Why Water Climbs Up a Narrow Glass Tube on Its Own

Water beads into round drops, climbs up a thin straw and slowly cools in a cup as heat flows away. Surface tension explains the first two; calorimetry and heat transfer explain the rest.

This lesson covers surface tension and excess pressure, capillary rise, calorimetry, and Newton's law of cooling with conduction, convection and radiation.

**Surface tension is the force per unit length along a liquid surface, equal to the surface energy per unit area, and it makes the pressure inside a curved surface higher than outside — by $\frac{2S}{R}$ for a drop and $\frac{4S}{R}$ for a soap bubble.

Definitions:

- Surface tension** $S=\frac{F}{L}$, in N m${}^{-1}$; molecules at the surface are pulled inwards, so the surface acts like a stretched membrane
- Surface energy — the work needed to increase the area by one unit, numerically equal to S, so $W=S\Delta A$

Excess pressure:

- Liquid drop or air bubble in a liquid (one surface): $\Delta p=\frac{2S}{R}$
- Soap bubble (two surfaces): $\Delta p=\frac{4S}{R}$

Worked example. A water drop of radius 1.0 mm, with $S=0.072$ N m${}^{-1}$, has

$$\Delta p=\frac{2\times 0.072}{1.0\times 10^{-3}}=144\text{ Pa}$$

An everyday example. Drops of oil spitting from a hot tadka are round, because surface tension pulls each drop into the smallest possible surface area.

The substance. Smaller drops have higher internal pressure — excess pressure varies as $1/R$, which is why tiny bubbles are hard to form.

**A liquid rises in a narrow tube to a height $h=\frac{2S\cos \theta }{r\rho g}$, where $\theta$ is the angle of contact, so the rise is greater in narrower tubes and for liquids that wet the tube.

Angle of contact. The angle between the liquid surface and the solid, measured inside the liquid — acute for water on clean glass, which rises, and obtuse for mercury, which is pushed down.

Derivation.** The upward pull of surface tension round the rim balances the weight of the raised column:

$$2\pi rS\cos \theta =\pi r^{2}h\rho g\Rightarrow h=\frac{2S\cos \theta }{r\rho g}$$

Worked example. Water with $S=0.072$ N m${}^{-1}$ and $\theta =0^{\circ }$ rises in a glass tube of radius 0.50 mm:

$$h=\frac{2\times 0.072\times 1}{0.50\times 10^{-3}\times 1000\times 9.8}=0.029\text{ m}=2.9\text{ cm}$$

An everyday example. Oil climbing the cotton wick of a diya and kerosene rising in a lantern wick are both capillary action.

The substance. A tube shorter than h does not overflow — the liquid stops at the top and its surface simply becomes flatter, so capillarity cannot power perpetual motion.

**Calorimetry uses the principle that, in an insulated system, heat lost by hot bodies equals heat gained by cold ones, with $Q=mc\Delta T$ for a temperature change and $Q=mL$ for a change of state.

Key relations:

- Specific heat capacity**: $Q=mc\Delta T$, with $c=4200$ J kg${}^{-1}$ K${}^{-1}$ for water
- Latent heat: $Q=mL$ at constant temperature, with $L_f=3.34\times 10^5$ J kg${}^{-1}$ for ice

Worked example 1 — mixing water. 0.20 kg of water at 80 °C is mixed with 0.30 kg at 20 °C:

$$0.20\times 4200\times (80-T)=0.30\times 4200\times (T-20)\Rightarrow T=44\text{ °C}$$

Worked example 2 — melting ice. Turning 0.50 kg of ice at 0 °C into water at 20 °C needs

$$Q=0.50\times 3.34\times 10^5+0.50\times 4200\times 20=1.67\times 10^5+4.2\times 10^4=2.09\times 10^5\text{ J}$$

An everyday example. Ice cubes in a glass of nimbu pani cool it far more than the same mass of cold water would, because melting ice absorbs latent heat without warming up.

The substance. Heat and temperature are different — a bucket of warm water holds more heat than a cup of boiling water, though its temperature is lower.

Newton's law of cooling states that a body's rate of heat loss is proportional to the difference between its temperature and its surroundings', for small differences, while heat travels by conduction in solids, convection in fluids and radiation through space.

Newton's law of cooling:

$$-\frac{dT}{dt}=k(T-T_s)\text{average form: }\frac{T_1-T_2}{t}=k\left(\frac{T_1+T_2}{2}-T_s\right)$$

Worked example. A body cools from 80 °C to 60 °C in 5.0 min in a room at 20 °C. How long does it take to cool from 60 °C to 40 °C?

- First interval: $\frac{20}{5.0}=k(70-20)$, so $k=0.080$ min${}^{-1}$
- Second interval: $\frac{20}{t}=0.080\times (50-20)=2.4$, so $t=8.3$ min

Modes of heat transfer:

- Conduction — through a solid by particle collisions, with no bulk movement; rate $\frac{KA(T_1-T_2)}{L}$
- Convection — by bulk movement of a fluid, as in sea breezes and land breezes
- Radiation — by electromagnetic waves, needing no medium, as sunlight crosses space

An everyday example. Steel vessels with copper bottoms and wooden handles use a good conductor to spread heat and a poor one to keep hands safe.

The substance. Cooling slows as a body nears room temperature — the rate depends on the temperature difference, so the second 20 °C drop takes longer than the first.