CR
Cryospheric Sciences

Snow, sub-zero temperatures for several days, and then back to long grey days of near-constant rain. A normal winter week in Gothenburg, south-west Sweden. Yet as I walk home in the evening, I can’t help but notice that piles of ice have survived. Using the equations that I normally need to investigate the demise of Greenland glaciers, I want to know: how hard can it be to melt this pile of ice by my door? In the image of this week, we will do the simplified maths to calculate this.

## Why should the ice melt faster when it rains?

The icy piles of snow are made of frozen freshwater. They will melt if they are in contact with a medium that is above their freezing temperature (0°C); in this case either the ambient air or the liquid rainwater.

How fast they will melt depends on the heat content of this medium. Bear with me now – maths is coming! The heat content of the medium per area of ice, $Q$, is a function of the density $\rho$ and specific heat capacity $c$ of the medium. Put it simply, the heat capacity is a measure of by how much something will warm when a certain amount of energy is added to it. $Q$ also depends on the temperature $T$ of the medium over the thickness $H$ of the boundary layer i.e. the thickness of the rain or air layer that directly impacts the ice.

Assuming that I have not scared you away yet, here comes the equation:

$Q = \rho c \int_{ice}^{H} T dz$

For liquid water (in this article, the rain): $\rho_{rain} \approx 1000 \: kg \: m^{-3}$, $c_{rain} \approx 3.9 \: kJ \: kg^{-1} \: K^{-1}$. For the ambient air: $\rho_{air} \approx 1.2 \: kg \: m^{-3}$, $c_{air} \approx 1.0 \: kJ \: kg^{-1} \: K^{-1}$. So we can plug those values into our equation to obtain the heat content $Q$ of the rain and of the air. We can consider the same temperature over the same $H$ (e.g. Byers et al., 1949), and hence we get $Q_{rain} \approx 3250 Q_{air}$.

Stepping away from the maths for a moment, this result means that the heat contained in the rain is more than 3000 times that of the ambient air. Reformulating, on a rainy day, the ice is exposed to 3000 times more heat than on a dry day!

The calculations have obviously been simplified. The thickness $H$ of the boundary layer is larger for the atmosphere than for the rain, i.e. larger than just a rain drop. At the same time, the rain does not act on the ice solely by bringing heat to it (this is the thermic energy), but also acts mechanically (kinematic energy): the rain falls on the ice and digs through it. For the sake of this blogpost however, we will keep it simple and concentrate on the thermic energy of the rain.

## How long will it take for the rain to melt this pile of ice then?

Promise, this will be the last equation of this blogpost! Reformulating the question, what is the melt rate of that ice? Be it for a high latitude glacier or a sad pile of snow on the side of a road, the melt rate $F_{melt}$ is the ratio of the heat flux from the rain $F_{Qrain}$ (or any other medium) over the heat needed to melt the ice. It indicates whether the rain brings enough heat to the ice surface to melt it, or whether the ice hardly feels it:

$F_{melt} = \frac{F_{Qrain}}{\rho_{ice}(L+c_{ice}\Delta T)}$

More parameters are involved

• $\rho_{ice} = 917 \: kg \: m^{-3}$ the density of the ice;
• $L = 335 \: kJ \: kg^{-1}$ the latent heat of fusion, defined as how much energy is needed to turn one kilogram of solid water into liquid water;
• $c_{ice} = 2.0 \: kJ \: kg^{-1} \: K^{-1}$ the heat capacity of the ice (see previous paragraph);
• $\Delta T$ the difference between the freezing temperature (0°C) and that of the interior of the ice (usually taken as -20°C).

But what is $F_{Qrain}$ I am glad you ask! This heat flux , i.e. $Q_{rain}/time$, is crucial: it not only indicates how much heat your medium has, but also how fast it brings it to the ice. After all, it does not matter whether you are really hot if you stay away from your target. I actually lied to you, here comes the final equation, defining the heat flux:

$F_{Qrain} = \rho_{rain}c_{rain}T_{rain}P$

We can consider that $T_{rain} \approx T_{air}$. We already gave $\rho_{rain}$ and $c_{rain}$ earlier. As for $P$, this is our precipitation, or how much water is falling on a surface over a certain time (given in mm/hour usually during weather bulletins). On 24th January 2018, as I was pondering why the ice had still not melted, my favourite weather forecast website indicated that $T_{air} = 5^{\circ}C$ (278.15 K) and $P = 1 \: mm/hour$.

Eventually putting all the numbers together, we obtain $F_{melt} \approx 3 \: mm/hour$. So that big pile on the picture that is about 1 m high will require constant rain for nearly 14 days – assuming that the temperature and precipitation do not change, and neglecting a lot of effects as already explained above. Or it would take just over one hour of the Wikipedia record rainfall of 300 mm/hour – but then ice would be the least of my worries.

The exact same equations apply to this small icy island, melted by the air and ocean [Credit: Monika Dragosics (distributed via imaggeo.egu.eu)]

In conclusion, liquid water contains a lot more heat than the air, but ice is very resilient. The mechanisms involved in melting ice are more complex than this simple calculation from only three equations, yet they are the same whether you are on fieldwork on an Antarctic ice shelf or just daydreaming on your way home.

### Other blogposts where ice melts…

Edited by Adam Bateson and Clara Burgard

(Dr) Céline Heuzé is an assistant professor in climatology at the University of Gothenburg, in southwest Sweden. She is a polar physical oceanographer primarily focussing on the transport of heat by global deep waters and their interaction with the cryosphere. To study them she uses mostly global coupled models, but also goes at sea to collect new measurements. She completed her PhD in 2015 from the University of East Anglia, UK, and before that studied hydrodynamics engineering in France. She tweets as @ClnHz and blogs on PolarFever when at sea.