Published by Patrick Mutisya · 14 days ago
Explain why an object placed in contact with an evaporating liquid becomes cooler.
For a mass \$m\$ of liquid that evaporates, the energy taken from the surroundings is
\$Q = m \, L_v\$
where \$L_v\$ is the latent heat of vaporisation (J kg⁻¹). This energy is supplied by:
When an object touches an evaporating liquid, heat flows from the object into the liquid to replace the energy lost by the evaporating molecules. The rate of heat loss from the object can be expressed as
\$\dot Q{\text{object}} = \frac{m{\text{evap}} \, L_v}{t}\$
where \$m_{\text{evap}}\$ is the mass of liquid evaporated in time \$t\$. Because the object supplies this heat, its internal energy decreases, leading to a drop in temperature according to
\$\Delta T = -\frac{\dot Q{\text{object}} \, \Delta t}{m{\text{obj}} \, c_{\text{obj}}}\$
(\$c_{\text{obj}}\$ = specific heat capacity of the object).
| Factor | How it Affects Evaporation Rate | Resulting Cooling Effect |
|---|---|---|
| Surface area of liquid | Greater area → more molecules can escape → higher \$m_{\text{evap}}\$ | Increased heat drawn from the object → stronger cooling |
| Temperature of liquid | Higher temperature → higher molecular kinetic energy → faster evaporation | More heat required for vaporisation → greater cooling |
| Air movement (wind) | Removes saturated vapour layer → maintains concentration gradient | Continues rapid evaporation → sustained cooling |
| Relative humidity | Low humidity → larger concentration gradient → faster evaporation | More heat taken from object → larger temperature drop |
| Pressure above liquid | Lower pressure reduces the energy needed for molecules to escape | Evaporation can occur at lower temperatures, still drawing heat from the object |
When a small amount of rubbing alcohol (\$L_v \approx 2.2 \times 10^6\ \text{J kg}^{-1}\$) is spread on the skin, the liquid evaporates rapidly. Assuming 0.5 g of alcohol evaporates in 5 s, the heat taken from the skin is
\$Q = 0.0005\ \text{kg} \times 2.2 \times 10^6\ \text{J kg}^{-1} = 1100\ \text{J}\$
If the skin region has a mass of 30 g and a specific heat capacity of \$c \approx 3500\ \text{J kg}^{-1}\text{K}^{-1}\$, the temperature fall is
\$\Delta T = -\frac{1100\ \text{J}}{0.03\ \text{kg} \times 3500\ \text{J kg}^{-1}\text{K}^{-1}} \approx -10.5^\circ\text{C}\$
This calculation illustrates why the skin feels noticeably cooler.