Pangong Tso lies in Ladakh at about 4200 m, a long blue lake between bare mountains. In January and February its surface freezes hard. People walk across it, and in February 2023 runners raced a half marathon of 21 km on its ice. Yet the ice does not grow without limit. Each week of cold adds less to it than the week before.
Why a lake freezes at the top
In the chapter on temperature you met the odd behaviour of water. It is densest at 4 °C, and it expands again as it cools from 4 °C to 0 °C. In autumn the surface water of a lake cools and sinks, until the whole lake is near 4 °C. Water cooled below 4 °C is lighter, so it stays at the top. The surface reaches 0 °C first and freezes. Ice is lighter still, so it floats as a lid on the lake.
Now the ice itself decides how fast the lake freezes. Fig 4.1 shows the temperatures. The top face of the ice is at the air's temperature, say \(-\theta\). The bottom face touches water at its freezing point, 0 °C. So a temperature difference \(\theta\) lies across the ice, and heat flows upward through it into the cold air.
Where does that heat come from? Water under the ice can freeze only by giving out its latent heat of fusion, \(L_f\). That heat is released at the lower face of the ice, where the new ice forms. It cannot go down, because the water below is warmer. It must be conducted up through the ice already formed. So the ice grows only as fast as the ice above can carry the latent heat away.
Ice covers a pond on a night when the air is at \(-15\,°\text{C}\). In which direction does heat flow through the ice, and where does it come from?
Explain why the water just under the ice of a frozen lake is at 0 °C, and not colder.
The water touching the ice is at its freezing point. If it lost any more heat it would freeze and join the ice, rather than cool below 0 °C. So the layer next to the ice stays at 0 °C. Below it the water is warmer, up to 4 °C at the bottom, because water is densest at 4 °C and sinks there.
Under air at −10 °C, the ice on a lake grows from nothing to 10 cm in about 19 hours. The air stays at −10 °C. How long does the ice take to grow from 10 cm to 20 cm?
Press play on the figure above. The thickness races up at first and then crawls. The arrow above the ice is the heat leaving each square metre, and it shrinks from over 1000 W to about 110 W. The ice reaches 10 cm after 19 h, and 20 cm only after 77 h. To see why, follow one thin new layer of ice.
\[ dQ = kA\,\dfrac{\theta}{x}\,dt \]
Why. Fourier's law for ice of thickness \(x\) and area \(A\), with \(\theta\) across it, during a short time \(dt\).
\[ dQ = (\rho A\,dx)\,L_f \]
Why. in that time a layer of thickness \(dx\), of mass \(\rho A\,dx\), freezes at the lower face and gives out its latent heat.
\[ x\,dx = \dfrac{k\theta}{\rho L_f}\,dt \]
Why. the heat conducted up is the latent heat released; equate the two lines and cancel \(A\).
\[ t = \dfrac{\rho L_f}{2k\theta}\,(x_2^2 - x_1^2) \]
Read it. integrate from \(x_1\) to \(x_2\) on the left and from 0 to \(t\) on the right; \(\int x\,dx = x^2/2\).
The air over a lake stays at \(-10\,°\text{C}\). Take for ice \(\rho = 917\,\text{kg/m}^3\), \(k = 2.2\,\text{W/(m K)}\) and \(L_f = 3.34 \times 10^5\,\text{J/kg}\). Find the time for the ice to grow from 0 to 10 cm, and from 10 cm to 20 cm.
The constant first: \(\rho L_f/(2k\theta) = 917 \times 3.34 \times 10^5/(2 \times 2.2 \times 10) = 6.96 \times 10^6\,\text{s/m}^2\).
From 0 to 10 cm: \(t = 6.96 \times 10^6 \times 0.1^2 = 69\,600\,\text{s}\), about 19.3 h.
From 10 cm to 20 cm: \(t = 6.96 \times 10^6 \times (0.04 - 0.01) = 2.09 \times 10^5\,\text{s}\), about 58.0 h.
The second 10 cm takes three times as long as the first, though it is the same thickness of ice.
The air over the lake of the example falls to \(-20\,°\text{C}\) instead. How long does the ice now take to grow from 0 to 10 cm, in hours?
For the lake of the example, find how fast the ice is thickening at the moment it is \(5\,\text{cm}\) thick. Give the answer in millimetres per hour.
Josef Stefan is best known for his law of radiation, met in the next chapter. In 1889 he took up a quieter question: how fast does ice grow on the polar seas? Explorers wintering in the Arctic had measured the ice week by week. Stefan showed that the thickness grows as the square root of the time, just as this lesson finds. Problems in which a boundary between ice and water moves are still called Stefan problems.
Each centimetre takes longer
Step 3 of the derivation gives the rate of growth, \(dx/dt = k\theta/(\rho L_f x)\). The temperature difference across the ice stays at \(\theta\), but the thickness \(x\) grows. So the heat current through each square metre, \(k\theta/x\), falls. Less heat carried away each second means less water frozen each second. The ice insulates the water below it, and so it slows its own growth.
Starting from open water, \(x_1 = 0\), the formula becomes \(x^2 = (2k\theta/\rho L_f)\,t\). So the square of the thickness grows in step with time. A graph of \(x^2\) against \(t\) is a straight line through the origin. Ice 2 cm thick after 4 hours will be 4 cm thick after 16 hours, not 8.
The slope of that straight line is \(2k\theta/(\rho L_f)\). For the lake of the example \(2k\theta = 44\) and \(\rho L_f = 3.06 \times 10^8\), so the slope is \(1.44 \times 10^{-7}\,\text{m}^2/\text{s}\). Measuring the thickness of the ice on a few days, and plotting its square, is a way to find the conductivity of ice.
On a cold night, ice \(2\,\text{cm}\) thick forms on a pond in \(4\) hours. The air temperature stays the same. How long will the ice take to grow the next \(2\,\text{cm}\), in hours?
Ice \(5\,\text{cm}\) thick forms in \(8\) hours when the air is at \(-10\,°\text{C}\). What thickness would form in \(8\) hours at \(-20\,°\text{C}\), in centimetres?
The formula assumes that the top face of the ice is at the air's temperature. In a real lake, still air over the ice and snow lying on it add their own thermal resistance. The top of the ice is then warmer than the air, and the ice grows more slowly than the formula says. Wind that sweeps the snow away brings the real ice closer to the formula.
The air over Pangong Tso stays at about \(-10\,°\text{C}\). Using the example's numbers, 10 cm of ice takes about 19 hours. Estimate how many days the ice takes to become \(50\,\text{cm}\) thick.
Now return to Pangong Tso. Under air at −10 °C, the lake gains its first 10 cm of ice in less than a day, enough to bear a walker. The next 10 cm take almost two and a half days more, and half a metre takes about three weeks. The ice grows ever more slowly, because each new layer must send its latent heat up through all the ice above it.
- Thinking each centimetre of ice takes the same time to form.The time grows as \(x_2^2 - x_1^2\); each centimetre takes longer than the one before.
- Thinking the ice's latent heat flows down into the water.The water below is warmer; the latent heat is conducted up through the ice to the colder air.
- Using the thickness in centimetres in \(t = \rho L_f(x_2^2 - x_1^2)/(2k\theta)\).Put \(x\) in metres, then convert the seconds to hours at the end.