Evaporative Cooling — Latent Heat at Work
A few grams of water absorb a startling amount of heat by evaporating. That single trick powers sweating, swamp coolers, and steam off your tea.
Show how phase changes compare to ordinary heating
The trick that breaks intuition
Why does sweat cool you off? Why does a swamp cooler work? Why is your tea steaming as it cools? They are all the same trick: when liquid water turns into vapor, each escaping molecule carries away a huge amount of energy compared to its mass. That energy is pulled out of whatever the water was sitting on, leaving it cooler.
The energy required is captured in a single equation:
Source: OpenStax College Physics 2e §14.3 — Phase Change and Latent Heat (eqn 14.18). is the latent heat of vaporization — the energy needed per unit mass to change liquid into vapor at constant temperature.
For water, the OpenStax reference gives:
- kJ/kg at the boiling point (100 °C)
- kJ/kg at body temperature (37 °C) — slightly higher because more energy is needed to break free from a lower-energy state
This concept linearly interpolates between those two anchor points, which matches steam tables to within a percent over the everyday temperature range.
The numbers are wild
Compare to ordinary "sensible" heating. Raising 1 kg of water by 1 °C takes about 4.186 kJ. Evaporating that same 1 kg of water — at no temperature change — takes around 2,400 kJ, or roughly 580× as much energy. Open the comparison view above to see fusion (melting ice) sitting between them; even melting takes 80× more than a 1-degree heat-up.
This is why the numbers in the visualization are so disproportionate. Five grams of sweat — about a single drop — absorbs roughly 12,000 J as it evaporates. That's the same energy as heating a litre of water from room temperature to 40 °C. From a single droplet.
Why this is the most efficient cooling mechanism in everyday life
The OpenStax §14.3 perspiration example puts it bluntly: "perspiration from the skin requires a heat input of 2428 kJ/kg... an effective cooling mechanism in hot weather." No machine you can build at the same scale and weight is more effective at moving heat per gram of working fluid. That's why:
- Athletes drink water — it's the working fluid for their built-in cooling system.
- Swamp coolers (evaporative coolers) work brilliantly in dry climates and pointlessly in humid ones — humid air cannot accept any more vapor, so collapses.
- Steam burns are far worse than hot-water burns at the same temperature: when steam condenses on your skin, it releases the same enormous in reverse, dumping ~2,300 kJ per kilogram into your skin in addition to whatever sensible heat the hot water also delivers.
What this concept doesn't model
Real evaporative cooling rate depends on a mass transfer — how quickly water can leave the surface — which in turn depends on humidity, air motion, and the surface's vapor pressure. We're not modelling that here; we're capturing only the energy bookkeeping: if mass of water evaporates from this surface, how much heat got pulled from it?
That's why the slider is m_w (mass available to evaporate) rather than time. The fraction f_evap lets you encode "only half of it actually got away" if humidity or pooling kept some water on the surface. The result is an idealized upper bound on cooling.
- Latent heat of vaporization of water — linear fit through OpenStax anchor points (37 °C → 2428 kJ/kg, 100 °C → 2256 kJ/kg)
- Latent heat in friendlier units (kJ/kg)
- L_v_kJ \;=\; L_v \;[\text{kJ/kg}] \;=\; \textcolor{#dc4124}{\left(2.429 \times 10^{6}\right)} \;[\text{kJ/kg}] \;=\; 2429\;\mathrm{kJ/kg}
- Mass of evaporating water in kg
- Mass that actually evaporates
- Heat absorbed by evaporation (OpenStax §14.3, eqn 14.18)
- Heat absorbed in kJ
- Resulting temperature drop in the cooled object
- Final temperature of the object after the evaporative cooling
5 g of sweat evaporating at body temperature (37 °C) absorbs Q = 0.005 · 2,428,620 J/kg ≈ 12,143 J. Spread that cooling over a 70 kg body with average specific heat 3,470 J/(kg·K) and the body cools by ΔT = 12,143 / (70·3,470) ≈ 0.0500 °C — a small drop per drop of sweat, but it adds up over hundreds of grams of perspiration in a hot afternoon.
| Quantity | Expected | Computed | Δ |
|---|---|---|---|
| L_v | 2.429 \times 10^{6} | 2.429 \times 10^{6} | 0 |
| Q | 12143 | 12143 | 0 |
| delta_T | 0.05001 | 0.04999 | 2.062 \times 10^{-5} |
2 g of water evaporating off a 250 g cup of tea at 80 °C absorbs Q = 0.002 · 2,310,800 J/kg ≈ 4,622 J. Pulled from the tea (c = 4,186 J/(kg·K)), that drops the cup by ΔT ≈ 4,622 / (0.25·4,186) ≈ 4.42 °C. The thin wisp of steam off your tea is doing more cooling than the visible volume suggests.
| Quantity | Expected | Computed | Δ |
|---|---|---|---|
| L_v | 2.311 \times 10^{6} | 2.311 \times 10^{6} | 0 |
| Q | 4622 | 4622 | 0 |
| delta_T | 4.416 | 4.416 | 7.462 \times 10^{-5} |
- [1] OpenStax College Physics 2e. Chapter 14.3 — Phase Change and Latent Heat (Eqn 14.18), eq. Q = m·L_v (vaporization). Water: L_v = 2256 kJ/kg at 100 °C; at body temperature (37 °C) the value rises to ~2428 kJ/kg, since slightly more energy is needed to evaporate water from a lower-energy state. (openstax.org)
- [2] OpenStax College Physics 2e. Chapter 14.3 — explicit perspiration cooling example: "perspiration from the skin requires a heat input of 2428 kJ/kg", eq. Sweat-driven body cooling: heat lost = m_sweat · 2428 kJ/kg, drawn from the skin/body. High humidity inhibits evaporation and the cooling fails. (openstax.org)
- [3] Wikipedia. Latent heat — temperature dependence of L_v(water), eq. L_v of water decreases roughly linearly with temperature from ~2501 kJ/kg at 0 °C down to 2257 kJ/kg at 100 °C. (en.wikipedia.org)
- All of the heat absorbed by evaporation comes from the cooled object — no heat enters from the surroundings during the cooling. In real settings the surroundings warm the object back up and the steady-state cooling is set by an evaporation-vs-convection balance.
- Latent heat L_v is treated as a temperature-dependent constant (linear interpolation between the OpenStax 37 °C and 100 °C anchor points). The actual variation is closer to linear over this range to within a percent.
- No heat is lost to the air via convection or radiation — only evaporation matters here. Real-world body cooling adds substantial radiative + convective losses on top of evaporation.
- The fraction f_evap captures the practical reality that not all water on a surface evaporates: pooled water that drips off, or saturated air above the surface, both reduce the effective mass that ever leaves the liquid phase.
- Pressure is ~1 atm. L_v changes with pressure (rises at lower pressure, drops at higher pressure) — the formulas here are not valid for vacuum or pressurized conditions.