Ice Pellets

Use areas on an atmospheric temperature graph to find how much of a falling ice pellet refreezes, then calculate the small temperature lag of liquid water.

4 Nov 2026 About 5 min read

Heat Transfer

JEE Advanced

Physics Olympiad

Before you begin

You should know latent heat, specific heat capacity and areas of triangles and trapeziums. The final temperature calculation uses a first-order differential equation and an exponential solution.

A falling ice pellet can melt in a warm layer and freeze again in colder air below. We can find how much refreezes without knowing its mass or fall speed. The key is to compare areas on the air-temperature graph. After melting is complete, the same heat-transfer law tells us why liquid water remains slightly cooler than the surrounding air.

The problem

A small spherical ice pellet falls at a constant terminal speed \(v\). Its mass \(m\) and size stay constant. Its temperature is uniform throughout, but may differ from the air temperature. Let \(h\) be height above the ground, \(\theta_a\) the air temperature and \(\theta_d\) the droplet temperature.

A qualitative illustration of one spherical ice pellet descending from a warm air layer into colder air. Layer thicknesses, pellet size and phase fractions are not quantitative. — Figure 1. A qualitative view of the falling pellet and warm and cold air layers. Layer thicknesses, pellet size and phase fractions are not represented quantitatively.

Join successive tabulated points by straight lines in each profile.

Height \(h\) in km	4	2	1	0
With inversion \(\theta_a\) in \(^{\circ}\mathrm{C}\)	0	4	\(-4\)	\(-2\)
Without inversion \(\theta_a\) in \(^{\circ}\mathrm{C}\)	0	4	6	8

At \(h_A=4\,\mathrm{km}\) the pellet is pure ice at \(0^{\circ}\mathrm{C}\). In the profile with inversion, it just finishes melting at the lower zero-temperature crossing \(h_B\). Ice and water coexist at \(0^{\circ}\mathrm{C}\), and freezing begins without supercooling or delay.

Heat enters the droplet at the rate

\[ \dot Q=H(\theta_a-\theta_d). \tag{1}\]

The coefficient \(H\), in \(\mathrm{W\,K^{-1}}\), is constant and has the same value in both profiles. Assume constant air density. Neglect evaporation, condensation, size changes and aerodynamic or viscous heating.

Use \(c_w=4.2\,\mathrm{kJ\,kg^{-1}\,K^{-1}}\) for water, \(c_i=2.1\,\mathrm{kJ\,kg^{-1}\,K^{-1}}\) for ice and \(L=334\,\mathrm{kJ\,kg^{-1}}\) for the latent heat of fusion.

Part a With inversion, what fraction of the original mass refreezes before reaching the ground?

Part b For the same droplet, speed and coefficient, find the ground-level droplet temperature without inversion. Account for its finite thermal response.

Turn graph areas into heat

The lower zero crossing is halfway between 1 km and 2 km, so \(h_B=1.5\,\mathrm{km}\). During melting or freezing the droplet stays at \(0^{\circ}\mathrm{C}\). Heat changes its phase rather than its temperature.

Figure 2. The solid curve is the inversion profile; the dashed curve is the alternative. Shaded areas refer only to the solid curve. They measure temperature difference times height and determine the heat transferred while the droplet is at its melting point.

The warm region is a triangle. Its area is

\[ A_+=\frac12(2.5)(4)=5.0\,\mathrm{K\,km}. \tag{2}\]

The cold region consists of a triangle and a trapezium:

\[ A_-=\frac12(0.5)(4)+\frac{4+2}{2}(1)=4.0\,\mathrm{K\,km}. \tag{3}\]

As the droplet descends, \(dh/dt=-v\). The time spent in a height interval is its vertical distance divided by \(v\). Integrating equation (1) while the droplet is at its melting point gives the positive heat magnitudes

\[ Q_+=\frac{H}{v}A_+,\qquad Q_-=\frac{H}{v}A_-. \tag{4}\]

Use \(A_+=5000\,\mathrm{K\,m}\) and \(A_-=4000\,\mathrm{K\,m}\) when \(v\) is in metres per second. The stated melting condition fixes the unknown combination of parameters:

\[ \frac{H}{v}A_+=mL. \tag{5}\]

If a fraction \(\eta\) refreezes, it releases \(\eta mL\). Dividing the two heat balances eliminates the unknown mass, speed and coefficient:

\[ \eta=\frac{Q_-}{mL}=\frac{A_-}{A_+}=\frac45=0.80. \tag{6}\]

Eighty per cent refreezes and twenty per cent remains liquid. Since freezing is incomplete, the droplet reaches the ground at \(0^{\circ}\mathrm{C}\) even though the air there is at \(-2^{\circ}\mathrm{C}\).

Find where melting ends without inversion

Without inversion the air warms at \(b=2.0\,\mathrm{K\,km^{-1}}\) as the droplet falls. The heat needed to melt the same pellet is unchanged. The accumulated area must again reach \(A_+\).

Let \(z_m\) be the distance descended from \(h_A\) when melting finishes. Then

\[ \frac12 b z_m^2=A_+,\qquad z_m=\sqrt{\frac{2A_+}{b}}=\sqrt5\,\mathrm{km}. \tag{7}\]

The corresponding height is

\[ h_m=h_A-z_m=(4-\sqrt5)\,\mathrm{km}\simeq1.764\,\mathrm{km}. \tag{8}\]

The droplet is still at \(0^{\circ}\mathrm{C}\) at that instant. The air temperature is

\[ \theta_m=2\sqrt5\,^{\circ}\mathrm{C}\simeq4.472\,^{\circ}\mathrm{C}. \tag{9}\]

Subsequent heat input warms liquid water.

Follow the temperature lag

Let \(x\) be the distance descended after melting finishes. Use metres from here, so \(b=0.0020\,\mathrm{K\,m^{-1}}\). The air temperature is

\[ \theta_a(x)=\theta_m+bx. \tag{10}\]

The heat-capacity equation is

\[ mc_w v\frac{d\theta_d}{dx}=H(\theta_a-\theta_d). \tag{11}\]

Define the thermal response length

\[ \ell=\frac{mc_w v}{H}. \tag{12}\]

Using the melting balance in equation (5) gives

\[ \ell=\frac{c_w A_+}{L}=\frac{4.2}{334}(5000)\,\mathrm{m}\simeq62.87\,\mathrm{m}. \tag{13}\]

The temperature lag \(\delta=\theta_a-\theta_d\) therefore obeys

\[ \frac{d\delta}{dx}=b-\frac{\delta}{\ell}. \tag{14}\]

Subtract the constant particular solution \(b\ell\). If \(y=\delta-b\ell\), then \(dy/dx=-y/\ell\), so \(y=y(0)e^{-x/\ell}\). At melting completion, \(\delta(0)=4.472136\,\mathrm{K}\). Hence

\[ \delta(x)=b\ell+[\delta(0)-b\ell]e^{-x/\ell}. \tag{15}\]

At the ground, \(x=h_m=1763.932\,\mathrm{m}\) and \(x/\ell\simeq28.05\). The exponential contribution is only \(2.84\times10^{-12}\,\mathrm{K}\). Thus

\[ \delta_g\simeq b\ell=0.1257485\,\mathrm{K}. \tag{16}\]

The ground-level air temperature is \(8.0^{\circ}\mathrm{C}\), so

\[ \theta_{d,g}=8.0^{\circ}\mathrm{C}-\delta_g\simeq7.87^{\circ}\mathrm{C}. \tag{17}\]

At the precision of the given data, the answer is approximately \(7.9^{\circ}\mathrm{C}\). Extra digits describe the ideal model, not measurement precision.

Without inversion, the droplet stays at zero degrees Celsius until height 1.764 kilometres, then warms below the air curve to 7.874 degrees at ground. An inset resolves the small temperature lag near ground. — Figure 3. Calculated air and droplet temperatures without inversion. The height axis decreases in the direction of descent. The droplet remains at its melting point until melting ends, then approaches the warming air with a small lag. The inset resolves the difference near ground.

After the initial transient, air and water warm at almost the same rate. The persistent lag \(b\ell\) provides the heat needed to warm the water. If \(\tau=mc_w/H\) is the response time, then \(\ell=v\tau\).

Read the result physically

Quantity	Result
Refrozen fraction with inversion	\(0.80\) of the original mass
Ground temperature with inversion	\(0^{\circ}\mathrm{C}\) with ice and water present
Melting height without inversion	\(1.764\,\mathrm{km}\)
Thermal response length	\(62.87\,\mathrm{m}\)
Ground temperature without inversion	\(7.87^{\circ}\mathrm{C}\) or about \(7.9^{\circ}\mathrm{C}\)

Internal uniformity does not imply equality with the air. High conductivity removes temperature differences inside the droplet. Heat exchange across its boundary still takes time, so its uniform temperature may differ from the air temperature.

Use latent heat only during phase change. Once all ice has melted, use \(mc_w\,d\theta_d\) for the heat input. The supplied ice heat capacity \(c_i\) is unused because the initial ice is already at its melting point.