In this video at 3:55, Mr. Wangchuck explains that the ice stupa doesn't melt because it has less surface area for a given volume compared to other figures in geometry (being a cone). I have theorized it happens due to it having a lesser exposed surface area and hence, allowing it to absorb less radiation and heat from the sun and give off less heat by Stefan Boltzmann's law. But is this the correct and complete explanation?

I still find this extremely weird and maybe even implausible that such a stupa made of ice can stand tall for a couple of months even if it has a lesser surface area for a given volume. The temperature in Ladakh in the month of March seems to be around 20 degrees celsius and about 13 degrees celsius in February (still way above the freezing point of water which is 0-degree celsius at 1 atm pressure) and he says the stupa stands from about January all the way to May! It honestly feels like all of it should have melted away come February and not have existed as late as June as shown in the video.

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    $\begingroup$ How did you arrive at the belief that it "should have melted away come February"? $\endgroup$
    – ACuriousMind
    Jun 7, 2020 at 14:58
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    $\begingroup$ Other factors to consider: reflectivity and emissivity of the ice, the ratio of surface area to volume, the average wind speed, average humidity, and the difference between average night temperatures and daytime temperatures. Perhaps you can write out the equations, answer your own question, and post the answer and equations here. $\endgroup$
    – S. McGrew
    Jun 7, 2020 at 15:00
  • $\begingroup$ I came to that conclusion since the temperatures exceed the melting point of ice by 10 degrees in february $\endgroup$ Jun 7, 2020 at 15:04
  • $\begingroup$ I don't really see how this ice stupa can withstand temperatures of more than 10 degrees for such a long period of time when a normal ice cube melts at even 1-degree centigrade! $\endgroup$ Jun 7, 2020 at 15:07
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    $\begingroup$ If you have a freezer and some plastic boxes in various sizes, you could make blocks of ice in a range of sizes. Time how long they take to melt and then try to extrapolate to larger sizes. $\endgroup$
    – badjohn
    Jun 7, 2020 at 15:13

3 Answers 3



An ice stupa such as the one in your video does not melt fast because it is huge. Even if you had an ice cube of equal mass, it would take about five months to melt. How did I arrive at this result?

The surface area to volume ratio of the shape matters, but the amount of energy absorbed by the stupa changes linearly with its area. As such, $t_{\text{melt}}\propto \text{area}/ \text{volume}$ for a fixed volume (and mass). In general, most shapes have this figure within the same order of magnitude and even an 'inefficient' shape would last a few months at such massive scale. Spheres have this value at $4.8$ while cubes are at 6.

Below I do the calculation and show that there is nothing mysterious happening with the stupa. I guess it would be a bit of a downer to say that in the youtube video though :)

Proper solution:

In general, answering these kinds of questions requires one to think about all the incoming and outgoing energy fluxes, and writing them as a differential equation. In this case we have two main pathways of heating:

  1. Convective heating due to the airflow around the stupa.
  2. Absorptive heating coming from the sunlight.

As an equation one can write this as:

$$\frac{dQ}{dt} = 4\pi R(t)^2k(T_o-T_s) + a\pi R(t)^2 P_{\textit{sun}} = \left(4\pi k (T_o - T_s) + a\pi P_{\textit{sun}} \right)R(t)^2, $$

where $R(t)$ is the radius of the spherical stupa (yes, this is the approximation we will be using), $(T_o-T_s)$ is a constant temperature difference between the outside at $7.5$C and the melting surface at $0$C, $a$ is the absorption of the ice at $50\%$ and $P_{\textit{sun}}$ is the average power of the sun rays impinging on the stupa at $250 W/m^2$.

Assuming all of these processes to be constant in time, and that all of them are melting the surface of the sphere uniformly (in reality one can guess the top to be melting away faster than the bottom).

The solution to the differential equation where the initial condition for a $2000$ ton stupa corresponds to $R(t=0)\approx 8 \text{m}$.

In order to solve the equation, we have to find the relationship between $\Delta{Q}$ and $R$. Writing the infinitesimal heat transfer $dQ$ required to melt a $dR$ thin layer of ice:

$$\frac{dQ}{dR} = 4\pi R^2 \rho_{\textit{ice}} C_{\text{lat}},$$

where $C_{\textit{lat}}$ is the latent heat of ice and $\rho_{\textit{ice}}$ is the density of ice. To make progress one writes the master flux equation as:

$$\frac{dQ}{dt} = \frac{dQ}{dR}\frac{dR}{dt} = 4\pi R^2 \rho_{\textit{ice}} C_{\text{lat}}\frac{dR}{dt} = \left(4\pi k (T_o - T_s) + a\pi P_{\textit{sun}} \right)R(t)^2. $$

The above equation simplifies to:

$$\frac{dR}{dt} = \frac{4\pi k (T_o - T_s) + a\pi P_{\textit{sun}}}{4\pi \rho_{\textit{ice}} C_{\text{lat}}},$$

yielding the solution:

$$R(t) = R(t=0) - \frac{4\pi k (T_o - T_s) + a\pi P_{\textit{sun}}}{4\pi \rho_{\textit{ice}} C_{\text{lat}}}t.$$

In order to estimate the time until the total sphere melts we set $R(t_{\text{melt}})=0$, and find that the ball melts at:

$$t_{\text{melt}}= R(t=0)\frac{4\pi \rho_{\textit{ice}} C_{\text{lat}}}{4\pi k (T_o - T_s) + a\pi P_{\textit{sun}}} \approx 2.3 \text{months}.$$

In many ways this is the lower estimate as I assumed:

  1. A stable temperature difference between the melting outer layer and the outside atmosphere to be 7.5 degrees celsius. In reality the average temperature gradient will be lower.
  2. I also assumed the whole stupa to be at $0$C initially, while it actually is at lower temperature and so more heat has to be involved to raise it to phase transition.
  3. I assumed moderately windy environment with $k= 50J/m^2 K$, with no wind this drops to $k= 10J/m^2 K$ and the melt time becomes about 9 months. It is a tricky parameter to guess, but usually it is between 10-100. I chose the average.

Nonetheless, this estimate shows that the stupa will be melting for a couple of months as asserted in the video. It is not much more than an order of magnitude estimate, but it proves that you do not need to look for any mysterious effects past usual heat transfers.

Reference for choice of temperatures: Temperatures in Leh throughout the year

enter image description here

  • $\begingroup$ Great work! I still think that, as seen by the average temperatures, that until march the stupa would be definitely at a lower temperature, but then again, this would give about 3 to 4 months. I think that the cone shape is also more efficient at retaining the "cold" because of convection, the top melts and protects the bottom as the cold air sinks, effectively shielding the stupa. $\endgroup$ Jun 14, 2020 at 9:35
  • $\begingroup$ It's a simple twenty minute model to show that a multi-month melting time is absolutely plausible. Of course there are many assumptions that could be much improved on if I was solving the equation numerically with real datasets. $\endgroup$
    – Akerai
    Jun 14, 2020 at 9:50

OK, it may go longer but go through it for once.

Look the stupa takes a month to form and can reach a maximum height of about $15$ to $50\,\mathrm m$ and can store up to $2$ million litres of water which is a lot.

If we calculate its mass it will be the same $2$ million kg.

Such a great mass at about $-30\mathrm{°C}$ needs a lot of energy to melt and flow in streams. If we calculate, it can reach to about 210 million joules of energy (with some approximations).

Now if we google about the average energy from the sun in Ladakh, we get this:

result (data in keh/m²/day)

We can take the average to be $6\,\mathrm{kW\,h/m^2/day}$ and this means $250\,\mathrm{J/m^2}$.

Now the curved surface area matters. Its radius ranges from $10\,\mathrm{m}$ to $20\,\mathrm{m}$ and height ranges from $20$ to $50\,\mathrm m$. So we can calculate the area as an average. And then calculate the total energy to the stupa. And from this we can get the result. I did the calculations and got $180$ days approximately. The days are calculated from January because it takes a month for the stupas to form to such a height.

So what the deciding factor in this case was it's mass and its temperature during the winters and rest of the months (shown in the figure).

Also the heat absorption happens from the warm air currents which blow in Ladakh mostly in the months of March to May also cause it to melt and as per Akerai's calculations it will take two months or more to melt it completely.

NOTE: if any query with the data then you can google it. Hope it helps.

Great thanks to Akerai for informing me about my mistake.

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    $\begingroup$ Sadly, the great approximation here is that all heating happens via absorption of sunlight. It's actually the case that the absorption of ice is pretty low. Most of the heat transfer is probably conductive - think warm air flowing past the stupa. I think your calculation misses the point. $\endgroup$
    – Akerai
    Jun 13, 2020 at 18:19

@Ankit Kumar provide a very insightful answer on the practical side of the phenomena. I would like to give some light on the theoretical reason behind Mr. Wangchuk's statement (and a simpler way to look at it).

The stupa is made up of ice, so the way that heat from the sun (or hot air) gets absorbed and transferred by the stupa is conductance. Let's say that the stupa absorbs heat $Q$ in an amount of time $t$. Now, the heat absorbed in that amount of time (rate of heat) is going to be: $$\frac{Q}{t} = k A \frac{T_o - T_i}{d}$$ where $A$ is the area of the stupa, $T_o$ is the temperature outside the stupa and $T_i$ is the temperature inside.

Now, as you increase the area of the stupa, the heat absorbed by it in a given amount of time goes up. enter image description here

You want to make sure that you maximize the mass of ice stored in the stupa and minimize the heat absorbed by it. This would make sure that it lasts longer. Since mass depends on volume ($M = \rho V$; $\rho$ is the density), you basically want to get the surface area to volume ratio as low as possible (to ensure lower surface area for absorption and conductance; and increase volume and mass).

Mathematically, the lowest surface area to volume ration is that of the sphere (or, as Mr. Wangchuk says, the hemisphere: you don't want to build a huge snowball :). But, when you consider how they construct the ice stupa, it makes sense: they release water from vertically-oriented pipe and when the water is exposed to the chilly climate of Ladakh, it freezes, eventually, forming a cone.

The cone has a surface area of $\pi r l$ (not considering the circular bottom surface) and a volume of $\frac{1}{3} \pi r^2 h$; that makes it one of the figures to have the lowest surface area to volume ratio (not as low as the sphere though).

To sum up: ice transfers heat through conducting it, and the way to minimize the heat absorbed and conducted is to minimize the surface area, and to maximize the amount of ice you need the maximum volume. So, a shape having less surface area for a large volume, in the case of this project the cone, is used. It is jus because of the conductance phenomena. You don't even need the Stephan-Boltzmann law for this one.


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