Cambridge Syllabus Notes

Cooling of an Object in Contact with an Evaporating Liquid

Key Concepts

When a liquid evaporates, it takes energy from its surroundings – this is called latent heat of evaporation.
The energy required to change a liquid into a gas is removed from the liquid itself and from any object in contact with it.
Because the liquid loses energy, its temperature drops, cooling the object it touches.
Evaporation is a phase change that occurs at a constant temperature for a pure substance.

Analogy: Sweat Cooling the Body ???

Think of your skin as a hot surface and sweat as a thin layer of liquid. As sweat evaporates, it removes heat from your skin, just like a cooling blanket that takes the warmth away. This is why you feel cooler when you’re sweating in a hot room.

Example: Ice in a Glass of Water ❄️💧

Place a block of ice in a glass of warm water. The ice melts, absorbing heat from the water. If the water is still warm, the surface of the ice will start to evaporate, taking additional heat away. The result? The water temperature drops faster than it would from melting alone.

Mathematical Expression

The heat removed by evaporation can be calculated as: $$Q = m \, L_v$$ where:

$m$ = mass of liquid that evaporates (kg)
$L_v$ = latent heat of evaporation (J kg⁻¹)

This heat $Q$ is taken from the surrounding object, lowering its temperature.

Table: Latent Heat of Evaporation for Common Liquids

Liquid	$L_v$ (kJ kg⁻¹)
Water	2260
Ethanol	840
Acetone	242

Exam Tips 📚

Remember that evaporation occurs at a constant temperature for a pure liquid.
Use the formula $Q = mL_v$ to calculate the heat removed by evaporation.
When asked about cooling effects, explain that the energy taken for the phase change comes from the object in contact.
Use real‑world examples (sweat, ice in water) to illustrate the concept clearly.
Check units carefully – $L_v$ is usually given in kJ kg⁻¹, so convert if necessary.

Explain the cooling of an object in contact with an evaporating liquid