Subfield: Thermodynamics

## 1 Ice pellets

An interesting weather phenomenon can occur when the temperature profile in the atmosphere shows an inversion. The solid blue line in figure 1 shows such a temperature profile. The inversion occurs at heights between $1 \mathrm{~km}$ and $2 \mathrm{~km}$.

Under these conditions snow falling through the atmosphere (partially) melts in the warmer layer and (partially) freezes again before reaching the ground in the form of "ice pellets".

<img_4520>

Figure 1: Atmospheric temperature $T$ vs. height $h$ above the ground.

Assume that a small, spherical ice droplet almost completely melts while falling through the atmospheric layer between $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$ where the temperature is above freezing point.
Context question:
a. Determine the mass fraction of the droplet that freezes before reaching the ground.
Context answer:
\boxed{$\frac{4}{5}$}


Question: b. Find, as precisely as possible, the temperature of the droplet at ground level if there were no inversion and the temperature profile followed the dashed line below a height of $2 \mathrm{~km}$.
Neglect evaporation, condensation and size changes of the droplet. Assume that water and ice have very high thermal conductivity and that the density of the atmosphere is constant with height.

Use $c_{\text {water }}=4.2 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ for the the specific heat of water and $c_{\text {ice }}=2.1 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ for that of ice. The specific latent heat for the melting of ice is $L=334 \mathrm{~kJ} \mathrm{~kg}^{-1}$.

Solution:
If the temperature profile follows the dashed line the droplet will completely melt and the heat transferred from the atmosphere will heat it up. Since the latent heat of melting is much bigger than the specific heat of water times some degrees of temperature variation the temperature of the liquid droplet will closely follow the temperature of the atmosphere eventually. Its temperature at ground level should therefore be close to $8{ }^{\circ} \mathrm{C}$.

For a better estimate let us introduce a new coordinate $x$ whose origin is at the height, where the droplet is completely molten (somewhat higher than $\left.h_{\mathrm{B}}\right)$, and which is oriented downwards. For the change in temperature of the droplet we then have

$$
m c_{\text {water }} \frac{\mathrm{d} T_{\mathrm{d}}}{\mathrm{d} t}=m c_{\text {water }} v \frac{\mathrm{d} T_{\mathrm{d}}}{\mathrm{d} x}=\kappa\left(T-T_{\mathrm{d}}\right), \tag{5}
$$

where the temperature of the atmosphere is given by $T(x)=T(x=0)+2.0^{\circ} \mathrm{Ckm}^{-1} x=: T(x=0)+b x$.
For the difference $\Delta T=T-T_{\mathrm{d}}$ between the atmospheric temperature and the droplet temperature we have:

$$
\frac{\mathrm{d} \Delta T}{\mathrm{~d} x}=b-\frac{\mathrm{d} T_{\mathrm{d}}}{\mathrm{d} x}=b-\frac{\kappa}{m c_{\text {water }} v} \Delta T . \tag{6}
$$

This differential equation is solved by

$$
\Delta T=b x_{0}+\text { const } \cdot \exp \left(-x / x_{0}\right), \tag{7}
$$

where $x_{0}=m c_{\text {water }} v / \kappa$. From equation (2) we find that $\frac{m v}{\kappa}=\frac{A}{L}$ such that $x_{0}=\frac{c_{\text {water }}}{L} A \approx 0.063 \mathrm{~km}$. Therefore the exponential factor in the above equation is negligible at ground level (the constant is close to $4^{\circ} \mathrm{C}$ ) and we arrive at a temperature difference between the droplet temperature and the atmospheric temperature at ground level of

$$
\Delta T \approx 0.13^{\circ} \mathrm{C}, \quad \text { and therefore } \quad T_{\mathrm{d}} \approx 7.9^{\circ} \mathrm{C} \text {. } \tag{8}
$$

Another phenomenon, which effectively shifts atmosphere temperature as it is "felt" by the droplet is due to viscous dissipation. It can be estimated by equating the droplet potential energy loss rate $m g v$ to thermal power carried away by the circumfluent air $\kappa \Delta T^{*}$. This gives $\Delta T^{*} \sim m g v / \kappa=g A / L=0.17^{\circ} \mathrm{C}$. If both corrections were taken into account, then the droplet temperature before hitting the ground would again be $T_{\mathrm{d}} \approx 8.0^{\circ} \mathrm{C}$.

Final answer: $8$ $^{\circ} \mathrm{C}$