Dip a mercury thermometer into a cup of hot tea and watch the column closely. For a moment it slips down a little. Then it climbs quickly and settles high up the scale. Nothing in the tea was cooler than the room.
The vessel expands too
A liquid has no shape of its own, so it must be measured inside a vessel. When the two are heated, the vessel grows as well. Lesson 4 showed that the space inside a vessel grows as if it were solid material. Part of the liquid's growth only fills this new space. The rest is what we see as a rise in level.
A glass flask is filled with coloured water up to a mark on its narrow neck. A flame heats it from below. What does the level in the neck do first?
This is the dip of the thermometer in hot tea. The glass bulb warms first and grows, and the mercury has not yet warmed. So the column slips down. A moment later the mercury warms and expands about seven times as much as the glass, and the column climbs.
Real and apparent expansion
The true growth of the liquid's volume is its real expansion. The growth we see, measured against the marks on the vessel, is its apparent expansion. The real growth is the apparent growth plus the growth of the vessel. For each kelvin, the coefficients add in the same way.
For mercury \(\gamma_\text{real} = 1.82 \times 10^{-4}\,\text{K}^{-1}\). For glass \(\alpha = 9 \times 10^{-6}\,\text{K}^{-1}\), so \(\gamma_\text{glass} = 2.7 \times 10^{-5}\,\text{K}^{-1}\). The apparent expansion of mercury in glass is therefore \(1.82 \times 10^{-4} - 2.7 \times 10^{-5}\), which is \(1.55 \times 10^{-4}\,\text{K}^{-1}\). This apparent value is what a mercury thermometer actually shows; the marks on its scale already allow for the glass.
If a liquid and its vessel had the same coefficient of volume expansion, \(\gamma_\text{app}\) would be zero. The level would stay at its mark at every temperature. A liquid with \(\gamma = 7.2 \times 10^{-5}\,\text{K}^{-1}\) in a metal vessel with \(\alpha = 2.4 \times 10^{-5}\,\text{K}^{-1}\) behaves this way.
A glass flask holds exactly \(250\,\text{cm}^3\) of mercury at \(20\,°\text{C}\), filled to the brim. It is heated to \(80\,°\text{C}\). How much mercury overflows? Take \(\gamma_\text{Hg} = 1.82 \times 10^{-4}\,\text{K}^{-1}\) and \(\alpha_\text{glass} = 9 \times 10^{-6}\,\text{K}^{-1}\).
\(\Delta T = 60\,\text{K}\). The flask's space grows with \(\gamma_\text{glass} = 3 \times 9 \times 10^{-6} = 2.7 \times 10^{-5}\,\text{K}^{-1}\).
Overflow \(= V\gamma_\text{app}\,\Delta T\), with \(\gamma_\text{app} = \gamma_\text{Hg} - \gamma_\text{glass} = 1.55 \times 10^{-4}\,\text{K}^{-1}\).
Overflow \(= 250 \times 1.55 \times 10^{-4} \times 60 = 2.33\,\text{cm}^3\).
About \(2.3\,\text{cm}^3\) of mercury overflows. Forgetting the glass would give \(2.7\,\text{cm}^3\), too much by a sixth.
A glass vessel (\(\gamma = 2.7 \times 10^{-5}\,\text{K}^{-1}\)) is filled with \(500\,\text{cm}^3\) of glycerine. When both are warmed by \(40\,\text{K}\), \(9.46\,\text{cm}^3\) of glycerine overflows. Find the real coefficient of expansion of glycerine, in units of \(10^{-4}\,\text{K}^{-1}\).
A glass vessel of volume \(1000\,\text{cm}^3\) (\(\gamma = 2.7 \times 10^{-5}\,\text{K}^{-1}\)) holds some mercury (\(\gamma = 1.82 \times 10^{-4}\,\text{K}^{-1}\)). How much mercury, in cm³, keeps the empty space above it the same at every temperature?
Density falls as a body warms
Heating does not change the mass of a body, but it raises its volume to \(V(1 + \gamma\,\Delta T)\). Density is mass divided by volume. So the density falls to \(\rho' = \rho/(1 + \gamma\,\Delta T)\). Because \(\gamma\,\Delta T\) is small, \(1/(1 + x) \approx 1 - x\), and \(\rho' \approx \rho(1 - \gamma\,\Delta T)\).
Kerosene has a density of \(800\,\text{kg/m}^3\) at \(20\,°\text{C}\) and \(\gamma = 1.0 \times 10^{-3}\,\text{K}^{-1}\). At \(45\,°\text{C}\) its density is \(800/1.025 = 780.5\,\text{kg/m}^3\). A litre bought on a hot afternoon holds about \(2.4\,\%\) less kerosene than a litre bought on a cool morning.
Upthrust in a warmer liquid
The chapter on Archimedes' principle gave the upthrust on a body wholly under a liquid. It equals the weight of liquid pushed aside, \(F = V\rho_l g\). Warm both. The body's volume grows to \(V(1 + \gamma_s\,\Delta T)\), and the liquid's density falls to \(\rho_l/(1 + \gamma_l\,\Delta T)\). Liquids usually expand far more than solids, so the upthrust falls.
\[ \dfrac{F'}{F} = \dfrac{1 + \gamma_s\,\Delta T}{1 + \gamma_l\,\Delta T} \approx 1 - (\gamma_l - \gamma_s)\,\Delta T \]
A floating body pushes aside just enough liquid to carry its weight, so the fraction \(f\) of its volume under the surface is \(\rho_s/\rho_l\). When both warm, this becomes \(f' = f(1 + \gamma_l\,\Delta T)/(1 + \gamma_s\,\Delta T)\). If the liquid expands more, the liquid becomes lighter faster, and the body floats deeper.
A steel ball of volume \(200\,\text{cm}^3\) hangs from a spring balance, wholly under kerosene at \(20\,°\text{C}\) (\(\rho = 800\,\text{kg/m}^3\), \(\gamma_l = 1.0 \times 10^{-3}\,\text{K}^{-1}\)). For steel \(\gamma_s = 3.6 \times 10^{-5}\,\text{K}^{-1}\). Both are warmed by \(40\,\text{K}\). Take \(g = 10\,\text{m/s}^2\). Find the upthrust before and after, and say how the balance reading changes.
(a) Before: \(F = V\rho g = 200 \times 10^{-6} \times 800 \times 10 = 1.6\,\text{N}\).
(b) After: \(F' = F \times \dfrac{1 + 3.6 \times 10^{-5} \times 40}{1 + 1.0 \times 10^{-3} \times 40} = 1.6 \times \dfrac{1.00144}{1.04} = 1.54\,\text{N}\). The upthrust falls by \(3.7\,\%\).
(c) The balance reads the weight minus the upthrust. The upthrust fell by \(0.06\,\text{N}\), so the reading rises by \(0.06\,\text{N}\).
A plastic block floats in kerosene with \(0.80\) of its volume under the surface. For the kerosene \(\gamma_l = 1.0 \times 10^{-3}\,\text{K}^{-1}\); for the plastic \(\gamma_s = 2.0 \times 10^{-4}\,\text{K}^{-1}\). Both are warmed by \(40\,\text{K}\). What fraction of the block is now under the surface?
Copper has a density of \(8900\,\text{kg/m}^3\) at \(0\,°\text{C}\) and \(\alpha = 1.7 \times 10^{-5}\,\text{K}^{-1}\). What is its density at \(100\,°\text{C}\), in kg/m³?
Water breaks the rule
Nearly every liquid expands steadily as it warms. Water does too above \(4\,°\text{C}\). But between \(0\,°\text{C}\) and \(4\,°\text{C}\), water contracts as it warms. Its volume is least, and its density greatest, at \(4\,°\text{C}\), where the density is \(999.97\,\text{kg/m}^3\), very nearly \(1000\,\text{kg/m}^3\). Between \(0\) and \(4\,°\text{C}\) its \(\gamma\) is negative. This is the anomalous expansion of water.
When water freezes, it expands again, by about \(9\,\%\). Ice has a density of about \(917\,\text{kg/m}^3\), less than any liquid water, so ice floats with about nine tenths of it below the surface. A sealed bottle or a water pipe filled with water can burst in a hard frost. The water needs more room as it freezes, while the steel or glass around it shrinks.
A lake in winter
Now follow a lake in Ladakh as the winter air cools its surface. Fig 5.1 runs the story beside the density curve of water.
Above \(4\,°\text{C}\), cooled surface water is denser than the water below it, so it sinks. Warmer water rises to take its place and is cooled in turn. This stirring goes on until the whole lake is at \(4\,°\text{C}\). Then the stirring stops. Water cooled below \(4\,°\text{C}\) is lighter, so it stays at the top. The surface cools to \(0\,°\text{C}\) and freezes, while the densest water, at \(4\,°\text{C}\), lies at the bottom.
The ice floats and covers the lake like a lid. Ice and still water conduct heat poorly, so the water below loses heat slowly. Most lakes never freeze to the bottom. Fish and plants live through the winter in water near \(4\,°\text{C}\).
Imagine the opposite. If water kept getting denser as it cooled, the coldest water would sink, and the lake would freeze from the bottom up. Each winter the ice at the bottom would be shielded from the summer sun by the water above it. Lakes in cold places would slowly turn into solid ice.
A deep lake is covered with ice in winter. What is the temperature of the water at the bottom?
A small glass ball has exactly the density of water at \(4\,°\text{C}\) and hangs just below the surface. The water is warmed to \(20\,°\text{C}\). What does the ball do?
Why would a thermometer filled with water instead of mercury be useless between \(0\,°\text{C}\) and \(8\,°\text{C}\)?
Between \(0\,°\text{C}\) and \(8\,°\text{C}\) the volume of water first falls, down to its least value at \(4\,°\text{C}\), and then rises again. So two temperatures, such as \(2\,°\text{C}\) and \(6\,°\text{C}\), give the same volume and the same column. One reading would mean two temperatures, and a thermometric property must have one value for one temperature only.
In 1805 the Edinburgh chemist Thomas Charles Hope stood a tall jar of water with one thermometer near the top and one near the bottom. He packed a band of ice and salt round the middle of the jar. As the middle cooled, the bottom thermometer fell to about \(4\,°\text{C}\) and stayed there. Only after that did the top thermometer fall, towards \(0\,°\text{C}\). The experiment showed that water at \(4\,°\text{C}\) is the densest and sinks.
In January the high lakes of Ladakh freeze over, and the ice on some of them grows thick enough to walk on. Air temperatures fall below \(-20\,°\text{C}\). Under the ice, the water near the bottom stays close to \(4\,°\text{C}\). Fish, insects and water plants live through the winter in it, because water is densest at \(4\,°\text{C}\) and ice floats.
- Forgetting that the vessel expands too.The overflow or the rise you see is the apparent expansion: \(\gamma_\text{app} = \gamma_\text{real} - 3\alpha_\text{vessel}\).
- Using \(\alpha\) in place of \(\gamma\) for a density.Density changes with volume, so use \(\gamma = 3\alpha\) for a solid.
- Thinking the upthrust rises in warmer water.Warmer liquid is less dense, so the weight of liquid pushed aside, and the upthrust, falls.
- Assuming water behaves like other liquids below \(4\,°\text{C}\).Water contracts as it warms from \(0\) to \(4\,°\text{C}\) and is densest at \(4\,°\text{C}\).