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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 15

Thermal Expansion - all with Video Answers

Educators

Chapter Questions

01:13

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 $1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.
$$
\Delta L=\alpha L_{0} \Delta T=\left(1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.80 \mathrm{~m})\left[(35-15){ }^{\circ} \mathrm{C}\right]=2.7 \times 10^{-4} \mathrm{~m}
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
03:30

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

Kajal Gautam
Kajal Gautam
Numerade Educator
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,
$$
\frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(-35^{\circ} \mathrm{C}\right)=-3.9 \times 10^{-4}=-0.039 \%
$$

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
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,
$$
\frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(-35^{\circ} \mathrm{C}\right)=-3.9 \times 10^{-4}=-0.039 \%
$$

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:47

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 $\operatorname{rod}\left(\alpha=2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$
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,
or
$$
\begin{array}{c}
\left(\alpha L_{0} \Delta T\right)_{\text {copper }}=\left(\alpha L_{0} \Delta T\right)_{\text {abminum }} \\
\left(1.70 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) L_{0} \Delta T=\left(2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(L_{0}-0.20 \mathrm{~m}\right) \Delta T
\end{array}
$$
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 $L_{0}=0.88 \mathrm{~m}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
02:53

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 $\left(\alpha=2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$ 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 $121^{\circ} \mathrm{C}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
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 $30^{\circ} \mathrm{C}$ ? $\alpha_{\text {stel }}=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} ; \alpha_{\text {copper }}=1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$
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 $90.01 \mathrm{~cm}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
02:55

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 $\beta_{\text {elass }}=3 \alpha_{\text {plass }}$ as a good approximation. The flask interior will expand just as though it were a solid piece of glass. Thus,
Volume of mercury above mark $=(\Delta V$ for mercury) $-(\Delta V$ for glass)
$$
\begin{array}{l}
=\beta_{m} V_{0} \Delta T-\beta_{g} V_{0} \Delta T=\left(\beta_{m}-\beta_{g}\right) V_{0} \Delta T \\
=\left[(182-27) \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right]\left(50.00 \mathrm{~cm}^{3}\right)\left[(38-18){ }^{\circ} \mathrm{C}\right] \\
=0.15 \mathrm{~cm}^{3}
\end{array}
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
03:01

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
$$
\begin{array}{l}
\rho_{0}=\text { Density of mercury at } 0{ }^{\circ} \mathrm{C} \\
\rho_{1}=\text { Density of mercury at } 50^{\circ} \mathrm{C} \\
V_{0}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 0{ }^{\circ} \mathrm{C} \\
V_{1}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 50^{\circ} \mathrm{C}
\end{array}
$$
Since the mass does not change, $m=\rho_{0} V_{0}=\rho_{1} V_{1}$, from which it follows that
$$
\rho_{1}=\rho_{0} \frac{V_{0}}{V_{1}}=\rho_{0} \frac{V_{0}}{V_{0}+\Delta V}=\rho_{0} \frac{1}{1+\left(\Delta V / V_{0}\right)}
$$
But
$$
\frac{\Delta V}{V_{0}}=\beta \Delta T=\left(1.82 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\right)\left(50.0^{\circ} \mathrm{C}\right)=0.00910
$$
Substitution into the first equation yields
$$
\rho_{1}=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right) \frac{1}{1+0.00910}=13.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
05:03

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,
$$
\Delta \rho=\rho-\rho_{0}=-\rho \beta \Delta T
$$
In practice, $\rho$ is close enough to $\rho_{0}$ so that we can say $\Delta \rho \approx-\rho_{0} \beta \Delta T$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
01:18

Problem 10

Solve Problem $15.8$ using the result of Problem $15.9 .$
We have
$$
\Delta \rho \approx-\rho_{0} \beta \Delta T
$$
Hence $\quad \Delta \rho \approx-\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(182 \times 10^{-6 \circ} \mathrm{C}^{-1}\right)\left(50.0^{\circ} \mathrm{C}\right)=-124 \mathrm{~kg} / \mathrm{m}^{3}$
and $\quad \rho_{50^{\circ} \mathrm{C}}=\rho_{0^{+} \mathrm{C}}-124 \mathrm{~kg} / \mathrm{m}^{3}=13.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Kajal Gautam
Kajal Gautam
Numerade Educator
03:43

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

Kajal Gautam
Kajal Gautam
Numerade Educator
01:39

Problem 12

When a building is constructed at $-10^{\circ} \mathrm{C}$, a steel beam (cross-sectional 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:
so
$$
\begin{array}{c}
\frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(35^{\circ} \mathrm{C}\right)=3.85 \times 10^{-4} \\
F=Y A \frac{\Delta L}{L_{0}}=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(45 \times 10^{-4} \mathrm{~m}^{2}\right)\left(3.85 \times 10^{-4}\right)=3.5 \times 10^{5} \mathrm{~N}
\end{array}
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
01:46

Problem 13

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

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:06

Problem 14

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?

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:38

Problem 15

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, $\alpha=1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$

Vishal Gupta
Vishal Gupta
Numerade Educator
03:36

Problem 16

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.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
01:54

Problem 17

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 aluminum is $22 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
02:26

Problem 18

A solid sphere of mass $m$ and radius $b$ is spinning freely on its axis with angular velocity $\omega_{0}$. 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
Kajal Gautam
Numerade Educator
01:15

Problem 19

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

Kajal Gautam
Kajal Gautam
Numerade Educator
01:26

Problem 20

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

Kajal Gautam
Kajal Gautam
Numerade Educator
01:59

Problem 21

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
Ma Ednelyn Lim
Numerade Educator
03:06

Problem 22

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 $97 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:59

Problem 23

The density 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 gold at $90.0^{\circ} \mathrm{C}$. [Hint: Take a look at Problem 15.9.]

Kajal Gautam
Kajal Gautam
Numerade Educator