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

2.2.3 Melting, Boiling and Evaporation

Objective

Explain why an object placed in contact with an evaporating liquid becomes cooler, and recall the basic facts about melting, boiling, condensation and solidification required by the Cambridge IGCSE/A‑Level syllabus (S 2.2.3 1‑5, 6‑9).

Quick Checklist – the four phase‑change processes

• Melting (solid → liquid)
• Freezing (liquid → solid)
• Boiling (liquid → gas, bulk)
• Condensation (gas → liquid)

Key Concepts

• All phase changes involve a latent heat – the amount of energy required to change the state of 1 kg of a substance without changing its temperature.
• Latent heat of fusion \(L_f\) (solid ↔ liquid) and latent heat of vapourisation \(L_v\) (liquid ↔ gas) are material properties.
• During a phase change the energy is taken from or released to the surroundings, so the temperature of the material (or any object in thermal contact) can remain constant or change.

Phase‑change Data for Water (required recall – S 2.2.3 2)

Phase change	Temperature (1 atm)	Latent heat (J kg⁻¹)
Melting / Freezing	0 °C	\(L_f \approx 3.34 \times 10^{5}\)
Boiling / Condensation	100 °C	\(L_v \approx 2.26 \times 10^{6}\)

Mnemonic (flash‑card style)

Fusion ≈ 3.3 × 10⁵ J kg⁻¹, Vapour ≈ 2.3 × 10⁶ J kg⁻¹ – remember the “3‑5‑3” and “2‑6‑2” patterns.

Overview of the Four Phase Changes (S 2.2.3 1‑5)

Process	Direction	Energy flow	Temperature change
Melting	Solid → Liquid	Absorbs \(L_f\) from surroundings	No change – stays at 0 °C until all solid has melted
Freezing	Liquid → Solid	Releases \(L_f\) to surroundings	No change – stays at 0 °C until all liquid has solidified
Boiling	Liquid → Gas (through the bulk)	Absorbs \(L_v\) from surroundings	No change – stays at 100 °C while boiling continues
Condensation	Gas → Liquid	Releases \(L_v\) to surroundings	No change – stays at 100 °C while condensation continues

Boiling vs Evaporation (S 2.2.3 6‑7)

Aspect	Boiling	Evaporation
Where it occurs	Through the whole volume once vapour pressure = external pressure.	Only at the free surface.
Temperature	Fixed at the boiling point (100 °C for water at 1 atm).	Can occur at any temperature; rate rises with temperature.
Energy source	Heat supplied to the bulk of the liquid.	Energy taken from surface molecules and, if the surface cools, from any object in thermal contact.
Syllabus wording	“Bulk phase change at a fixed temperature.”	“Escape of more‑energetic particles from the surface.”

Key exam sentence: “Boiling = bulk change at a fixed temperature; evaporation = surface change that can occur at any temperature.”

Why Evaporation Cools an Object (S 2.2.3 5)

• At the liquid surface only the most energetic molecules have enough kinetic energy to overcome intermolecular forces and escape as vapour.
• Each escaping molecule requires the latent heat of vapourisation \(L_v\). That energy is taken from the liquid itself and, if the liquid is in contact with another body, from that body as well.
• Consequently the temperature of the liquid and of the contacting object fall – the familiar cooling effect.

Energy Balance for Evaporation

For a mass \(m_{\text{evap}}\) that evaporates, the energy removed from the surroundings is

\[ Q = m_{\text{evap}}\,L_v \]

Typical sources of this energy (ordered from most to least important):

1. Heat stored in the liquid.
2. Heat conducted from any object in direct contact with the liquid.
3. Heat supplied by the surrounding air (if the surface is exposed).

Temperature Drop of the Contacting Object

If an object of mass \(m_{\text{obj}}\) and specific heat capacity \(c_{\text{obj}}\) supplies the heat, its temperature change over a short interval \(\Delta t\) is

\[ \Delta T = -\frac{Q}{m_{\text{obj}}\,c_{\text{obj}}} = -\frac{m_{\text{evap}}\,L_v}{m_{\text{obj}}\,c_{\text{obj}}} \]

The negative sign shows that the object loses internal energy and therefore cools.

Important Caveat (higher‑level problem solving)

The above equations assume **no heat loss to the surrounding air or container**. In real situations a small fraction of the absorbed heat may be replenished from the environment, reducing the net temperature drop.

Factors that Influence the Rate of Evaporation (and thus the cooling) (S 2.2.3 8‑9)

Factor	Effect on evaporation rate	Resulting cooling
Surface area of the liquid	More area → more molecules can escape (doubling the area roughly doubles the rate).	Higher heat draw → stronger cooling.
Temperature of the liquid (or object)	Higher kinetic energy → faster escape.	More \(L_v\) taken per unit time → greater cooling.
Air movement (wind)	Removes the saturated vapour layer, maintaining the concentration gradient.	Continues rapid evaporation → sustained cooling.
Relative humidity	Low humidity → larger gradient between surface vapour pressure and ambient.	Faster evaporation → larger temperature drop.
Ambient pressure	Lower pressure reduces the energy needed for molecules to escape.	Evaporation can be vigorous even at lower temperatures, still extracting heat from the object.

Exam tip: When answering a multi‑part question, list the factors in order of importance for the given situation (e.g., “surface area → temperature → airflow → humidity → pressure”).

Practical Example – Sweating (water evaporating from skin)

Assume 1 g of water evaporates from a sweaty forearm in 10 s.

\[ Q = 0.001\;\text{kg}\times 2.26\times10^{6}\;\text{J kg}^{-1}=2260\;\text{J} \]

If the skin region has a mass of 40 g and an effective specific heat capacity of \(c\approx3500\;\text{J kg}^{-1}\text{K}^{-1}\), the temperature fall is

\[ \Delta T = -\frac{2260}{0.040\times3500}\approx -16^{\circ}\text{C} \]

The calculated drop explains why we feel a marked cooling sensation when sweat evaporates.

Summary

• Melting, boiling, condensation and solidification are phase changes that involve the absorption or release of latent heat (\(L_f\) or \(L_v\)) without a temperature change – “melting and boiling occur without a change in temperature”.
• For water the memorised values are: 0 °C (melting), 100 °C (boiling), \(L_f\approx3.3\times10^{5}\) J kg⁻¹, \(L_v\approx2.3\times10^{6}\) J kg⁻¹ (use the mnemonic above).
• Evaporation is a surface phenomenon; the most energetic molecules escape, taking \(L_v\) from the liquid and any object in contact, producing cooling.
• The rate of evaporation – and therefore the magnitude of cooling – depends on surface area, temperature, airflow, humidity and ambient pressure. Remember to order these factors when answering exam questions.
• Condensation and solidification are the reverse processes; they release the same amount of latent heat back to the surroundings.