Thermal expansion

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 10-1Core

   A detached house has a $6.0\,\text{m}$ long gutter made of coated steel ($\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$). On a summer day the sun heats the gutter from $15\,{}^\circ\text{C}$ in the morning to $55\,{}^\circ\text{C}$ in the afternoon.

   By how much does the gutter get longer? Give the answer in millimetres.

   mmCheck
   Answer

   $\Delta L \approx 2.9\,\text{mm}$

   Show the solution

   Given

   steel gutter, $L_0 = 6.0\,\text{m}$, $T_0 = 15\,{}^\circ\text{C}$, $T = 55\,{}^\circ\text{C}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the change in length of the gutter $\Delta L$ in millimetres

   Principle

   The vibration of the particles of a solid gets stronger when it warms up and the average distance between the particles grows. The gutter is not jammed and its state of matter does not change, so the change in length is proportional to the initial length and the change in temperature.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, where $\Delta T = T - T_0$. The quantity asked for, $\Delta L$, is already alone on the left side.

   Conversions

   The change in temperature is the same in degrees Celsius and in kelvins:

   $\Delta T$$= 55\,{}^\circ\text{C} - 15\,{}^\circ\text{C}$

   $= 40\,\text{K}$

   The result comes out in metres, and $1\,\text{m} = 1000\,\text{mm}$.

   Substitution

   $\Delta L$$= 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 6.0\,\text{m} \cdot 40\,\text{K}$

   $= 0.00288\,\text{m}$

   $= 2.88\,\text{mm}$

   Answer

   $\Delta L \approx 2.9\,\text{mm}$

   Two significant figures. The limiting data are the length $6.0\,\text{m}$ and the coefficient $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (both have two figures); the temperature difference $\Delta T = 40\,\text{K}$, as a difference of whole degrees, also has two figures, so it does not limit the result any further.

   Reasonableness

   A rough estimate:

   $10^{-5}\,\tfrac{1}{\text{K}} \cdot 6\,\text{m} \cdot 40\,\text{K}$

   $\approx 2.4\,\text{mm}$

   so the result is of the right order of magnitude. $\Delta L$ is positive because the gutter warms up. The change is only $\alpha \Delta T = 0.048\,\%$ of the length, but the fastenings must still let the gutter slide.

2. Exercise 10-2Core

   An aluminium overhead wire ($\alpha = 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) between two electricity poles is $50.0\,\text{m}$ long at $18\,{}^\circ\text{C}$ on a summer evening. On a frosty night the temperature of the wire drops to $-27\,{}^\circ\text{C}$. Calculate the change in the length of the wire $\Delta L$ with its sign. Give the answer in millimetres.

   mmCheck
   Answer

   $\Delta L \approx -52\,\text{mm}$

   Show the solution

   Given

   aluminium wire, $L_0 = 50.0\,\text{m}$, $T_0 = 18\,{}^\circ\text{C}$, $T = -27\,{}^\circ\text{C}$, $\alpha = 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the change in length $\Delta L$ with its sign in millimetres

   Principle

   When it cools, the vibration of the particles gets weaker and their average distance decreases, so the wire gets shorter. The state of matter of the wire does not change, so the same formula $\Delta L = \alpha L_0 \Delta T$ describes both warming and cooling, and a negative $\Delta T$ gives a negative $\Delta L$.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, where $\Delta T = T - T_0$ (final minus initial).

   Conversions

   The change in temperature is the same in kelvins:

   $\Delta T$$= -27\,{}^\circ\text{C} - 18\,{}^\circ\text{C}$

   $= -45\,\text{K}$

   The result comes out in metres, and $1\,\text{m} = 1000\,\text{mm}$.

   Substitution

   $\Delta L$$= 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 50.0\,\text{m} \cdot (-45\,\text{K})$

   $= -0.05175\,\text{m}$

   $= -51.75\,\text{mm}$

   Answer

   $\Delta L \approx -52\,\text{mm}$

   Two significant figures. The limiting data are $\alpha = 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ and the difference of whole degrees $\Delta T = -45\,\text{K}$ (both have two figures); the length $50.0\,\text{m}$ has three figures and does not limit the result.

   Reasonableness

   A rough estimate:

   $2 \cdot 10^{-5}\,\tfrac{1}{\text{K}} \cdot 50\,\text{m} \cdot 45\,\text{K}$

   $\approx 45\,\text{mm}$

   The sign is correct because the temperature falls. The wire gets shorter by about $0.10\,\%$, so it must be installed slack in summer so that it does not become too tight in winter.

3. Exercise 10-3

   In a physics lesson a copper tube ($\alpha = 17 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) is cooled with ice water to $0\,{}^\circ\text{C}$, and its length is measured to be $500.0\,\text{mm}$. One end of the tube is fixed in a stand, and a dial gauge is placed against the other end.

   Then steam from boiling water is passed through the tube, and the tube warms up to $100\,{}^\circ\text{C}$.

   By how much does the tube get longer, that is, by how much does the reading of the dial gauge change? Give the answer in millimetres.

   mmCheck
   Answer

   $\Delta L = 0.85\,\text{mm}$

   Show the solution

   Given

   copper tube, $L_0 = 500.0\,\text{mm}$ at the temperature $T_0 = 0\,{}^\circ\text{C}$, $T = 100\,{}^\circ\text{C}$, $\alpha = 17 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the change in length of the tube $\Delta L$ in millimetres, that is, the change in the reading of the dial gauge

   Principle

   One end of the tube is fixed, so the whole change in length shows at the free end and in the dial gauge. The copper warms up and its state of matter does not change, so the change in length is proportional to the initial length and the change in temperature.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, where $\Delta T = T - T_0$. The new length is $L = L_0 + \Delta L$.

   Conversions

   $\Delta T$$= 100\,{}^\circ\text{C} - 0\,{}^\circ\text{C}$

   $= 100\,\text{K}$

   the same in kelvins. No unit conversion is needed: $L_0$ is in millimetres, so $\Delta L$ also comes out in millimetres.

   Substitution

   $\Delta L$$= 17 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 500.0\,\text{mm} \cdot 100\,\text{K}$

   $= 0.85\,\text{mm}$

   and

   $L$$= 500.0\,\text{mm} + 0.85\,\text{mm}$

   $= 500.85\,\text{mm}$

   Answer

   $\Delta L = 0.85\,\text{mm}$

   Two significant figures. The limiting datum is $\alpha = 17 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (two figures); the length $500.0\,\text{mm}$ and the difference of whole degrees $\Delta T = 100\,\text{K}$ are more precise.

   Reasonableness

   The relative change is

   $\alpha \Delta T$$= 17 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 100\,\text{K}$

   $= 0.17\,\%$

   and $0.17\,\%$ of $500\,\text{mm}$ is $0.85\,\text{mm}$. The tube gets longer because it warms up, and a change of less than a millimetre shows on a dial gauge but not on a ruler.

4. Exercise 10-4

   On the roof of a summer cottage sauna there is a black $200.0\,\textsf{l}$ tank from which shower water is drawn. In the morning the temperature of the water is $12\,{}^\circ\text{C}$, and in the afternoon the sun has heated it to $32\,{}^\circ\text{C}$. The coefficient of volume expansion of water is $\gamma = 0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$.

   By how many litres does the volume of the water grow?

   lCheck
   Answer

   $\Delta V = 0.84\,\textsf{l}$

   Show the solution

   Given

   water, $V_0 = 200.0\,\textsf{l}$, $T_0 = 12\,{}^\circ\text{C}$, $T = 32\,{}^\circ\text{C}$, $\gamma = 0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the increase in the volume of the water $\Delta V$ in litres

   Principle

   A liquid expands in all directions, so the change in its volume is proportional to the initial volume and the change in temperature. Water stays liquid below $100\,{}^\circ\text{C}$, so the coefficient of volume expansion $\gamma$ describes the situation. The tank does not limit the expansion, because the question is only about the growth of the volume of the water.

   Formula

   $\Delta V = \gamma V_0 \Delta T$, where $\Delta T = T - T_0$.

   Conversions

   $\Delta T$$= 32\,{}^\circ\text{C} - 12\,{}^\circ\text{C}$

   $= 20\,\text{K}$

   the same in kelvins. The litres do not need converting, because $V_0$ is in litres and $\Delta V$ also comes out in litres.

   Substitution

   $\Delta V$$= 0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}} \cdot 200.0\,\textsf{l} \cdot 20\,\text{K}$

   $= 0.84\,\textsf{l}$

   Answer

   $\Delta V = 0.84\,\textsf{l}$

   Two significant figures. The limiting datum is $\gamma = 0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ (two figures); the volume $200.0\,\textsf{l}$ has four figures and the difference of whole degrees $\Delta T = 20\,\text{K}$ has two figures, so neither limits the result any further.

   Reasonableness

   The relative growth is

   $\gamma \Delta T$$= 0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}} \cdot 20\,\text{K}$

   $= 0.42\,\%$

   and $0.42\,\%$ of $200.0\,\textsf{l}$ is $0.84\,\textsf{l}$. The growth is positive because the water warms up. Almost a litre more would not fit in a tank filled to the brim and closed, so the tank has an overflow or an air space.

5. Exercise 10-5

   The living room window has a $1.5\,\text{m}$ wide pane of ordinary glass ($\alpha = 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$). On a summer day the sunshine heats the glass from $20\,{}^\circ\text{C}$ to $45\,{}^\circ\text{C}$.

   By how much does the width of the glass grow? Give the answer in millimetres.

   mmCheck
   Answer

   $\Delta L \approx 0.34\,\text{mm}$

   Show the solution

   Given

   ordinary glass, width $L_0 = 1.5\,\text{m}$, $T_0 = 20\,{}^\circ\text{C}$, $T = 45\,{}^\circ\text{C}$, $\alpha = 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the change in width $\Delta L$ in millimetres

   Principle

   Glass is a solid, and it expands in all directions in the same proportion when it warms up. The glass does not melt, so the change in width is proportional to the initial width and the change in temperature.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, where $\Delta T = T - T_0$.

   Conversions

   $\Delta T$$= 45\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 25\,\text{K}$

   the same in kelvins. The result comes out in metres, and $1\,\text{m} = 1000\,\text{mm}$.

   Substitution

   $\Delta L$$= 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 1.5\,\text{m} \cdot 25\,\text{K}$

   $= 0.0003375\,\text{m}$

   $= 0.3375\,\text{mm}$

   Answer

   $\Delta L \approx 0.34\,\text{mm}$

   Two significant figures. There are three limiting data: $\alpha = 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, the width $1.5\,\text{m}$ and the difference of whole degrees $\Delta T = 25\,\text{K}$ all have two figures.

   Reasonableness

   The relative change is $9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25\,\text{K} = 0.0225\,\%$, and $1500\,\text{mm} \cdot 0.000225 = 0.34\,\text{mm}$. The same situation with steel ($\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) would give about $0.45\,\text{mm}$, so glass expands less than metal, as its smaller $\alpha$ promises.

6. Exercise 10-6

   The steel lid of a glass jar that is stuck tight often comes loose more easily when hot water is run over the lid for a moment.

   What is the reason?

   Choose an option.

   • Hot water makes the metal lid shrink, because the particles of the metal move closer to each other when they warm up.
   • The mouth of the jar gets smaller, because the particles of the glass grow when they warm up and fill the gap.
   • The steel lid expands more than the glass when it warms up, so the lid becomes loose around the mouth of the jar.
   • Glass and steel expand by the same amount, but hot water makes the edge of the lid slippery.

   Answer
   I have done my solution.
   Answer

   The steel lid expands more than the glass when it warms up, so the lid becomes loose around the mouth of the jar.

   Show the solution

   Find

   why hot water helps to loosen a tight steel lid from a glass jar

   Principle

   When a solid warms up, the average distance between its particles grows, so the object expands, and the expansion is proportional to the temperature coefficient $\alpha$. The coefficient of steel, $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, is greater than that of glass, $9 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$. The hot water is run over the lid, so the lid warms up first and the most.

   Model answer

   The steel lid expands more than the glass when it warms up, so the lid becomes loose around the mouth of the jar and comes off more easily. The lid does not shrink, because when it warms up the distances between the particles grow. The particles do not grow either, and different materials expand by different amounts because their temperature coefficients are different.

7. Exercise 10-7

   Three samples of the same size, a piece of steel, a glass of water and a balloon full of air (at constant pressure), warm up from $20\,{}^\circ\text{C}$ to $30\,{}^\circ\text{C}$.

   Which sample expands the most in relative terms, and why?

   Choose an option.

   • Water, because the particles of a liquid can move freely from place to place.
   • All equally, because the change in temperature is the same for all the samples.
   • Air, because only weak interactions bind the particles of a gas to each other.
   • Steel, because the particles of a solid vibrate the most strongly when they warm up.

   Answer
   I have done my solution.
   Answer

   Air, because only weak interactions bind the particles of a gas to each other.

   Show the solution

   Find

   which of the three samples (steel, water, air) expands the most in relative terms when the temperature rises by $10\,\text{K}$, and why

   Principle

   The relative growth in volume is $\dfrac{\Delta V}{V_0} = \gamma \Delta T$, and $\Delta T$ is the same for all the samples, so the difference comes from the temperature coefficient $\gamma$. The size of the coefficient tells how tightly the particles are bound to each other: most tightly in a solid and most loosely in a gas.

   Model answer

   Air expands the most, because only weak interactions bind the particles of a gas to each other and the distances between the particles grow easily. According to the table, the $\gamma = 3.4 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ of air is about 16 times the $0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ of water and about 90 times the $\gamma \approx 3\alpha = 36 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ of steel. So the same change in temperature does not expand everything equally, and the solid expands the least.

8. Exercise 10-8

   A student calculated how much a $2.0\,\text{m}$ long aluminium rod ($\alpha = 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) gets longer when its temperature rises from $15\,{}^\circ\text{C}$ to $65\,{}^\circ\text{C}$. The student wrote $\Delta L = 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 2.0\,\text{m} \cdot 65 = 3.0\,\text{mm}$ (units left out).

   There is an error here. Calculate the change in the length of the rod correctly. Give the answer in millimetres.

   mmCheck
   Answer

   $\Delta L = 2.3\,\text{mm}$

   Show the solution

   Given

   aluminium rod, $L_0 = 2.0\,\text{m}$, $T_0 = 15\,{}^\circ\text{C}$, $T = 65\,{}^\circ\text{C}$, $\alpha = 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the change in the length of the rod $\Delta L$ in millimetres (the error in the student's calculation)

   Principle

   The change in length is proportional to the initial length and to the change in temperature, not to the final temperature. The student put the final temperature $65$ into the formula instead of the change in temperature.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, where $\Delta T = T - T_0$. The quantity asked for, $\Delta L$, is already alone on the left side.

   Conversions

   The change in temperature is the same in degrees Celsius and in kelvins:

   $\Delta T$$= 65\,{}^\circ\text{C} - 15\,{}^\circ\text{C}$

   $= 50\,\text{K}$

   The length is in metres, so $\Delta L$ comes out in metres; finally $1\,\text{m} = 1000\,\text{mm}$.

   Substitution

   $\Delta L$$= 23 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 2.0\,\text{m} \cdot 50\,\text{K}$

   $= 0.0023\,\text{m}$

   $= 2.3\,\text{mm}$

   Answer

   $\Delta L = 2.3\,\text{mm}$

   Two significant figures like the given data.

   Reasonableness

   The student's result $3.0\,\text{mm}$ came from the factor $65$ instead of $50$, so it is $\tfrac{65}{50} = 1.3$ times as large, that is $30\,\%$ too big. If the rod had been heated from $0\,{}^\circ\text{C}$ to $65\,{}^\circ\text{C}$, the student's number would have been correct. The relative change in the length of the rod, $\alpha \Delta T = 0.12\,\%$, is typical for a metal.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 10-9Core

   A flexible bellows section is installed at the end of a straight section of a district heating pipe, and it can take up a change in length of at most $50.0\,\text{mm}$. The steel pipe ($\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) is installed at $10\,{}^\circ\text{C}$, and in use its temperature is $90\,{}^\circ\text{C}$.

   What is the greatest length the straight section of the pipe may have? Give the answer in metres.

   mCheck
   Answer

   $L_0 \approx 52\,\text{m}$

   Show the solution

   Given

   steel pipe, maximum allowance of the bellows section $\Delta L = 50.0\,\text{mm}$, $T_0 = 10\,{}^\circ\text{C}$, $T = 90\,{}^\circ\text{C}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the greatest length $L_0$ of the straight section in metres

   Principle

   The change in the length of the pipe is proportional to its length and the change in temperature, and the bellows section takes up at most $50.0\,\text{mm}$. Steel stays solid, so the same expansion formula applies. The formula is rearranged for the initial length, because the change is known and the length is asked for.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, from which

   $L_0$$= \dfrac{\Delta L}{\alpha \Delta T}$

   Conversions

   $\Delta L$$= 50.0\,\text{mm}$

   $= 0.0500\,\text{m}$

   so that the result comes out in metres.

   $\Delta T$$= 90\,{}^\circ\text{C} - 10\,{}^\circ\text{C}$

   $= 80\,\text{K}$

   the same in kelvins.

   Substitution

   $L_0$$= \dfrac{0.0500\,\text{m}}{12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 80\,\text{K}}$

   $= \dfrac{0.0500\,\text{m}}{9.6 \cdot 10^{-4}}$

   $= 52.083\,\text{m}$

   Answer

   $L_0 \approx 52\,\text{m}$

   Two significant figures. The limiting data are $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ and the difference of whole degrees $\Delta T = 80\,\text{K}$ (both have two figures); the allowed change $50.0\,\text{mm}$ has three figures and does not limit the result.

   Reasonableness

   Check forwards:

   $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 52\,\text{m} \cdot 80\,\text{K}$

   $= 0.0499\,\text{m}$

   $\approx 50\,\text{mm}$

   A larger allowed change or a smaller temperature difference would allow a longer pipe, so the result goes in the right direction. If the millimetres were not converted to metres, the result would be a thousand times too large.

2. Exercise 10-10Core

   A steel kettlebell with a volume of $2.0\,\text{dm}^3$ has been in a garage at $-22\,{}^\circ\text{C}$ on a frosty night. It is brought inside, where it warms to $20\,{}^\circ\text{C}$. The coefficient of linear expansion of steel is $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$.

   By how many cubic centimetres does the volume of the kettlebell grow? Hint: for a solid $\gamma \approx 3\alpha$.

   cm³Check
   Answer

   $\Delta V \approx 3.0\,\text{cm}^3$

   Show the solution

   Given

   steel kettlebell, $V_0 = 2.0\,\text{dm}^3$, $T_0 = -22\,{}^\circ\text{C}$, $T = 20\,{}^\circ\text{C}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the increase in volume $\Delta V$ in cubic centimetres

   Principle

   The kettlebell expands in all three directions in the same proportion, so its volume grows in relative terms about three times as much as its length: $\gamma \approx 3\alpha$. This holds for a solid whose state of matter does not change. The change in volume is proportional to the initial volume and the change in temperature.

   Formula

   $\gamma \approx 3\alpha$ and $\Delta V = \gamma V_0 \Delta T$, that is, $\Delta V = 3\alpha V_0 \Delta T$.

   Conversions

   $V_0$$= 2.0\,\text{dm}^3$

   $= 2000\,\text{cm}^3$

   because $1\,\text{dm}^3 = 1000\,\text{cm}^3$.

   $\Delta T$$= 20\,{}^\circ\text{C} - (-22\,{}^\circ\text{C})$

   $= 42\,\text{K}$

   the same in kelvins.

   Substitution

   $\gamma$$\approx 3 \cdot 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   $= 36 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   and

   $\Delta V$$= 36 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 2000\,\text{cm}^3 \cdot 42\,\text{K}$

   $= 3.024\,\text{cm}^3$

   Answer

   $\Delta V \approx 3.0\,\text{cm}^3$

   Two significant figures. There are three limiting data: the volume $2.0\,\text{dm}^3$, the coefficient $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ and the difference of whole degrees $\Delta T = 42\,\text{K}$ all have two figures. The factor $3$ ($\gamma \approx 3\alpha$) is an exact number and does not limit the result.

   Reasonableness

   The relative growth is $36 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 42\,\text{K} = 0.15\,\%$, and $0.15\,\%$ of $2000\,\text{cm}^3$ is $3.0\,\text{cm}^3$. The volume grows because the kettlebell warms up. The change is so small that you cannot notice it by hand.

3. Exercise 10-11

   On an old-style railway, the steel rails ($\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) are $25.0\,\text{m}$ long. When the track is built at $12\,{}^\circ\text{C}$, a gap of $7.2\,\text{mm}$ is left between the rails. Assume that the gap closes when a rail has lengthened by the width of the gap.

   At what temperature does the gap close?

   °CCheck
   Answer

   $T = 36\,{}^\circ\text{C}$

   Show the solution

   Given

   steel rail, $L_0 = 25.0\,\text{m}$, $T_0 = 12\,{}^\circ\text{C}$, gap $\Delta L = 7.2\,\text{mm}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the temperature $T$ at which the gap closes

   Principle

   The rail gets longer when it warms up, in proportion to its length and its change in temperature. According to the assumption in the problem, the gap closes when the rail has lengthened by the width of the gap, so $\Delta L$ is set equal to the gap and the required change in temperature is solved.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, from which

   $\Delta T$$= \dfrac{\Delta L}{\alpha L_0}$

   and finally $T = T_0 + \Delta T$.

   Conversions

   $\Delta L$$= 7.2\,\text{mm}$

   $= 0.0072\,\text{m}$

   so that the lengths are in the same unit. The change in temperature $\Delta T$ comes out in kelvins, and it is the same in degrees Celsius, so it is added directly to the temperature $T_0$.

   Substitution

   $\Delta T$$= \dfrac{0.0072\,\text{m}}{12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25.0\,\text{m}}$

   $= \dfrac{0.0072\,\text{m}}{3.0 \cdot 10^{-4}\,\text{m/K}}$

   $= 24\,\text{K}$

   and

   $T$$= 12\,{}^\circ\text{C} + 24\,\text{K}$

   $= 36\,{}^\circ\text{C}$

   Answer

   $T = 36\,{}^\circ\text{C}$

   Two significant figures. The figures of the change in temperature $\Delta T = 24\,\text{K}$ are limited by the gap $7.2\,\text{mm}$ and the coefficient $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (both have two figures; the length $25.0\,\text{m}$ has three). The sum $12\,{}^\circ\text{C} + 24\,\text{K}$ is in whole degrees, so $36\,{}^\circ\text{C}$ keeps those two figures.

   Reasonableness

   Check:

   $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25.0\,\text{m} \cdot 24\,\text{K}$

   $= 7.2\,\text{mm}$

   so the rail fills the gap. The temperature is above the initial temperature, because the rail gets longer only when it warms up. The neighbouring rail also gets longer, but each rail lengthens in both directions: each edge of the gap moves $\tfrac{1}{2}\Delta L$, so in total the lengthening of one rail, $\Delta L$, is directed towards the gap, and the assumption of the problem holds. A dark rail easily heats above $36\,{}^\circ\text{C}$ in a heatwave.

4. Exercise 10-12

   A metal beam used in a bridge structure, with a length of $5.00\,\text{m}$ at $20\,{}^\circ\text{C}$, is heated and the change in its length is measured. The graph shows the change in length $\Delta L$ as a function of temperature.

   Calculate the coefficient of linear expansion $\alpha$ of the material of the beam in the unit $10^{-6}\,\tfrac{1}{\text{K}}$ and use the table in the lesson to work out which metal the beam is made of (steel 12, aluminium 23, copper 17, all in the unit $10^{-6}\,\tfrac{1}{\text{K}}$). Read the values you need from the graph.

   10⁻⁶ 1/KCheck
   Answer

   $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   Show the solution

   Given

   metal beam, $L_0 = 5.00\,\text{m}$ at $20\,{}^\circ\text{C}$; change in length as a function of temperature in the graph; from the table $\alpha_\text{steel} = 12$, $\alpha_\text{aluminium} = 23$, $\alpha_\text{copper} = 17$ (in the unit $10^{-6}\,\tfrac{1}{\text{K}}$)

   Find

   the coefficient of linear expansion $\alpha$ in the unit $10^{-6}\,\tfrac{1}{\text{K}}$ and, based on it, the metal

   Principle

   The beam gets longer when it warms up, in proportion to its length and its change in temperature, and the coefficient $\alpha$ is a constant characteristic of the material. The points of the graph give the change in temperature and the change in length, and from them $\alpha$, which can be compared with the table.

   Reading the graph

   The line starts at the grid crossing $(20\,{}^\circ\text{C}$; $0\,\text{mm})$ and ends at the crossing $(70\,{}^\circ\text{C}$; $3.0\,\text{mm})$. So the change in temperature is

   $\Delta T$$= 70\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 50\,\text{K}$

   and the change in length is $\Delta L = 3.0\,\text{mm}$.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, from which

   $\alpha$$= \dfrac{\Delta L}{L_0 \Delta T}$

   Conversions

   $\Delta L$$= 3.0\,\text{mm}$

   $= 0.0030\,\text{m}$

   so that the units cancel. The change in temperature is the same in degrees Celsius and in kelvins. The result is given in the unit $10^{-6}\,\tfrac{1}{\text{K}}$, so the factor $10^{-6}$ is left in the unit.

   Substitution

   $\alpha$$= \dfrac{0.0030\,\text{m}}{5.00\,\text{m} \cdot 50\,\text{K}}$

   $= \dfrac{0.0030}{250}\,\tfrac{1}{\text{K}}$

   $= 0.000012\,\tfrac{1}{\text{K}}$

   $= 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   Answer

   $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   So the beam is steel. Two significant figures, because the change in length $3.0\,\text{mm}$ was read with two figures. The values are read exactly from the grid, so $11.6\ldots12.4$ is accepted.

   Reasonableness

   Units:

   $\dfrac{\text{m}}{\text{m} \cdot \text{K}}$$= \tfrac{1}{\text{K}}$

   Comparison with the table: an aluminium beam would lengthen

   $23 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 5.00\,\text{m} \cdot 50\,\text{K}$

   $= 5.75\,\text{mm}$

   and a copper beam $4.25\,\text{mm}$, so only steel gives the $3.0\,\text{mm}$ that was read. A fraction of a millimetre per metre of length is typical of metals.

5. Exercise 10-13

   Example data: the table gives the volume of $1.000\,\text{kg}$ of water at different temperatures. The volume calculated from a density table is given to six significant figures, because the changes are small.

   a) At which temperature in the table does water take up the least space, and what does that tell you about the density of water at that temperature?

   b) Calculate the temperature coefficient of the volume of water $\gamma$ over the temperature range $4\,{}^\circ\text{C} \ldots 10\,{}^\circ\text{C}$ in the unit $10^{-6}\,\tfrac{1}{\text{K}}$. Use the volume at $4\,{}^\circ\text{C}$ as the initial volume.

   c) Compare the result with the value $\gamma = 210 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ valid at $20\,{}^\circ\text{C}$ and explain why water differs from other liquids over the temperature range $0\,{}^\circ\text{C} \ldots 4\,{}^\circ\text{C}$ and why the temperature of the water at the bottom of a deep lake in winter is about $4\,{}^\circ\text{C}$.

   The answer field takes the result of part b).

   Volume of $1.000\,\text{kg}$ of water (example data)

   $T\,({}^\circ\text{C})$	$V\,(\text{dm}^3)$
   0	1.00016
   2	1.00006
   4	1.00003
   6	1.00006
   8	1.00015
   10	1.00030

   10⁻⁶ 1/KCheck
   Answer

   $\gamma \approx 45 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   Show the solution

   Given

   $1.000\,\text{kg}$ of water: the volume at $4\,{}^\circ\text{C}$ is $V_4 = 1.00003\,\text{dm}^3$ and at $10\,{}^\circ\text{C}$ $V_{10} = 1.00030\,\text{dm}^3$; the smallest volume in the table is $1.00003\,\text{dm}^3$

   Find

   a) the temperature at which water takes up the least space and what that tells about the density; b) $\gamma$ over $4\,{}^\circ\text{C} \ldots 10\,{}^\circ\text{C}$ in the unit $10^{-6}\,\tfrac{1}{\text{K}}$; c) the comparison with the value $210 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ and the unusual behaviour of water

   Principle

   The mass of a substance is constant, so the smallest volume means the greatest density $\rho = m/V$. The temperature coefficient is obtained from the change in volume: when $\Delta V = \gamma V_0 \Delta T$ holds, then $\gamma = \dfrac{\Delta V}{V_0 \Delta T}$.

   Reading the table

   The volume in the table decreases from $1.00016\,\text{dm}^3$ at $0\,{}^\circ\text{C}$ to $1.00003\,\text{dm}^3$ at $4\,{}^\circ\text{C}$ and then grows to $1.00030\,\text{dm}^3$ at $10\,{}^\circ\text{C}$. So the smallest volume is at $4\,{}^\circ\text{C}$.

   Formula

   $\Delta V = \gamma V_0 \Delta T$, so

   $\gamma$$= \dfrac{\Delta V}{V_0 \Delta T}$

   $\Delta V = V_{10} - V_4$

   Conversions

   The volumes are in the same unit, so $\Delta V$ and $V_0$ cancel in the same unit. The change in temperature is

   $\Delta T$$= 10\,{}^\circ\text{C} - 4\,{}^\circ\text{C}$

   $= 6\,\text{K}$

   the same in kelvins. The result is given in the unit $10^{-6}\,\tfrac{1}{\text{K}}$.

   Substitution

   $\Delta V$$= 1.00030\,\text{dm}^3 - 1.00003\,\text{dm}^3$

   $= 0.00027\,\text{dm}^3$

   and

   $\gamma$$= \dfrac{0.00027\,\text{dm}^3}{1.00003\,\text{dm}^3 \cdot 6\,\text{K}}$

   $= 0.000045\,\tfrac{1}{\text{K}}$

   $= 45 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   Answer

   b) $\gamma \approx 45 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, two significant figures, because the change in volume $0.00027\,\text{dm}^3$ is known to two figures. Part a): the smallest volume is at $4\,{}^\circ\text{C}$, so the density is greatest there. Model answer c): the $\gamma$ of water is not constant; it grows with temperature and is close to zero near $4\,{}^\circ\text{C}$. So the table value $210 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ cannot be used over this range. Unlike most liquids, water shrinks when it warms up over $0\,{}^\circ\text{C} \ldots 4\,{}^\circ\text{C}$. In winter the surface water of a lake cools, becomes denser and sinks until all the water is at about $4\,{}^\circ\text{C}$. After that the colder water is lighter and stays at the surface, so the water at the bottom stays at about $4\,{}^\circ\text{C}$ and the lake freezes from the surface.

   Reasonableness

   The coefficient $45 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ is clearly smaller than the comparison value of part c), about a fifth of it, because $\gamma$ grows with temperature. Over $0\,{}^\circ\text{C} \ldots 4\,{}^\circ\text{C}$ the volume decreases, so the coefficient there is negative.

6. Exercise 10-14

   Example data: a group used a dial gauge to measure the change in length $\Delta L$ of a $0.800\,\text{m}$ long metal rod when the rod was heated from $20\,{}^\circ\text{C}$. The reading accuracy of the dial gauge is $0.01\,\text{mm}$.

   a) Calculate the temperature coefficient of the length $\alpha$ of the material of the rod in the unit $10^{-6}\,\tfrac{1}{\text{K}}$ using the first and the last point of the table.

   b) Work out from the table in the lesson (steel 12, aluminium 23, copper 17, all in the unit $10^{-6}\,\tfrac{1}{\text{K}}$) which metal the rod is made of, and explain why the measurement points do not usually lie exactly on the same straight line.

   The answer field takes the result of part a).

   Change in the length of a metal rod (example data)

   $T\,({}^\circ\text{C})$	$\Delta L\,(\text{mm})$
   20	0.00
   30	0.19
   40	0.38
   50	0.55
   60	0.75
   70	0.92
   80	1.10

   10⁻⁶ 1/KCheck
   Answer

   $\alpha \approx 22.9 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   Show the solution

   Given

   metal rod, $L_0 = 0.800\,\text{m}$ at $20\,{}^\circ\text{C}$; first point $(20\,{}^\circ\text{C}$; $0.00\,\text{mm})$ and last point $(80\,{}^\circ\text{C}$; $1.10\,\text{mm})$; from the table $\alpha_\text{steel} = 12$, $\alpha_\text{aluminium} = 23$, $\alpha_\text{copper} = 17$ (in the unit $10^{-6}\,\tfrac{1}{\text{K}}$)

   Find

   a) $\alpha$ in the unit $10^{-6}\,\tfrac{1}{\text{K}}$; b) the material of the rod and why the points are not exactly on a straight line

   Principle

   The rod gets longer in proportion to the change in temperature and the initial length, and the coefficient $\alpha$ is characteristic of the material. A straight line through the points tells the change per kelvin, and the same property gives the coefficient.

   Reading the table

   From the first point to the last, the temperature changes by $\Delta T = 80\,{}^\circ\text{C} - 20\,{}^\circ\text{C} = 60\,\text{K}$ and the length by $\Delta L = 1.10\,\text{mm}$.

   Formula

   $\Delta L = \alpha L_0 \Delta T$, so

   $\alpha$$= \dfrac{\Delta L}{L_0 \Delta T}$

   Conversions

   $\Delta L$$= 1.10\,\text{mm}$

   $= 0.00110\,\text{m}$

   so that the units cancel. The result is given in the unit $10^{-6}\,\tfrac{1}{\text{K}}$.

   Substitution

   $\alpha$$= \dfrac{0.00110\,\text{m}}{0.800\,\text{m} \cdot 60\,\text{K}}$

   $= \dfrac{0.00110}{48}\,\tfrac{1}{\text{K}}$

   $= 22.9 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$

   Answer

   a) $\alpha \approx 22.9 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, three significant figures, because the change in length $1.10\,\text{mm}$ is given to three figures. Model answer b): the coefficient is closest to the value for aluminium, $23 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, so the rod is aluminium. The value for steel, $12$, and for copper, $17$, are clearly further away. The measurement points usually do not lie exactly on the same straight line because of measurement uncertainty: every reading has an uncertainty, and the rod does not warm up completely evenly. So the variation of the steps $0.17\ldots 0.20\,\text{mm}$ is to be expected, and a straight line fitted through all the points gives the same slope as the end points, $\alpha \approx 22.9 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$.

   Reasonableness

   The expected change in the length of an aluminium rod over the whole range is

   $23 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 0.800\,\text{m} \cdot 60\,\text{K}$

   $= 1.10\,\text{mm}$

   which matches the measured value. For steel it would be $0.58\,\text{mm}$ and for copper $0.82\,\text{mm}$, which differ from the measured value clearly more than the accuracy of the dial gauge, $0.01\,\text{mm}$.

7. Exercise 10-15

   A group is planning an experiment in which the temperature coefficient of the length of steel ($\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) is measured from a steel tube. At $20\,{}^\circ\text{C}$ the tube is $1.00\,\text{m}$ long, and it is heated with steam to $100\,{}^\circ\text{C}$. The group intends to measure the length with a tape measure whose reading accuracy is $0.5\,\text{mm}$.

   a) Calculate how much the tube gets longer.

   b) Explain why the tape measure is not suitable for this measurement, and suggest an instrument with which a small change in length can be measured reliably.

   c) How long should the steel tube be so that the reading accuracy of the tape measure is at most $5.0\,\%$ of the change in length, when the change in temperature is the same? Give the answer in metres.

   The answer field takes the result of part c).

   mCheck
   Answer

   $L_0 \approx 10.4\,\text{m}$

   Show the solution

   Given

   steel tube, $L_0 = 1.00\,\text{m}$, $T_0 = 20\,{}^\circ\text{C}$, $T = 100\,{}^\circ\text{C}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem); reading accuracy of the tape measure $0.5\,\text{mm}$

   Find

   a) the change in length $\Delta L$; b) why the tape measure is not suitable and which instrument is; c) the length of the tube $L_0$ when the reading accuracy is at most $5.0\,\%$ of the change

   Principle

   The change in length is proportional to the initial length and the change in temperature, $\Delta L = \alpha L_0 \Delta T$. A measurement is reliable only if the change to be measured is much greater than the reading accuracy of the instrument. In part c) the condition is written for the change and the length is solved from it.

   Formula

   a) $\Delta L = \alpha L_0 \Delta T$. c) The condition $0.5\,\text{mm} \le 0.050 \cdot \Delta L$ gives $\Delta L \ge 10\,\text{mm}$, and $\Delta L = \alpha L_0 \Delta T$ gives

   $L_0$$= \dfrac{\Delta L}{\alpha \Delta T}$

   Conversions

   $\Delta T$$= 100\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 80\,\text{K}$

   the same in kelvins. The change in length $10\,\text{mm} = 0.010\,\text{m}$, so that $L_0$ comes out in metres.

   Substitution

   a)

   $\Delta L$$= 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 1.00\,\text{m} \cdot 80\,\text{K}$

   $= 0.00096\,\text{m}$

   $= 0.96\,\text{mm}$

   c) The required change is

   $\Delta L$$= \dfrac{0.5\,\text{mm}}{0.050}$

   $= 10\,\text{mm}$

   $= 0.010\,\text{m}$

   so

   $L_0$$= \dfrac{0.010\,\text{m}}{12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 80\,\text{K}}$

   $= 10.4\,\text{m}$

   Answer

   c) $L_0 \approx 10.4\,\text{m}$, three significant figures as calculated; the reading accuracy $0.5\,\text{mm}$ is a one-figure value, so in practice the length is about $10\,\text{m}$. Model answer b): the change in part a), $0.96\,\text{mm}$, is only about twice the reading accuracy of the tape measure, $0.5\,\text{mm}$, so the measurement uncertainty could be over half of the result ($52\,\%$) and the coefficient $\alpha$ could not be obtained reliably. A small change is measured with an instrument that reads to a hundredth of a millimetre, such as a dial gauge whose probe is against the end of the tube. A longer tube or a greater change in temperature would also reduce the relative uncertainty.

   Reasonableness

   A check:

   $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 10.4\,\text{m} \cdot 80\,\text{K}$

   $= 10\,\text{mm}$

   and $5.0\,\%$ of that is $0.5\,\text{mm}$. A ten-metre steam pipe is impractical in a school laboratory, so a dial gauge is a more sensible solution than a longer tube. The reading accuracy of a dial gauge, $0.01\,\text{mm}$, would be only $1\,\%$ of the change $0.96\,\text{mm}$.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 10-16

   In a farm machine workshop a steel gear is fitted on the axle of a tractor with a shrink fit. The diameter of the central hole of the gear is $75.000\,\text{mm}$ at $20\,{}^\circ\text{C}$ and the diameter of the axle is $75.090\,\text{mm}$. For steel $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$.

   To what temperature must the gear be heated at least so that it fits on the axle? Also justify why the hole gets bigger and not smaller when the gear warms up, and why the gear is held tightly in place after it has cooled.

   °CCheck
   Answer

   $T = 120\,{}^\circ\text{C}$

   Show the solution

   Given

   steel gear, diameter of the hole $d_0 = 75.000\,\text{mm}$ at the temperature $T_0 = 20\,{}^\circ\text{C}$, diameter of the axle $d = 75.090\,\text{mm}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the lowest temperature $T$ at which the hole is at least as big as the axle

   Principle

   An object expands in all directions in the same proportion, as when a photograph is enlarged, and so does a hole. The diameter of the hole is a dimension of the object itself, so its change follows the same formula $\Delta d = \alpha d_0 \Delta T$ as a change in length. So the hole does not get smaller even though the material around it expands. When it cools, the hole tries to return to the size $75.000\,\text{mm}$, but the axle is in the way, and the gear is pressed tightly in place.

   Formula

   $\Delta d = \alpha d_0 \Delta T$, from which

   $\Delta T$$= \dfrac{\Delta d}{\alpha d_0}$

   and $T = T_0 + \Delta T$. Here $\Delta d = d - d_0$.

   Conversions

   All the dimensions are in millimetres, so $\Delta d$ and $d_0$ cancel in the same unit and no conversion is needed. $\Delta T$ comes out in kelvins and is the same as in degrees Celsius.

   Substitution

   $\Delta d$$= 75.090\,\text{mm} - 75.000\,\text{mm}$

   $= 0.090\,\text{mm}$

   and

   $\Delta T$$= \dfrac{0.090\,\text{mm}}{12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 75.000\,\text{mm}}$

   $= \dfrac{0.090}{9.0 \cdot 10^{-4}}\,\text{K}$

   $= 100\,\text{K}$

   So

   $T$$= 20\,{}^\circ\text{C} + 100\,\text{K}$

   $= 120\,{}^\circ\text{C}$

   Answer

   $T = 120\,{}^\circ\text{C}$

   Two significant figures. The difference of the diameters $\Delta d = 0.090\,\text{mm}$ has two figures, because the dimensions are given to the nearest thousandth of a millimetre ($75.000$ and $75.090$), and the coefficient $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ also has two figures; they limit the result $\Delta T = 100\,\text{K}$ and thus the temperature $120\,{}^\circ\text{C}$.

   Reasonableness

   Check:

   $75.000\,\text{mm} \cdot (1 + 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 100\,\text{K})$

   $= 75.090\,\text{mm}$

   The temperature is clearly higher than room temperature but below the melting of steel. In practice the gear is heated a little above $120\,{}^\circ\text{C}$ so that a clearance is left between the hole and the axle.

2. Exercise 10-17

   The thermostat of an old-style electric sauna heater has a bimetallic strip in which copper and steel strips, each $50.0\,\text{mm}$ long, are joined together (copper $\alpha = 17.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, steel $\alpha = 12.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$).

   As the sauna warms up, the temperature of the strip rises from $20\,{}^\circ\text{C}$ to $100\,{}^\circ\text{C}$.

   How much more would the copper strip lengthen than the steel strip if they were separate? Give the answer in micrometres, that is, thousandths of a millimetre. Justify which way the joined strip bends.

   µmCheck
   Answer

   $20\,\mu\text{m}$

   Show the solution

   Given

   copper and steel strips, $L_0 = 50.0\,\text{mm}$, $T_0 = 20\,{}^\circ\text{C}$, $T = 100\,{}^\circ\text{C}$, $\alpha_\text{Cu} = 17.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$, $\alpha_\text{Fe} = 12.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the difference of the lengthenings $\Delta L_\text{Cu} - \Delta L_\text{Fe}$ in micrometres and the direction in which the strip bends

   Principle

   Both strips are solids and lengthen according to the formula $\Delta L = \alpha L_0 \Delta T$. The strips have the same initial length and the same change in temperature, so the difference in length comes only from the difference of the coefficients. Copper lengthens more because its $\alpha$ is greater.

   Formula

   $\Delta L_\text{Cu} - \Delta L_\text{Fe}$$= (\alpha_\text{Cu} - \alpha_\text{Fe}) L_0 \Delta T$

   where $\Delta T = T - T_0$.

   Conversions

   $\Delta T$$= 100\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 80\,\text{K}$

   the same in kelvins. $L_0$ is in millimetres, so the difference comes out in millimetres, and $1\,\text{mm} = 1000\,\mu\text{m}$.

   Substitution

   $\Delta L_\text{Cu}$$= 17.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 50.0\,\text{mm} \cdot 80\,\text{K}$

   $= 0.068\,\text{mm}$

   $\Delta L_\text{Fe}$$= 12.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 50.0\,\text{mm} \cdot 80\,\text{K}$

   $= 0.048\,\text{mm}$

   The difference is

   $0.068\,\text{mm} - 0.048\,\text{mm}$

   $= 0.020\,\text{mm}$

   $= 20\,\mu\text{m}$

   Answer

   $20\,\mu\text{m}$

   Two significant figures. The limiting datum is the difference of the coefficients $17.0 - 12.0 = 5.0$ (in the unit $10^{-6}\,\tfrac{1}{\text{K}}$), which has two figures because the coefficients are given to the nearest tenth; $\Delta T = 80\,\text{K}$ also has two figures. If the coefficients were given only as whole numbers, the difference $17 - 12 = 5$ would have one figure.

   Reasonableness

   The same result with a single multiplication:

   $(17.0 - 12.0) \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 50.0\,\text{mm} \cdot 80\,\text{K}$

   $= 0.020\,\text{mm}$

   The difference is positive because copper lengthens more. The strips are fastened to each other, so the longer copper stays on the outer edge of the curve and the strip bends towards the steel side. The end of the strip moves much more than $20\,\mu\text{m}$ and opens the electric circuit.

3. Exercise 10-18

   A $20.0\,\textsf{l}$ steel can for a motorboat is filled to the brim at a filling station with petrol at $10\,{}^\circ\text{C}$ and closed.

   On the jetty the sun heats the can and the petrol to $35\,{}^\circ\text{C}$. For petrol $\gamma = 0.95 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ and for steel $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$. The inner volume of the can grows in the same way as the volume of a steel object of the same size.

   How many cubic centimetres of petrol would try to overflow if the cap is opened? Justify why a can is not filled to the brim even though the can also expands.

   cm³Check
   Answer

   $\Delta V_\text{over} \approx 460\,\text{cm}^3$

   Show the solution

   Given

   petrol $V_0 = 20.0\,\textsf{l}$, $T_0 = 10\,{}^\circ\text{C}$, $T = 35\,{}^\circ\text{C}$, $\gamma_\text{petrol} = 0.95 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$; steel can $\alpha_\text{steel} = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem)

   Find

   the volume of petrol that overflows $\Delta V_\text{over}$ in cubic centimetres and a justification for why the can is not filled to the brim

   Principle

   Both the petrol and the inner space of the can expand, and the inner volume of the can grows like that of a steel object of the same size ($\gamma \approx 3\alpha$). The can was full, so what overflows is the difference between the growth of the petrol and the growth of the inner volume. Both volumes are calculated with the same formula.

   Formula

   $\Delta V = \gamma V_0 \Delta T$ for both, $\gamma_\text{steel} \approx 3\alpha_\text{steel}$ and $\Delta V_\text{over} = \Delta V_\text{petrol} - \Delta V_\text{can}$.

   Conversions

   $V_0$$= 20.0\,\textsf{l}$

   $= 20\,000\,\text{cm}^3$

   because $1\,\textsf{l} = 1000\,\text{cm}^3$.

   $\Delta T$$= 35\,{}^\circ\text{C} - 10\,{}^\circ\text{C}$

   $= 25\,\text{K}$

   the same in kelvins.

   Substitution

   $\Delta V_\text{petrol}$$= 0.95 \cdot 10^{-3}\,\tfrac{1}{\text{K}} \cdot 20\,000\,\text{cm}^3 \cdot 25\,\text{K}$

   $= 475\,\text{cm}^3$

   $\Delta V_\text{can}$$= 3 \cdot 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 20\,000\,\text{cm}^3 \cdot 25\,\text{K}$

   $= 18\,\text{cm}^3$

   So

   $\Delta V_\text{over}$$= 475\,\text{cm}^3 - 18\,\text{cm}^3$

   $= 457\,\text{cm}^3$

   Answer

   $\Delta V_\text{over} \approx 460\,\text{cm}^3$

   Two significant figures. The growth of the petrol $475\,\text{cm}^3$ is limited by the coefficient $\gamma = 0.95 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ (two figures), so it is known only to the nearest tens of cubic centimetres. In the difference $475 - 18$ the less precise number determines the decimals, so the result is rounded to tens: $460\,\text{cm}^3$. The $\alpha$ of steel and the volume $20.0\,\textsf{l}$ affect only the small part $18\,\text{cm}^3$.

   Reasonableness

   The growth of the petrol is $0.95 \cdot 10^{-3}\,\tfrac{1}{\text{K}} \cdot 25\,\text{K} = 2.4\,\%$ and that of the can only $0.09\,\%$, so the difference is almost the whole growth of the petrol, about $0.46\,\textsf{l}$. The expansion of the can is not enough to compensate, because the $\gamma$ of petrol is more than 25 times the $\gamma$ of steel. That is why an air space is left in the can.

4. Exercise 10-19

   Example data: steel rails ($\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$) that are $25.0\,\text{m}$ long are to be installed on a track. The table gives the temperature of the rail at the time of installation and the lowest and highest temperatures of the year (the sun heats the rail hotter than the air). Assume, as is usual with railway rails, that the change in the width of the gap is equal to the change in the length of one rail. The gap must not close in summer, and it must not exceed $30\,\text{mm}$ in winter.

   a) Calculate the smallest installation gap at which the gap does not close.

   b) Calculate the greatest installation gap at which the gap does not exceed $30\,\text{mm}$ in winter.

   c) Justify with a calculation why an installation gap of $15\,\text{mm}$ is acceptable when installing at the temperature in the table but not when installing in a winter frost.

   The answer field takes the result of part a) in millimetres.

   Temperatures of the rail (example data)

   Situation	$T\,({}^\circ\text{C})$
   winter minimum	−32
   installation	8
   summer maximum	52

   mmCheck
   Answer

   $g_\text{min} = 13.2\,\text{mm}$

   Show the solution

   Given

   steel rails, $L_0 = 25.0\,\text{m}$, $\alpha = 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ (given in the problem); temperature of the rail: winter minimum $T_\text{min} = -32\,{}^\circ\text{C}$, installation $T_\text{inst} = 8\,{}^\circ\text{C}$, summer maximum $T_\text{max} = 52\,{}^\circ\text{C}$ (example data); the gap must not close, and it must not exceed $30\,\text{mm}$ in winter

   Find

   a) the smallest installation gap $g_\text{min}$; b) the greatest installation gap $g_\text{max}$; c) why a gap of $15\,\text{mm}$ is acceptable at the installation temperature but not in a winter frost

   Principle

   The change in the width of the gap is equal to the change in the length of one rail, $\Delta L = \alpha L_0 \Delta T$. In summer the rail gets longer and the gap narrows, so the installation gap must be at least this change. In winter the rail gets shorter and the gap widens, so the installation gap plus the change in winter must be at most $30\,\text{mm}$.

   Reading the table

   The temperatures are read from the table. The change in temperature from installation to summer is $\Delta T_\text{summer} = 52\,{}^\circ\text{C} - 8\,{}^\circ\text{C} = 44\,\text{K}$. The change from installation to winter is $\Delta T_\text{winter} = 8\,{}^\circ\text{C} - (-32\,{}^\circ\text{C}) = 40\,\text{K}$.

   Formula

   $g_\text{min} = \alpha L_0 \Delta T_\text{summer}$ and $g_\text{max} = g_\text{limit} - \alpha L_0 \Delta T_\text{winter}$, where $g_\text{limit}$ is the greatest gap allowed in winter

   Conversions

   The change in temperature is the same in degrees Celsius and in kelvins. The length is in metres, so the changes come out in metres; finally $1\,\text{m} = 1000\,\text{mm}$.

   Substitution

   a)

   $g_\text{min}$$= 12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25.0\,\text{m} \cdot 44\,\text{K}$

   $= 0.0132\,\text{m}$

   $= 13.2\,\text{mm}$

   b) The change in winter is

   $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25.0\,\text{m} \cdot 40\,\text{K}$

   $= 0.0120\,\text{m}$

   $= 12.0\,\text{mm}$

   so

   $g_\text{max}$$= 30\,\text{mm} - 12.0\,\text{mm}$

   $= 18\,\text{mm}$

   Answer

   a) $g_\text{min} = 13.2\,\text{mm}$, in practice $13\,\text{mm}$, and b) $g_\text{max} = 18\,\text{mm}$. The coefficient $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ is a two-figure value, so the results are acceptable rounded to two figures. Model answer c): the gap $15\,\text{mm}$ is in the range $13.2\,\text{mm} \ldots 18\,\text{mm}$, so it is acceptable when installing at $8\,{}^\circ\text{C}$: in summer a gap of $15\,\text{mm} - 13.2\,\text{mm} = 1.8\,\text{mm}$ is left, and in winter it is $15\,\text{mm} + 12.0\,\text{mm} = 27\,\text{mm} < 30\,\text{mm}$. If the rails were installed in a frost at $-32\,{}^\circ\text{C}$ with the same gap, the rail would warm up to summer by $\Delta T = 84\,\text{K}$ and lengthen by $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25.0\,\text{m} \cdot 84\,\text{K} = 25.2\,\text{mm}$. That is more than the gap of $15\,\text{mm}$, so the gap would close and the rails would be pressed against each other. So the installation gap depends on the installation temperature.

   Reasonableness

   The change over the whole year is

   $12 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 25.0\,\text{m} \cdot 84\,\text{K}$

   $= 25.2\,\text{mm}$

   and it is the sum $13.2\,\text{mm} + 12.0\,\text{mm}$, so the width of the range

   $g_\text{max} - g_\text{min}$$= 18\,\text{mm} - 13.2\,\text{mm}$

   $= 4.8\,\text{mm}$

   $= 30\,\text{mm} - 25.2\,\text{mm}$

   matches. In practice the gap is chosen from the middle of the range, about $15\,\text{mm}$, and nowadays long rails are welded together without a gap.

5. Exercise 10-20

   Example data: the glass bulb of a liquid thermometer contains ethanol, whose volume at $10\,{}^\circ\text{C}$ is $1.00\,\text{cm}^3$ and which fills the bulb exactly. The ethanol rises from the bulb into a glass capillary tube with an inner radius of $0.30\,\text{mm}$. Use $\gamma = 1.1 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ for ethanol and $\alpha = 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$ for glass, and assume that the coefficients are constant over the temperature range $10\,{}^\circ\text{C} \ldots 30\,{}^\circ\text{C}$. The inner volume of the bulb grows in the same way as an equal piece of glass, and the cross-sectional area of the capillary is assumed to be constant.

   a) Calculate how many millimetres the column of ethanol rises in the capillary when the temperature rises from $10\,{}^\circ\text{C}$ to $30\,{}^\circ\text{C}$.

   b) By what percentage would the rise be greater if the expansion of the bulb were ignored?

   c) Calculate how much water would rise in the same device, and use it to justify why water is not suitable as the liquid of a liquid thermometer.

   The answer field takes the result of part a) in millimetres.

   mmCheck
   Answer

   $h \approx 75.9\,\text{mm}$

   Show the solution

   Given

   ethanol $V_0 = 1.00\,\text{cm}^3$ at $10\,{}^\circ\text{C}$, $\gamma_\text{ethanol} = 1.1 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$; glass bulb $\alpha_\text{glass} = 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}}$; inner radius of the capillary $r = 0.30\,\text{mm}$; temperature $10\,{}^\circ\text{C} \rightarrow 30\,{}^\circ\text{C}$ (example data)

   Find

   a) the rise of the column $h$ in millimetres; b) how much greater the rise would be without the expansion of the bulb; c) the rise with water and a justification of why water is not suitable

   Principle

   Ethanol expands more than the glass bulb that contains it. Of the growth in the volume of the ethanol, the part that does not fit in the expanded bulb rises into the capillary. The inner volume of the glass bulb grows like an equal piece of glass, so $\gamma_\text{glass} \approx 3\alpha$. The volume that moves into the tube is the volume of a cylinder $A h$, where $A = \pi r^2$.

   Formula

   $\Delta V_\text{ethanol} = \gamma V_0 \Delta T$, $\Delta V_\text{bulb} = 3\alpha V_0 \Delta T$ and $A h = \Delta V_\text{ethanol} - \Delta V_\text{bulb}$, so

   $h$$= \dfrac{\Delta V_\text{ethanol} - \Delta V_\text{bulb}}{\pi r^2}$

   Conversions

   $V_0$$= 1.00\,\text{cm}^3$

   $= 1000\,\text{mm}^3$

   so that the volumes and the area of the capillary are in the same unit, and

   $\Delta T$$= 30\,{}^\circ\text{C} - 10\,{}^\circ\text{C}$

   $= 20\,\text{K}$

   The area of the capillary is

   $A$$= \pi \cdot (0.30\,\text{mm})^2$

   $= 0.283\,\text{mm}^2$

   Substitution

   $\Delta V_\text{ethanol}$$= 1.1 \cdot 10^{-3}\,\tfrac{1}{\text{K}} \cdot 1000\,\text{mm}^3 \cdot 20\,\text{K}$

   $= 22.0\,\text{mm}^3$

   $\Delta V_\text{bulb}$$= 3 \cdot 9.0 \cdot 10^{-6}\,\tfrac{1}{\text{K}} \cdot 1000\,\text{mm}^3 \cdot 20\,\text{K}$

   $= 0.54\,\text{mm}^3$

   So

   $h$$= \dfrac{22.0\,\text{mm}^3 - 0.54\,\text{mm}^3}{0.283\,\text{mm}^2}$

   $= 75.9\,\text{mm}$

   Answer

   a) $h \approx 75.9\,\text{mm}$, three significant figures as calculated; the coefficients $1.1 \cdot 10^{-3}$ and $9.0 \cdot 10^{-6}$ are values with 2–3 figures, so in practice the rise is about $76\,\text{mm}$. Model answer b): without the expansion of the bulb the rise would be $\dfrac{22.0\,\text{mm}^3}{0.283\,\text{mm}^2} = 77.8\,\text{mm}$, that is $\dfrac{77.8\,\text{mm} - 75.9\,\text{mm}}{75.9\,\text{mm}} = 2.5\,\%$ greater. So the expansion of the bulb is a small but not negligible correction. Model answer c): with water $\gamma = 0.21 \cdot 10^{-3}\,\tfrac{1}{\text{K}}$ the growth in volume would be $4.2\,\text{mm}^3$, of which the bulb's share $0.54\,\text{mm}^3$ leaves $3.66\,\text{mm}^3$, that is a rise of only $12.9\,\text{mm}$. That is less than a fifth of the rise of the ethanol, so the scale would be dense and the uncertainty of reading large. In addition, water shrinks over $0\,{}^\circ\text{C} \ldots 4\,{}^\circ\text{C}$ and freezes, so the column would not work correctly in the cold.

   Reasonableness

   The density of the scale is

   $\dfrac{75.9\,\text{mm}}{20\,\text{K}}$$= 3.8\,\tfrac{\text{mm}}{\text{K}}$

   so reading one degree is possible, and the length of the column, $76\,\text{mm}$, fits in an ordinary thermometer. The relative growth in the volume of the ethanol is $1.1 \cdot 10^{-3}\,\tfrac{1}{\text{K}} \cdot 20\,\text{K} = 2.2\,\%$ and that of the bulb only $0.054\,\%$, so the effect of the bulb is small, as the result shows.