Set an electric iron to "cotton" and listen. After a minute or so it gives a faint click and its little lamp goes out. A while later it clicks again and the lamp comes back on. Nobody touched the dial. The iron keeps itself near the temperature you chose.
Two metals, one strip
Take a strip of brass and a strip of steel of the same length and bond them face to face along their whole length. This is a bimetallic strip. Heat it. Brass expands more than steel, since \(\alpha\) is \(1.9 \times 10^{-5}\,\text{K}^{-1}\) for brass and \(1.2 \times 10^{-5}\,\text{K}^{-1}\) for steel. But the two strips cannot slide past each other. The only way a longer strip and a shorter strip can stay bonded is to bend into an arc.
On an arc the outer edge is longer than the inner edge. So the metal that expands more, brass, goes on the outside of the curve. The strip curls towards the steel. When the strip is cooled below the temperature at which it was straight, the brass shrinks more and becomes the inner side. The strip then curls the other way.
How much does it bend?
Let each strip be \(d\) thick and let the strip, straight at first, be warmed by \(\Delta T\). It bends into an arc of radius \(R\), measured to the joint between the metals, through an angle \(\phi\). The middle line of the outer strip is then at radius \(R + d/2\), and that of the inner strip at \(R - d/2\).
\[ L(1 + \alpha_1\,\Delta T) = \left(R + \tfrac{d}{2}\right)\phi, \qquad L(1 + \alpha_2\,\Delta T) = \left(R - \tfrac{d}{2}\right)\phi \]
Why. each layer takes its new length, and an arc of radius \(r\) through an angle \(\phi\) has length \(r\phi\). Metal 1, which expands more, is outside.
\[ \dfrac{1 + \alpha_1\,\Delta T}{1 + \alpha_2\,\Delta T} = \dfrac{R + d/2}{R - d/2} \]
Why. divide the two equations; \(L\) and \(\phi\) cancel.
\[ 1 + (\alpha_1 - \alpha_2)\,\Delta T \approx 1 + \dfrac{d}{R} \]
Why. for small quantities \((1 + a)/(1 + b) \approx 1 + a - b\) on both sides, with \(d \ll R\).
\[ R = \dfrac{d}{(\alpha_1 - \alpha_2)\,\Delta T} \]
Read it. compare the small terms. A larger difference in \(\alpha\) or a larger rise in temperature gives a sharper bend.
The thermostat of an iron is a brass-steel strip \(5.0\,\text{cm}\) long, each layer \(0.5\,\text{mm}\) thick. Take \(\alpha_\text{brass} = 1.9 \times 10^{-5}\,\text{K}^{-1}\) and \(\alpha_\text{steel} = 1.2 \times 10^{-5}\,\text{K}^{-1}\). Find the radius of the bend and the movement of the tip when the strip is \(100\,\text{K}\) above room temperature.
\(d = 0.5 \times 10^{-3}\,\text{m}\), \(\alpha_1 - \alpha_2 = 0.7 \times 10^{-5}\,\text{K}^{-1}\), \(\Delta T = 100\,\text{K}\), \(L = 0.050\,\text{m}\).
\(R = \dfrac{d}{(\alpha_1 - \alpha_2)\,\Delta T}\) and \(y = \dfrac{L^2}{2R}\).
\(R = \dfrac{0.5 \times 10^{-3}}{0.7 \times 10^{-5} \times 100} = 0.71\,\text{m}\). \(y = \dfrac{0.050^2}{2 \times 0.71} = 1.75 \times 10^{-3}\,\text{m}\).
The strip bends to a radius of \(0.71\,\text{m}\), and its tip moves \(1.75\,\text{mm}\), enough to open a contact.
In Fig 6.1 the tip leaves its contact at \(60\,\text{K}\) above room temperature, when \(R = 1.19\,\text{m}\). The heater stops and the strip cools a little. The tip touches the contact again, the heater restarts, and the strip warms. The strip keeps the iron near its set temperature. That is the click you hear. A thermostat of this kind also runs geysers, electric kettles and fire alarms.
A thermostat strip has steel above and brass below, clamped at one end. Which way does its free end move when it is heated?
A strip of brass bonded to invar has layers \(1.0\,\text{mm}\) thick. The difference in their coefficients is \(1.8 \times 10^{-5}\,\text{K}^{-1}\). What is the radius of the bend, in metres, when the strip is warmed by \(50\,\text{K}\)?
A clock that feels the weather
A pendulum clock counts the swings of a pendulum. The chapter on simple harmonic motion shows that a pendulum of length \(L\) has a period \(T = 2\pi\sqrt{L/g}\); here we use the result. In this topic \(T\) is the period, and \(\Delta\theta\) is the change in temperature. The period grows as the square root of the length.
Let the rod warm by \(\Delta\theta\). Its length grows by the fraction \(\alpha\,\Delta\theta\). A square root halves a small fractional change, since \(\sqrt{1 + x} \approx 1 + x/2\). So the period grows by the fraction \(\tfrac12\alpha\,\Delta\theta\).
A longer period means fewer swings in a day. The clock counts swings, so it shows less time than has passed: a warm clock runs slow. Take a brass rod, \(\alpha = 1.9 \times 10^{-5}\,\text{K}^{-1}\), that is \(20\,\text{K}\) warmer than when it was set. The period is longer by \(\tfrac12 \times 1.9 \times 10^{-5} \times 20 = 1.9 \times 10^{-4}\) of itself. In a day of \(86\,400\,\text{s}\) the clock loses \(1.9 \times 10^{-4} \times 86\,400 = 16.4\,\text{s}\). A clock colder than its setting runs fast by the same rule.
A pendulum clock loses \(10\,\text{s}\) a day at \(35\,°\text{C}\) and gains \(5\,\text{s}\) a day at \(20\,°\text{C}\). At what temperature does it keep correct time? Find \(\alpha\) for its rod.
Let the clock be correct at \(\theta\). At \(35\,°\text{C}\) it is \((35 - \theta)\) too warm; at \(20\,°\text{C}\) it is \((\theta - 20)\) too cold.
\(10 = \tfrac12\alpha(35 - \theta) \times 86\,400\) and \(5 = \tfrac12\alpha(\theta - 20) \times 86\,400\).
Dividing: \(2 = \dfrac{35 - \theta}{\theta - 20}\), so \(3\theta = 75\) and \(\theta = 25\,°\text{C}\). Adding: \(15 = \tfrac12\alpha \times 15 \times 86\,400\), so \(\alpha = 2.3 \times 10^{-5}\,\text{K}^{-1}\).
The clock is correct at \(25\,°\text{C}\), and its rod has \(\alpha \approx 2.3 \times 10^{-5}\,\text{K}^{-1}\), close to aluminium.
A pendulum clock with a steel rod (\(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\)) keeps correct time at \(25\,°\text{C}\). It is moved to a hill station at \(15\,°\text{C}\). How many seconds does it gain in a day?
A pendulum clock loses \(8.64\,\text{s}\) a day when its temperature rises by \(10\,\text{K}\). Find \(\alpha\) for its rod, in units of \(10^{-5}\,\text{K}^{-1}\).
A warm rule reads short
A steel rule or measuring tape is itself a solid, and it expands like any other. Its marks are correct at the temperature at which it was made, usually \(20\,°\text{C}\). On a hot day every centimetre division is a little longer than a true centimetre. A length therefore spans fewer divisions, and the rule reads too little.
Each division of the warm rule is \((1 + \alpha\,\Delta\theta)\) true units long, where \(\Delta\theta\) is the rise above the marking temperature. So a reading \(x\) is really \(x(1 + \alpha\,\Delta\theta)\) true units. In Fig 6.2 each division is \(1 + 1.2 \times 10^{-5} \times 25 = 1.0003\) true centimetres. So the reading means \(29.991 \times 1.0003 = 30.000\,\text{cm}\).
A surveyor measures a plot in Jaipur at \(45\,°\text{C}\) with a steel tape marked at \(20\,°\text{C}\). The tape reads \(50.000\,\text{m}\). What is the true length? Take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\).
(a) The tape is \(25\,\text{K}\) warmer than when it was marked, so each metre division is \(1 + 1.2 \times 10^{-5} \times 25 = 1.0003\) true metres.
(b) True length \(= 50.000 \times 1.0003 = 50.015\,\text{m}\). The plot is \(1.5\,\text{cm}\) longer than the tape says.
Suppose the object expands too, with coefficient \(\alpha_1\), while the rule has \(\alpha_2\). At \(\Delta\theta\) above the marking temperature the reading is \(x = L[1 + (\alpha_1 - \alpha_2)\,\Delta\theta]\). Here \(L\) is the object's length at the marking temperature. A brass rod measured with a brass rule reads its marking-temperature length on any day.
A steel metre rule marked at \(20\,°\text{C}\) is used at \(40\,°\text{C}\) and reads \(100.000\,\text{cm}\) for a rod. What is the true length of the rod at \(40\,°\text{C}\)?
When expansion is blocked
Suppose the rails of Lesson 4 were laid end to end with no gaps. On the hot afternoon each rail tries to grow by \(7.0\,\text{mm}\), and its neighbours stop it. The chapter on Elasticity showed what happens to a rod held between rigid supports. It is squeezed back by the whole expansion it was denied, and it carries a thermal stress \(Y\alpha\,\Delta\theta\), where \(Y\) is Young's modulus. For steel, \(Y = 2 \times 10^{11}\,\text{Pa}\), and a rise of \(45\,\text{K}\) gives a stress of \(2 \times 10^{11} \times 1.2 \times 10^{-5} \times 45\) pascals. That is \(1.08 \times 10^8\,\text{Pa}\). A stress of about \(108\,\text{MPa}\) can bend a track sideways into a kink. The gap at the rail joint is a stress that was never allowed to appear.
Long stretches of modern track are welded into one continuous rail, with no gaps. The rail is laid and clamped on a warm day, at the middle of its yearly range of temperature. Why?
In 1726 the English carpenter and clockmaker John Harrison built a pendulum out of nine parallel rods of brass and steel, the gridiron pendulum. The steel rods let the bob down as they grew, and the brass rods lifted it up. He chose their lengths so that the two growths cancelled, since brass expands about one and a half times as much as steel. The bob stayed at the same height in summer and winter. His clocks lost less than a second a month.
The miniature circuit breakers in the distribution box of an Indian home each hold a small bimetallic strip. The current of the circuit flows through it and warms it. If too much current flows for too long, the strip bends far enough to release a spring, and the switch snaps off. A thermal stress, a blocked expansion and a bent strip all come from one number, \(\alpha\).
- Putting the metal with the smaller \(\alpha\) on the outside of the bend.The metal that expands more becomes longer and must take the longer, outer side of the arc.
- Using \(\alpha\,\Delta\theta\) for the change in period.\(T \propto \sqrt{L}\), so the period changes by half the fractional change in length, \(\tfrac12\alpha\,\Delta\theta\).
- Saying a warm clock gains time.A warm pendulum is longer and swings more slowly, so the clock counts fewer seconds and loses time.
- Correcting a warm rule's reading downward.A warm rule has longer divisions and reads short; the true length is the reading times \((1 + \alpha\,\Delta\theta)\).