The Workbook

1. Each scale is linear between the two fixed points. So the fraction of the way from the ice point to the steam point is the same on every scale.

   \(\dfrac{C - 0}{100 - 0} = \dfrac{F - 32}{212 - 32} = \dfrac{T - 273}{373 - 273}\), that is \(\dfrac{C}{100} = \dfrac{F - 32}{180} = \dfrac{T - 273}{100}\).

2. \(F = \dfrac{9}{5}C + 32 = 72 + 32 = 104^\circ\)F. \(T = C + 273 = 313\) K.

3. \(C = \dfrac{5}{9}(F - 32) = \dfrac{5}{9} \times 45 = 25^\circ\)C.

A Celsius degree and a kelvin are the same size: both are \(\dfrac{1}{100}\) of the interval between the ice point and the steam point. So a rise of \(20^\circ\)C is a rise of \(20\) K.

The same interval is \(180\) Fahrenheit degrees, so one Celsius degree is \(1.8\) Fahrenheit degrees. The rise is \(20 \times 1.8 = 36^\circ\)F.

The two scales differ only in where zero is placed. A difference does not depend on the zero.

Put \(F = C\) in \(F = \dfrac{9}{5}C + 32\): \(C = \dfrac{9}{5}C + 32\), so \(-\dfrac{4}{5}C = 32\) and \(C = -40\). The scales agree at \(-40^\circ\).

Put \(F = 2C\): \(2C = \dfrac{9}{5}C + 32\), so \(\dfrac{1}{5}C = 32\) and \(C = 160^\circ\)C. Check: \(160^\circ\)C is \(320^\circ\)F.

\(F = \dfrac{9}{5}(-273.15) + 32 = -491.67 + 32 = -459.67^\circ\)F.

The Kelvin scale places its zero at absolute zero, the lowest temperature possible. No body can be colder than that, so no reading can be negative.

\(C = \dfrac{5}{9}(98.6 - 32) = \dfrac{5}{9} \times 66.6 = 37^\circ\)C. \(T = 37 + 273 = 310\) K.

1. \(\dfrac{X - 20}{220 - 20} = \dfrac{C}{100}\). At \(50^\circ\)C, \(X = 20 + 200 \times 0.5 = 120^\circ\)X.

2. \(C = 100 \times \dfrac{65 - 20}{200} = 22.5^\circ\)C.

3. \(200\) degrees of \(X\) cover \(100\) Celsius degrees, so one degree \(X\) is \(0.5^\circ\)C.

1. At \(t = 0\) the line gives \(X = 10\), and at \(t = 100\) it gives \(X = 210\).

2. Slope \(= \dfrac{210 - 10}{100} = 2\) and intercept \(10\), so \(X = 2t + 10\).

3. Put \(X = t\): \(t = 2t + 10\), so \(t = -10\). Both scales read \(-10\) there.

1. The faulty scale has \(90\) divisions between the fixed points. True temperature \(= \dfrac{59 - 5}{95 - 5} \times 100 = 60^\circ\)C.

2. Reading \(= 5 + \dfrac{20}{100} \times 90 = 23^\circ\)C.

3. Let the reading equal the true temperature \(t\): \(t = 5 + 0.9t\), so \(0.1t = 5\) and \(t = 50^\circ\)C.

1. There are \(105\) divisions between the fixed points. Reading \(= -2 + \dfrac{60}{100} \times 105 = -2 + 63 = 61^\circ\)C.

2. \(t = -2 + 1.05t\), so \(0.05t = 2\) and \(t = 40^\circ\)C.

\(\dfrac{A - 30}{90 - 30} = \dfrac{B + 10}{110 + 10}\), so \(B + 10 = 2(A - 30)\) and \(B = 2A - 70\).

At \(A = 60\): \(B = 120 - 70 = 50^\circ\)B.

\(\dfrac{R}{80} = \dfrac{C}{100}\), so \(C = \dfrac{100}{80} \times 60 = 75^\circ\)C.

\(F = \dfrac{9}{5} \times 75 + 32 = 135 + 32 = 167^\circ\)F.

The volume of the gas is kept fixed and the pressure is measured. For a gas at constant volume the pressure is proportional to the absolute temperature: \(\dfrac{T}{273} = \dfrac{P}{P_0}\), where \(P_0\) is the pressure at the ice point.

\(T = 273 \times \dfrac{80}{60} = 364\) K, which is \(364 - 273 = 91^\circ\)C.

1. All three lines, when extended, cut the temperature axis at the same point, \(-273^\circ\)C. There the pressure of every gas would be zero. This temperature is absolute zero.

2. The pressure rises linearly: \(P = P_0(1 + \beta t)\) with \(\beta = \dfrac{1.366 - 1}{100} = 0.00366\) per \(^\circ\)C. \(P = 0\) when \(t = -\dfrac{1}{\beta} = -\dfrac{100}{0.366} = -273^\circ\)C.

1. The resistance rises linearly with temperature, so \(t = \dfrac{R_t - R_0}{R_{100} - R_0} \times 100\).

2. \(t = \dfrac{12.4 - 10}{14 - 10} \times 100 = \dfrac{2.4}{4} \times 100 = 60^\circ\)C.

3. \(R_{100} = R_0(1 + 100\alpha)\), so \(\alpha = \dfrac{14 - 10}{10 \times 100} = 0.004\) per \(^\circ\)C.

1. At \(t = 0\), \(R_0 = 5\ \Omega\). At \(t = 100\), \(R_{100} = 7\ \Omega\).

2. \(\alpha = \dfrac{R_{100} - R_0}{R_0 \times 100} = \dfrac{2}{500} = 0.004\) per \(^\circ\)C.

3. \(t = \dfrac{6.5 - 5}{7 - 5} \times 100 = 75^\circ\)C.

1. When the gas is heated it pushes the mercury in the left limb down. The right limb is raised (or mercury is added) until the left level is back at the mark \(A\). The gas then fills the same volume as before.

2. The mercury in the right limb stands \(h\) higher, so \(P = P_0 + h\rho g\).

3. At low pressure all gases give the same reading, so the scale does not depend on the gas chosen. A gas thermometer also works over a very wide range, from about \(-270^\circ\)C to over \(1000^\circ\)C.

\(T = 273.16 \times 2 = 546.32\) K.

The triple point is one exact state: ice, water and vapour together at one fixed pressure. It is easy to reproduce anywhere. The ice point depends on the air pressure and on dissolved air in the water, so it varies slightly.

(i) The length of the mercury column. (ii) The pressure of the gas. (iii) The electrical resistance of the platinum wire. (iv) The emf across the junctions.

The property must change steadily with temperature, and one value of the property must correspond to one temperature only.

The length \(l\) is the thermometric property: \(t = \dfrac{l - l_0}{l_{100} - l_0} \times 100 = \dfrac{8 - 2}{22 - 2} \times 100 = 30^\circ\)C.

At \(45^\circ\)C, \(l = 2 + \dfrac{45}{100} \times 20 = 11\) cm.

\(27 + 273 = 300\) K and \(-73 + 273 = 200\) K. Ratio \(= \dfrac{300}{200} = 1.5\).

The Celsius ratio \(\dfrac{27}{-73}\) means nothing. The Celsius zero is an arbitrary point, so a Celsius reading is not proportional to how hot a body is. Only absolute temperatures can be compared as ratios.

Rise \(= 200 - 20 = 180^\circ\)C \(= 180\) K \(= 180 \times 1.8 = 324^\circ\)F.

Final temperature \(= \dfrac{9}{5} \times 200 + 32 = 392^\circ\)F.

At \(-50^\circ\)C mercury is solid, so the column cannot move. At \(400^\circ\)C mercury is a vapour, so there is no column.

For \(-50^\circ\)C use an alcohol thermometer (alcohol freezes at \(-115^\circ\)C). For \(400^\circ\)C use a platinum resistance thermometer or a gas thermometer.

\(\alpha\) is the fractional increase in length per degree rise in temperature: \(\alpha = \dfrac{\Delta L}{L\,\Delta T}\), so \(\Delta L = \alpha L\Delta T\).

\(\Delta L = 1.2 \times 10^{-5} \times 2 \times 50 = 1.2 \times 10^{-3}\) m \(= 1.2\) mm.

\(\alpha\) has the unit per \(^\circ\)C (the same as per kelvin) and the dimensions \([\mathrm{K}^{-1}]\), because \(\Delta L / L\) is a pure number.

\(\Delta L = \alpha L\Delta T = 1.2 \times 10^{-5} \times 20 \times 40 = 9.6 \times 10^{-3}\) m \(= 9.6\) mm. The gap must be at least \(9.6\) mm.

Without a gap the rails push against each other as they expand. The thermal stress bends them sideways and the track buckles.

\(\dfrac{\Delta L}{L} = \alpha\Delta T = 2 \times 10^{-5} \times 100 = 2 \times 10^{-3}\), that is \(0.2\%\).

The fraction depends only on \(\alpha\) and \(\Delta T\), not on the length. A long rod and a short rod expand by the same percentage.

1. \(L_0 = 100.0\) cm at \(0^\circ\)C and \(L = 100.2\) cm at \(100^\circ\)C.

2. \(\alpha = \dfrac{\Delta L}{L_0\Delta T} = \dfrac{0.2}{100 \times 100} = 2 \times 10^{-5}\) per \(^\circ\)C.

3. The line is straight, so \(\Delta L = 2 \times 10^{-5} \times 100 \times 250 = 0.5\) cm and \(L = 100.5\) cm.

The difference stays the same only if both rods expand by the same amount: \(\alpha_s L_s\Delta T = \alpha_b L_b\Delta T\).

\(1.2 L_s = 1.8 L_b\), so \(L_s = 1.5 L_b\). With \(L_s - L_b = 10\) cm, \(0.5 L_b = 10\) cm. So \(L_b = 20\) cm and \(L_s = 30\) cm.

Each side becomes \(a(1 + \alpha\Delta T)\). The area becomes \(a^2(1 + \alpha\Delta T)^2 = a^2(1 + 2\alpha\Delta T + \alpha^2\Delta T^2)\).

\(\alpha\Delta T\) is very small, so its square can be dropped: \(A' = A(1 + 2\alpha\Delta T)\). Comparing with \(A' = A(1 + \beta\Delta T)\) gives \(\beta = 2\alpha\).

For a cube the volume is \(a^3(1 + \alpha\Delta T)^3 \approx a^3(1 + 3\alpha\Delta T)\), so \(\gamma = 3\alpha\).

\(\beta = 2\alpha = 4 \times 10^{-5}\) per \(^\circ\)C. \(\Delta A = \beta A\Delta T = 4 \times 10^{-5} \times 0.5 \times 100 = 2 \times 10^{-3}\) m\(^2 = 20\) cm\(^2\).

\(\gamma = 3\alpha = 7.5 \times 10^{-5}\) per \(^\circ\)C. \(V = 10^3 = 1000\) cm\(^3\).

\(\Delta V = \gamma V\Delta T = 7.5 \times 10^{-5} \times 1000 \times 200 = 15\) cm\(^3\). New volume \(= 1015\) cm\(^3\).

1. \(V' = x(1 + \alpha_x\Delta T)\,y(1 + \alpha_y\Delta T)\,z(1 + \alpha_z\Delta T)\). Multiplying and dropping products of small terms: \(V' = V[1 + (\alpha_x + \alpha_y + \alpha_z)\Delta T]\). So \(\gamma = \alpha_x + \alpha_y + \alpha_z\).

2. \(\gamma = (1 + 2 + 3) \times 10^{-5} = 6 \times 10^{-5}\) per \(^\circ\)C. The \(xy\) face has \(\beta = \alpha_x + \alpha_y = 3 \times 10^{-5}\) per \(^\circ\)C.

1. Larger. Every distance in the plate grows by the same fraction, including the distance across the hole. The hole expands exactly as a brass disc filling it would.

2. \(\Delta d = \alpha d\Delta T = 2 \times 10^{-5} \times 2 \times 100 = 4 \times 10^{-3}\) cm \(= 0.04\) mm.

3. \(\dfrac{\Delta A}{A} = \beta\Delta T = 2\alpha\Delta T = 4 \times 10^{-3} = 0.4\%\).

1. \(\Delta d = 10.01 - 10.00 = 0.01\) cm.

2. \(\Delta T = \dfrac{\Delta d}{\alpha d} = \dfrac{0.01}{1.25 \times 10^{-5} \times 10} = 80^\circ\)C. The ring must be heated to \(20 + 80 = 100^\circ\)C.

3. As the ring cools it tries to shrink back to \(10.00\) cm. The shaft stops it, so the ring grips the shaft tightly. This is how iron tyres are fitted on cart wheels.

Yes. The glass around the cavity expands outward in every direction, so the cavity grows as if it were filled with glass.

\(\gamma = 3\alpha = 2.4 \times 10^{-5}\) per \(^\circ\)C. \(\Delta V = 2.4 \times 10^{-5} \times 1000 \times 100 = 2.4\) cm\(^3\). Volume at \(100^\circ\)C \(= 1002.4\) cm\(^3\).

The mass does not change. \(\rho' = \dfrac{m}{V'} = \dfrac{m}{V(1 + \gamma\Delta T)} = \dfrac{\rho}{1 + \gamma\Delta T}\).

For small \(x\), \(\dfrac{1}{1 + x} \approx 1 - x\), so \(\rho' \approx \rho(1 - \gamma\Delta T)\).

\(\gamma\Delta T = 5 \times 10^{-4} \times 20 = 0.01\). \(\rho' = \dfrac{1000}{1.01} = 990\) kg/m\(^3\) (the approximate form also gives \(990\)).

1. The volume is least at \(4^\circ\)C. The same mass in the least volume means the density is greatest at \(4^\circ\)C.

2. From \(0^\circ\)C to \(4^\circ\)C the volume falls as the temperature rises, so water contracts on heating. \(\gamma = \dfrac{\Delta V}{V\Delta T}\) is negative there.

3. \(1\) g fills \(1.0000\) cm\(^3\) at \(4^\circ\)C, so the density is \(1\) g/cm\(^3 = 1000\) kg/m\(^3\).

1. As the lake cools, the water at \(4^\circ\)C is the densest and sinks to the bottom. Water colder than \(4^\circ\)C is lighter and stays above it. So the bottom layer stays at \(4^\circ\)C.

2. The coldest water, at \(0^\circ\)C, is the least dense and floats at the top. It freezes first, so ice forms at the surface.

3. Water expands when it freezes, so ice is less dense than water (about \(0.92\) g/cm\(^3\)). The lighter ice floats. The ice layer also insulates the water below, so the lake does not freeze solid.

Between \(0^\circ\)C and \(8^\circ\)C the volume of water first falls and then rises. Two different temperatures, such as \(2^\circ\)C and \(6^\circ\)C, give the same volume, so one reading could mean either temperature. A thermometric property must change in one direction only.

Below \(4^\circ\)C water expands on cooling. The water needs more room, but the steel bottle contracts slightly on cooling. The water presses hard on the walls and the bottle may burst. This is why water pipes crack in a hard frost.

At \(40^\circ\)C every division of the scale is longer than \(1\) mm, so the scale reads too few divisions. The true length is more than \(100\) cm.

True length \(= 100(1 + \alpha\Delta T) = 100(1 + 1.2 \times 10^{-5} \times 20) = 100.024\) cm.

\(\gamma = 3\alpha = 3.6 \times 10^{-5}\) per \(^\circ\)C. \(\Delta V = 3.6 \times 10^{-5} \times 2 \times 50 = 3.6 \times 10^{-3}\) m\(^3 = 3.6\) litres.

\(\Delta L = \alpha L\Delta T = 1.7 \times 10^{-5} \times 5 \times (-40) = -3.4 \times 10^{-3}\) m. The wire contracts by \(3.4\) mm, to \(4.9966\) m.

\(\Delta L = \alpha L\Delta T\), so \(\dfrac{\Delta L_1}{\Delta L_2} = \dfrac{\alpha_1 L_1}{\alpha_2 L_2} = \dfrac{2 \times 1}{1 \times 2} = 1\). The ratio is \(1 : 1\).

Looser. Both the lid and the mouth of the jar expand, but \(\alpha\) for aluminium is three times that of glass. The lid grows more than the jar, so the grip loosens. This is why a stuck lid is run under hot water.

The real expansion is the true increase in the volume of the liquid. The apparent expansion is the increase in the level seen in the vessel.

The vessel expands as well, so part of the real expansion only fills the extra space in the vessel. Hence \(\gamma_{app} = \gamma_{real} - \gamma_{vessel}\), and the apparent expansion is smaller than the real one.

1. The heat reaches the glass first. The flask expands before the liquid has warmed, so the same liquid fills a larger flask and the level dips.

2. Once the liquid is heated it expands too. Its \(\gamma\) is much larger than that of glass, so the liquid soon fills the extra space and rises above the mark.

3. After the first dip the level would return to the mark and stay there. The liquid and the flask would expand by the same fraction.

1. \(\gamma = 3\alpha = 2.4 \times 10^{-5}\) per \(^\circ\)C.

2. \(\gamma_{app} = \gamma_{real} - \gamma_{glass} = 1.8 \times 10^{-4} - 0.24 \times 10^{-4} = 1.56 \times 10^{-4}\) per \(^\circ\)C.

3. Overflow \(= \gamma_{app}V\Delta T = 1.56 \times 10^{-4} \times 500 \times 100 = 7.8\) cm\(^3\). (The mercury expands by \(9\) cm\(^3\) and the flask by \(1.2\) cm\(^3\).)

\(\gamma_{app} = \dfrac{\text{overflow}}{V\Delta T} = \dfrac{3.52}{1000 \times 20} = 1.76 \times 10^{-4}\) per \(^\circ\)C.

\(\gamma_{real} = \gamma_{app} + \gamma_{vessel} = 1.76 \times 10^{-4} + 0.24 \times 10^{-4} = 2.0 \times 10^{-4}\) per \(^\circ\)C.

\(\gamma_{vessel} = 3\alpha = 7.5 \times 10^{-5}\) per \(^\circ\)C, the same as \(\gamma\) of the liquid. So \(\gamma_{app} = 0\): the liquid and the vessel expand by the same fraction, and the level does not move.

The empty space stays the same if the liquid expands by exactly the same volume as the vessel: \(\gamma_l V_l\Delta T = \gamma_v V\Delta T\).

\(V_l = V\dfrac{\gamma_v}{\gamma_l} = 1000 \times \dfrac{2.4 \times 10^{-5}}{2.4 \times 10^{-4}} = 100\) cm\(^3\).

1. Brass. It has the larger \(\alpha\), so it becomes longer than the steel. The longer strip must lie on the outside of the curve, where the arc is longer.

2. On cooling the brass contracts more and becomes the shorter strip. The strip bends the other way, with brass on the inside.

3. A thermostat in an iron, a geyser or a refrigerator; a fire alarm; a bimetallic thermometer.

The outer strip has length \(L(1 + \alpha_1\Delta T) = (R + \tfrac{d}{2})\theta\). The inner strip has length \(L(1 + \alpha_2\Delta T) = (R - \tfrac{d}{2})\theta\).

Dividing: \(\dfrac{1 + \alpha_1\Delta T}{1 + \alpha_2\Delta T} = \dfrac{R + d/2}{R - d/2}\).

For small quantities the left side is \(1 + (\alpha_1 - \alpha_2)\Delta T\) and the right side is \(1 + \dfrac{d}{R}\). So \(R = \dfrac{d}{(\alpha_1 - \alpha_2)\Delta T}\).

\(R = \dfrac{d}{(\alpha_b - \alpha_s)\Delta T} = \dfrac{10^{-3}}{(2 - 1.2) \times 10^{-5} \times 100} = \dfrac{10^{-3}}{8 \times 10^{-4}} = 1.25\) m.

\(R \propto \dfrac{1}{\Delta T}\), so doubling the rise halves the radius: \(R = 0.625\) m. The strip bends more sharply.

1. Upward. Brass expands more than steel, so the brass side becomes longer and must lie on the outside of the bend. With brass below, the free end lifts.

2. The free end leaves the contact, the circuit breaks and the heater stops.

3. The strip straightens, the free end rests on the contact again and the heater works again. The room temperature stays close to the set value.

\(T \propto \sqrt{L}\), so \(\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta L}{L}\) for a small change.

\(\dfrac{\Delta L}{L} = \alpha\Delta\theta\). Hence \(\dfrac{\Delta T}{T} = \dfrac{1}{2}\alpha\Delta\theta\).

\(\dfrac{\Delta T}{T} = \dfrac{1}{2} \times 2 \times 10^{-5} \times 10 = 10^{-4}\).

The rod is longer, so each swing takes longer and the clock counts fewer seconds. It runs slow.

Loss per day \(= 10^{-4} \times 86400 = 8.64\) s.

At \(10^\circ\)C the rod is shorter than at \(20^\circ\)C. The period is smaller, the clock counts more seconds, and it runs fast.

\(\dfrac{\Delta T}{T} = \dfrac{1}{2} \times 2 \times 10^{-5} \times 10 = 10^{-4}\), so the gain is \(8.64\) s per day.

Fractional loss \(= \dfrac{4.32}{86400} = 5 \times 10^{-5} = \dfrac{1}{2}\alpha\Delta\theta\).

\(\alpha = \dfrac{2 \times 5 \times 10^{-5}}{10} = 10^{-5}\) per \(^\circ\)C.

1. The line crosses zero gain at \(20^\circ\)C. The clock is correct there.

2. The line is straight: \(8.64\) s per \(10^\circ\)C. At \(40^\circ\)C, \(20^\circ\)C above the correct temperature, the loss is \(2 \times 8.64 = 17.28\) s per day.

3. Per \(10^\circ\)C the fractional change is \(\dfrac{8.64}{86400} = 10^{-4} = \dfrac{1}{2}\alpha \times 10\), so \(\alpha = 2 \times 10^{-5}\) per \(^\circ\)C.

The mass is fixed and the volume becomes \(V(1 + \gamma\Delta T)\). So \(\rho' = \dfrac{\rho}{1 + \gamma\Delta T} \approx \rho(1 - \gamma\Delta T)\), using \(\dfrac{1}{1 + x} \approx 1 - x\).

\(\gamma = 3\alpha = 3 \times 10^{-5}\) per \(^\circ\)C. \(\dfrac{\Delta\rho}{\rho} = -\gamma\Delta T = -3 \times 10^{-3}\), a fall of \(0.3\%\).

1. Floating: weight \(=\) buoyancy, so \(\rho_s V g = \rho_l (fV) g\) and \(f = \dfrac{\rho_s}{\rho_l}\).

   After heating, \(\rho_s' = \dfrac{\rho_s}{1 + \gamma_s\Delta T}\) and \(\rho_l' = \dfrac{\rho_l}{1 + \gamma_l\Delta T}\). So \(f' = \dfrac{\rho_s'}{\rho_l'} = f\,\dfrac{1 + \gamma_l\Delta T}{1 + \gamma_s\Delta T}\).

2. \(f' > f\) when \(\gamma_l > \gamma_s\). The liquid becomes lighter faster than the block, so the block must displace more liquid and sinks deeper. Liquids usually expand more than solids, so this is the common case.

3. \(f' = 0.8 \times \dfrac{1 + 0.05}{1 + 0.005} = 0.8 \times \dfrac{1.05}{1.005} = 0.836\). About \(84\%\) of the block is now below the surface.

Buoyant force \(F = V\rho_l g\). After heating, \(V' = V(1 + \gamma_s\Delta T)\) and \(\rho_l' = \dfrac{\rho_l}{1 + \gamma_l\Delta T}\).

\(\dfrac{F'}{F} = \dfrac{1 + \gamma_s\Delta T}{1 + \gamma_l\Delta T} \approx 1 - (\gamma_l - \gamma_s)\Delta T = 1 - 2.7 \times 10^{-4} \times 100 = 1 - 0.027\).

The buoyant force falls by \(2.7\%\).

\(\gamma\Delta T = 10^{-3} \times 10 = 0.01\). \(\rho' = \dfrac{800}{1.01} = 792\) kg/m\(^3\).

Sinks. Water is densest at \(4^\circ\)C, so at \(20^\circ\)C the water is less dense than before. The glass ball expands very little, so its density is almost unchanged and is now greater than that of the water.

\(\gamma = 3\alpha = 3.75 \times 10^{-5}\) per \(^\circ\)C. \(\gamma\Delta T = 3.75 \times 10^{-3}\).

\(\rho' = \dfrac{8000}{1.00375} = 7970\) kg/m\(^3\). The approximate form \(8000(1 - 0.00375) = 7970\) agrees.

For mercury \(\gamma = 1.8 \times 10^{-4}\) per \(^\circ\)C, for glass only \(2.4 \times 10^{-5}\) per \(^\circ\)C. Mercury expands about \(7.5\) times more than the bore, so the extra volume must move up the bore.

The scale reads the rise of the column, which is the apparent expansion, \(\gamma_{app} = 1.56 \times 10^{-4}\) per \(^\circ\)C. The bore's own expansion is already allowed for when the scale is marked.