If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$, break into $N$ small droplets each of radius $r$. Estimate the drop in temperature.

If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a

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Concept understanding — Capillary Action

Capillary Action: The Physics of Water Defying Gravity

You've seen it happen a hundred times. Dip the corner of a paper towel into a spill, and watch the water climb upward into the towel — against gravity. Or take a thin glass tube (a capillary), put it in water, and the water rises inside it, higher than the outside surface. That's capillary action.

The name comes from capillus, Latin for "hair" — because the effect is strongest in tubes as thin as a hair.

The Intuition: Two Forces at War

Think of water molecules as tiny magnets. They attract each other strongly (cohesion), and they also stick to the walls of a container (adhesion). In a narrow tube, adhesion to the glass pulls the water upward along the walls. The water molecules, clinging to each other, drag the entire column up with them.

But gravity pulls down. The water rises only until the upward pull from adhesion is exactly balanced by the weight of the water column. That's the equilibrium height.

Note

If adhesion is stronger than cohesion (water-glass), the liquid rises. If cohesion is stronger (mercury-glass), the liquid is depressed — it falls below the outside level. Mercury doesn't wet glass.

The Precise Physics: What Determines the Height?

For a liquid that wets the tube (contact angle $\theta < 90^\circ$ ), the rise height $h$ is given by the Jurin's law:

h = \frac{2\gamma \cos\theta}{\rho g r}

Where:

$\gamma$ = surface tension of the liquid (N/m)
$\theta$ = contact angle between liquid and tube wall
$\rho$ = density of the liquid (kg/m³)
$g$ = acceleration due to gravity (9.8 m/s²)
$r$ = radius of the tube (m)

h = \frac{2\gamma \cos\theta}{\rho g r}

The key insight: $h$ is inversely proportional to $r$ . Halve the tube radius, and the water rises twice as high. That's why the effect is only noticeable in very narrow tubes — in a wide pipe, $h$ is negligible.

Why Does Surface Tension Pull Upward?

The surface tension $\gamma$ acts along the circumference of the water-air interface inside the tube. The total upward force is:

F_{\text{up}} = (2\pi r) \times \gamma \cos\theta

The weight of the water column (height $h$ , density $\rho$ ) is:

F_{\text{down}} = (\pi r^2 h) \times \rho g

Set them equal, cancel $\pi r$ , and you get Jurin's law.

Tip

For water in clean glass, $\theta \approx 0^\circ$ (perfect wetting), so $\cos\theta = 1$ . Then $h \approx \frac{2\gamma}{\rho g r}$ . For water at room temperature, $\gamma \approx 0.073$ N/m, so $h \approx \frac{0.015}{r}$ (with $r$ in metres). A tube of radius 0.1 mm gives a rise of about 15 cm.

Real-World Examples

Paper towels and sponges: The fibres form millions of tiny capillary channels. Water rises through them, soaking the towel. …

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Breaking a drop into many smaller droplets increases the total surface area, and creating extra surface area costs energy — energy that is drawn from the liquid's own internal (thermal) energy. So the droplets end up cooler than the original drop, by $\Delta T=\dfrac{3S}{\rho c}\left(\dfrac1r-\dfrac1R\right)$ .

Why breaking a drop costs energy

Every liquid surface carries extra energy (surface energy) because molecules at the surface have fewer neighbours to bond with than molecules in the bulk — this shortfall is exactly what surface tension $S$ measures, as energy per unit area. Creating more surface area means moving more molecules to that higher-energy surface state, and since the process happens quickly with no external heat source, that energy has to come from the liquid's own thermal energy — cooling it down.

Volume conservation

A drop of radius $R$ breaks into $N$ droplets of radius $r$ . Since the liquid is incompressible, total volume is conserved:

\frac43\pi R^3 = N\times\frac43\pi r^3 \;\Longrightarrow\; N=\frac{R^3}{r^3}

Change in surface area

Initial surface area: $A_i=4\pi R^2$ . Final total surface area of all droplets:

A_f = N\times4\pi r^2 = \frac{R^3}{r^3}\times4\pi r^2 = 4\pi\frac{R^3}{r}

Increase in surface area:

\Delta A = A_f-A_i = 4\pi R^2\left(\frac{R}{r}-1\right)

Surface energy increase

Surface tension $S$ is the energy per unit area of surface, so the surface energy increases by

\Delta U_{surface} = S\Delta A = 4\pi SR^2\left(\frac{R}{r}-1\right)

This energy is drawn from thermal energy …

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Q.If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius RRR, break into NNN small droplets each of radius rrr. Estimate the drop in temperature.

Why breaking a drop costs energy

Volume conservation

Change in surface area

Surface energy increase

This energy is drawn from thermal energy …

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Q.If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$ , break into $N$ small droplets each of radius $r$ . Estimate the drop in temperature.

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