of radius \(r=0.10 \mathrm{~mm} .\) Determine the initial volume of mercury that allows this change. Then find the radius of a spherical bulb that contains this volume of mercury.
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