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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 15

Thermal Expansion - all with Video Answers

Educators

Chapter Questions

01:51

Problem 1

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 $1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$

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03:08

Problem 2

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 $\Delta L=\alpha L \Delta T$,
$$
\Delta T=\frac{\Delta L}{\alpha L_{0}}=\frac{0.00030 \mathrm{~cm}}{\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.99970 \mathrm{~cm})}=27^{\circ} \mathrm{C}
$$
The temperature of the plate must be $30+27=57^{\circ} \mathrm{C}$.

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02:15

Problem 3

A steel tape is calibrated at $20^{\circ} \mathrm{C}$. On a cold day when the temperature is $-15^{\circ} \mathrm{C}$, what will be the percent error in the tape? $\alpha_{\text {steel }} 1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$

For a temperature change from $20^{\circ} \mathrm{C}$ to $-15{ }^{\circ} \mathrm{C}$, we have $\Delta T-35$ ${ }^{\circ} \mathrm{C}$. Then,
$$
\frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(-35^{\circ} \mathrm{C}\right)=-3.9 \times 10^{-4}=-0.039 \%
$$

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03:23

Problem 4

A copper rod $\left(\alpha=1.70 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$ is $20 \mathrm{~cm}$ longer than an aluminum rod $\left(\alpha=2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$. How long should the copper rod be if the difference in their lengths is to be independent of temperature? For their difference in lengths not to change with temperature, $\Delta L$ must be the same for both rods under the same temperature change. That is,where $L_{0}$ is the length of the copper rod, and $\Delta T$ is the same for both rods. Solving for the original length yields $L_{0}=0.88 \mathrm{~m}$.

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03:42

Problem 5

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 $(\alpha=$ $2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ ) 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 $20.0^{\circ} \mathrm{C}$, the diameters of the hole and of the ball should be equal: $$ \begin{array}{l} 0.9000 \mathrm{~cm}+(0.9000 \mathrm{~cm})\left(1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) \Delta T \\ \quad=0.8990 \mathrm{~cm}+(0.8990 \mathrm{~cm})\left(2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) \Delta T \end{array} $$
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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03:20

Problem 6

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 ${ }^{\circ} \mathrm{C} ? \alpha_{\text {steel }}=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} ; \alpha_{\text {steel }}=1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.
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 Using the approximation
$$
\frac{1}{1+x} \approx 1=x
$$
for $x$ small compared to 1 , we have
$$
\begin{aligned}
90.00 \frac{1+3.4 \times 10^{-4}}{1+2.2 \times 10^{-4}} & \approx 90.0001+3.4 \times 10^{-4},\left(1-2.2 \times 10^{-4}\right) \approx 90.000\left(1+3.4 \times 10^{-4}-2.2 \times 10^{-4}\right) \\
&=90.00+0.0108
\end{aligned}
$$
The tape will read $90.01 \mathrm{~cm}$.

Kajal Gautam
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03:52

Problem 7

A glass flask is filled "to the mark" with $50.00 \mathrm{~cm}^{3}$ of mercury at $18^{\circ} \mathrm{C}$. If the flask and its contents are heated to $38^{\circ} \mathrm{C}$, how much mercury will be above the mark? $\alpha_{\text {glass }}=9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and $\beta_{\text {mercury }}=182 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$
We shall take $\beta_{\text {glass }}=3 \alpha_{\text {glass }}$ as a good approximation. The flask interior will expand just as though it were a solid piece of glass. Thus,$$ \begin{aligned} \text { Volume of mercury above mark } &=(\Delta V \text { for mercury) }-(\Delta V \text { for glass) }\\ &=\beta_{m} V_{0} \Delta T-\beta_{g} V_{0} \Delta T=\left(\beta_{m}-\beta_{g}\right) V_{0} \Delta T \\ &=\left[(182-27) \times 10^{-6}{ }^{-1} \mathrm{C}^{-1}\right]\left(50,00 \mathrm{~cm}^{3}\right)\left[(38-18){ }^{\circ} \mathrm{Cl}\right.\\ &=0.15 \mathrm{~cm}^{3} \end{aligned} $$

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04:47

Problem 8

The density of mercury at exactly $0{ }^{\circ} \mathrm{C}$ is $13600 \mathrm{~kg} / \mathrm{m}^{3}$, and its volume expansion coefficient is $1.82 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. Calculate the density of mercury at $50.0{ }^{\circ} \mathrm{C}$.
Let $$\begin{array}{c} \rho_{0}=\text { Density of mercury at } 0^{\circ} \mathrm{C} \\ \rho_{1}=\text { Density of mercury at } 50^{\circ} \mathrm{C} \\ V_{0}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 0{ }^{\circ} \mathrm{C} \\ V_{1}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 50^{\circ} \mathrm{C} \end{array} $$ Since the mass does not change, $m=\rho_{0} V_{0}=\rho_{1} V_{1}$, from which itfollows that But $$ \begin{array}{c} \rho_{1}=\rho_{0} \frac{V_{0}}{V_{1}}=\rho_{0} \frac{V_{0}}{V_{0}+\Delta V}=\rho_{0} \frac{1}{1+\left(\Delta V / V_{0}\right)} \\
\frac{\Delta V}{V_{0}}=\beta \Delta T=\left(1.82 \times 10^{-4 \circ} \mathrm{C}^{-1}\right)\left(50.0^{\circ} \mathrm{C}\right)=0.00910
\end{array}
$$
Substitution into the first equation yields
$$
\rho_{1}=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right) \frac{1}{1+0.00910}=13.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$

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03:00

Problem 9

Show that the density of a liquid or solid changes in the following way with temperature: $\Delta \rho=-\rho \beta \Delta T \approx-\rho_{0} \beta \Delta T$.
Consider a mass $m$ of liquid having a volume $V_{0}$ for which $\rho_{0}=$ $m / V_{0}$. After a temperature change $\Delta T$, the volume will be
$$
V=V_{0}+V_{0} \beta \Delta T
$$
and the density will be
$$
\rho=\frac{m}{V}=\frac{m}{V_{0}(1+\beta \Delta T)}
$$
But $m / V_{0}=\rho_{0}$, and so this can be written as
$$
\rho(1+\beta \Delta T)=\rho_{0}
$$
Thus,
$$
\Delta \rho=\rho-\rho_{0}=-\rho \beta \Delta T \text { . }
$$
In practice, $\rho$ is close enough to $\rho_{0}$ so that we can say $\Delta \rho \approx-\rho_{0} \beta \Delta T .$

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02:39

Problem 10

Solve Problem $15.8$ using the result of Problem $15.9 .$ We have $$ \Delta \rho \approx-\rho_{0} \beta \Delta T $$
Hence $\Delta \rho \approx-\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(182 \times 10^{-60} \mathrm{C}^{-1}\right)\left(50.0^{\circ} \mathrm{C}\right)=-124 \mathrm{~kg} / \mathrm{m}^{3}$ and $\rho_{50^{\circ} \mathrm{C}}=\rho_{0^{\circ} \mathrm{C}}-124 \mathrm{~kg} / \mathrm{m}^{3}=13.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Ma Ednelyn Lim
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03:32

Problem 11

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 $=F=\frac{\gamma A \Delta L}{L_{0}}=\frac{\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.0 \times 10^{-6} \mathrm{~m}^{2}\right)\left(6.6 \times 10^{-4} \mathrm{~m}\right)}{1.50 \mathrm{~m}}=176 \mathrm{~N}=0.18 \mathrm{kN}$
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.

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03:36

Problem 12

When a building is constructed at $-10{ }^{\circ} \mathrm{C}$, a steel beam (crosssectional area $45 \mathrm{~cm}^{2}$ ) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force on the beam when the temperature is $25^{\circ} \mathrm{C}$ ? For this kind of steel, $\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ and $Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$.
Proceed much as in Problem 15.11:
$$
\begin{array}{c}
\frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{-5} \mathrm{C}^{-1}\right)\left(35^{\circ} \mathrm{C}\right)=3.85 \times 10^{-4} \\
F=Y A \frac{\Delta L}{L_{0}}=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(45 \times 10^{-4} \mathrm{~m}^{2}\right)\left(3.85 \times 10^{-4}\right)=3.5 \times 10^{5} \mathrm{~N}
\end{array}
$$

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01:41

Problem 13

Create an equation to convert Fahrenheit degrees into Celsius degrees. [Hint: Refer to Eq. (15.1).]

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00:52

Problem 14

Water boils at $212{ }^{\circ} \mathrm{F}$. Use the equation you created in Problem $15.13$ to compute the corresponding Celsius temperature.

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01:14

Problem 15

Aluminum melts at $660^{\circ} \mathrm{C}$. How much is that in kelvins?

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00:55

Problem 16

Dry ice freezes at a temperature of $-109.3^{\circ} \mathrm{F}$. What is that in Celsius?

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01:19

Problem 17

Lead melts at $621^{\circ} \mathrm{F}$. What temperature is that in kelvins?

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01:29

Problem 18

A gold wire $20 \mathrm{~m}$ long has its temperature lowered by $25.0{ }^{\circ} \mathrm{C}$. Assume the linear coefficient of expansion is constant over that range of temperatures. Calculate the change in length of the wire. [Hint: Use Table 15-1.]

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02:13

Problem 19

A Pyrex glass rod $200.0 \mathrm{~cm}$ long has its temperature raised from $10.0^{\circ} \mathrm{C}$ to $50.0^{\circ} \mathrm{C}$. Will it end up longer or shorter and by how much? Assume the linear coefficient of expansion is constant over that range of temperatures. [Hint: Use Table 15-1.]

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02:37

Problem 20

A stainless steel wire is $150 \mathrm{~cm}$ long at $20.0{ }^{\circ} \mathrm{C}$. An electric current is passed along it, and it expands to $151.2 \mathrm{~cm}$. What is its new temperature? [Hint: Use Table 15-1.]

Ma Ednelyn Lim
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01:46

Problem 21

Compute the increase in length of $50 \mathrm{~m}$ of copper wire when its temperature changes from $12^{\circ} \mathrm{C}$ to $32^{\circ} \mathrm{C}$. For copper, $\alpha=1.7 \times$ $10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

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02:06

Problem 22

A rod $3.0 \mathrm{~m}$ long is found to have expanded $0.091 \mathrm{~cm}$ in length after a temperature rise of $60^{\circ} \mathrm{C}$. What is $\alpha$ for the material of the rod?

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02:43

Problem 23

At $15.0^{\circ} \mathrm{C}$, a bare wheel has a diameter of $30.000 \mathrm{~cm}$, and the inside diameter of its steel rim is $29.930 \mathrm{~cm}$. To what temperature must the rim be heated so as to slip over the wheel? For this type of steel, $\alpha=1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

Ma Ednelyn Lim
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01:48

Problem 24

An ordinary glass sphere has a volume of $2000 \mathrm{~cm}^{3}$ at a temperature of $0.00^{\circ} \mathrm{C}$. Determine its approximate volume change when raised to $100^{\circ} \mathrm{C}$. [Hint: Use Table $15-2 .$ Be careful converting from $\mathrm{cm}^{3}$ to $\mathrm{m}^{3}$.

Ma Ednelyn Lim
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03:36

Problem 25

An iron ball has a diameter of $6 \mathrm{~cm}$ and is $0.010 \mathrm{~mm}$ too large to pass through a hole in a brass plate when the ball and plate are at a temperature of $30^{\circ} \mathrm{C}$. At what temperature (the same for ball and plate) will the ball just pass through the hole? $\alpha=1.2 \times 10^{-5}$ ${ }^{\circ} \mathrm{C}^{-1}$ and $1.9 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ for iron and brass, respectively.

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02:27

Problem 26

(a) An aluminum measuring rod, which is correct at $5.0^{\circ} \mathrm{C}$, measures a certain distance as $88.42 \mathrm{~cm}$ at $35.0^{\circ} \mathrm{C}$. Determine the error in measuring the distance due to the expansion of the rod. $(b)$ If this aluminum rod measures a length of steel as $88.42$ $\mathrm{cm}$ at $35.0{ }^{\circ} \mathrm{C}$, what is the correct length of the steel at $35^{\circ} \mathrm{C}$ ? The coefficient of linear expansion of that sample of aluminum is $22 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

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02:26

Problem 27

A solid sphere of mass $m$ and radius $b$ is spinning freely on its axis with angular velocity $\omega$. When heated by an amount $\Delta T$, its angular velocity changes to $\omega$. Find $\omega_{0} / \omega$ if the linear expansion coefficient for the material of the sphere is $\alpha$.

Kajal Gautam
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01:32

Problem 28

Calculate the increase in volume of $100 \mathrm{~cm}^{3}$ of mercury when its temperature changes from $10^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. Take the volume coefficient of expansion of that mercury to be $0.00018{ }^{\circ} \mathrm{C}^{-1}$.

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02:24

Problem 29

If a glass specific gravity bottle holds $50.000 \mathrm{~mL}$ at $15^{\circ} \mathrm{C}$, find its capacity at $25^{\circ} \mathrm{C}$. Take the coefficient of linear expansion of the glass to be $9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

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01:59

Problem 30

Determine the change in volume of a block of cast iron $5.0 \mathrm{~cm} \times$ $10 \mathrm{~cm} \times 6.0 \mathrm{~cm}$, when the temperature of the block is made to change from $15^{\circ} \mathrm{C}$ to $47^{\circ} \mathrm{C}$. The coefficient of linear expansion of cast iron is $0.000010^{\circ} \mathrm{C}^{-1}$.

Ma Ednelyn Lim
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03:06

Problem 31

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 that glass is $9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$; the coefficient of volume expansion of turpentine is $97 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

Ma Ednelyn Lim
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02:32

Problem 32

The density of a particular sample of gold is $19.30 \mathrm{~g} / \mathrm{cm}^{3}$ at $20.0$ ${ }^{\circ} \mathrm{C}$, and the coefficient of linear expansion is $14.3 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$. Compute the density of that sample at $90.0^{\circ} \mathrm{C}$. [Hint: Take a look at Problem 15.9.]

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