Abstract
Sprinkling water is thought to work better cold. This paper shows that what cools is not the temperature of the water but its evaporation—a $20$-degree change in water temperature moves the effect by $\mathbf{3.3882}$ per cent. No new theorem or law is claimed.
First, one litre removes $\mathbf{4.1860\times10^{4}}$ J by temperature difference (Section 2).
Second, it removes $\mathbf{2.4500\times10^{6}}$ J by evaporation, $\mathbf{58.53}$ times as much (Section 2).
Third, this is the core of the paper. Water at $5 ^\circ$C beats water at $25$ by only $\mathbf{3.3882}$ per cent (Section 2).
Fourth, one litre evaporating cools $100$ m$^3$ of air by $\mathbf{20.3151}$ degrees (Section 3).
Fifth, at $100$ per cent humidity the cooling is exactly $\mathbf{0.000000000000}$ degrees (Section 3).
Sixth, the separator is whether the cooling is sensible or latent (Section 4).
Keywords: sprinkling water, latent heat, evaporative cooling, sensible heat.
Scope (what is not claimed)
No new theorem or law is claimed—the distinction between sensible and latent heat, the latent heat of vaporisation and evaporative cooling are all standard. No practice is recommended—when, where and how much to sprinkle is not the subject, nor are hygiene and safety. Air movement is excluded—where the cooled air goes is not the subject; a closed $100$ m$^3$ is simply assumed. Evaporation rate is excluded—how fast the water evaporates is not the subject; only the heat once it has. Humidity is treated crudely—the evaporable fraction is taken as $1-\mathrm{RH}$; in reality it depends on the saturation deficit and air movement. Relation to earlier papers: Paper 463 showed that only a tenth of the geometric promise arrives—there too, the conspicuous quantity and the acting one differed. Paper 300 showed that common root and distinct root can be decided—“cooling by temperature difference” and “by evaporation” are distinct roots by that criterion. Paper 429 showed that the same air crosses the same height and comes out warmer—there too, latent heat held the answer. Paper 460 showed that the limit of heat is set with humidity—there too, humidity decided the room for evaporation. What is added is setting sensible against latent at $58.53$, giving the effect of water temperature as $3.3882$ per cent, giving the cooling of $100$ m$^3$ as $20.3151$ degrees, showing the linear dependence on humidity at four values, placing $100$ per cent humidity as the boundary, and placing the separator at sensible versus latent.
1. Notation and the Question
Setting: A mass $m$ of water removes, by a temperature difference $\Delta T$ (sensible heat),
$$Q_{\mathrm{sens}}=m\,c_{w}\,\Delta T$$and by evaporation (latent heat),
$$Q_{\mathrm{lat}}=m\,L_{v}$$Below, $c_{w}=4186$ J/(kg K), $L_{v}=2.45\times10^{6}$ J/kg, one litre of water and ground at $30 ^\circ$C. Air is taken as $\rho_{a}=1.2$ kg/m$^3$ and $c_{p}=1005$ J/(kg K).
Observation 1.1 Sprinkling water is thought to work better cold. Put something cold down and things cool—a natural thought.
Observation 1.2 But water evaporates. And in doing so carries off a great deal of heat without changing temperature—there are two routes.
Question: Which one is doing the work.
2. Latent Heat Is 58.53 Times Sensible
Computation 2.1: Take both quantities of heat for one litre.
| Route | Heat removed [J] |
|---|---|
| Temperature difference ($20 ^\circ$C water on $30 ^\circ$C ground) | $\mathbf{4.1860\times10^{4}}$ |
| Evaporation | $\mathbf{2.4500\times10^{6}}$ |
Observation 2.2 Evaporation is $\mathbf{58.53}$ times larger. The temperature difference contributes $1.68$ per cent of the total—evaporation alone accounts for almost all of it.
Computation 2.3: Vary the water temperature and take the total.
| Water temperature | Total heat [J] |
|---|---|
| $5 ^\circ$C | $\mathbf{2.5546\times10^{6}}$ |
| $25 ^\circ$C | $\mathbf{2.4709\times10^{6}}$ |
Observation 2.4 Twenty degrees of water temperature buy $\mathbf{3.3882}$ per cent. Iced water beats tap water by three per cent—not worth the trouble.
Observation 2.5 Latent heat is large because it is the work of breaking hydrogen bonds. Raising a temperature and changing a phase differ by orders of magnitude—the same word “heat” with different destinations.
3. One Litre Cools 100 Cubic Metres by 20.3151 Degrees
Computation 3.1: Take the cooling of $100$ m$^3$ of air by one litre evaporating.
$\mathbf{20.3151}$ degrees.
Observation 3.2 Enough to move the air around a house. From one bucket—which is why sprinkling is felt at all.
Computation 3.3: Take the evaporable fraction as $1-\mathrm{RH}$ and tabulate the cooling.
| Humidity [%] | Evaporable fraction | Cooling [$^\circ$C] |
|---|---|---|
| $30$ | $0.70$ | $\mathbf{14.2206}$ |
| $60$ | $0.40$ | $8.1260$ |
| $90$ | $0.10$ | $2.0315$ |
| $100$ | $0.00$ | $\mathbf{0.000000000000}$ |
Observation 3.4 At $100$ per cent it is exactly zero. Not approximately, but identically in this model—with no evaporation, the latent route closes.
Observation 3.5 What remains is the sensible heat, $1.68$ per cent of the total. Sprinkle on a humid day—and almost nothing happens.
Observation 3.6 The cooling is linear in humidity. The cooling at $60$ per cent is exactly $4/7$ of that at $30$—because it is proportional to $1-\mathrm{RH}$.
Observation 3.7 So whether sprinkling works is decided by humidity, not water temperature. $3.3882$ per cent against everything from zero to a hundred—a wholly different range of influence.
4. The Separator
The separator: is the cooling sensible, or latent.
Observation 4.1 Sensible, and the temperature difference acts. Colder is better—and humidity is irrelevant.
Observation 4.2 Latent, and whether evaporation is possible acts. The water temperature hardly matters—and humidity is decisive.
Observation 4.3 These are distinct roots by the criterion of Paper 300. Both are called “cooling with water”—one a quantity of temperature difference, one of phase change.
Observation 4.4 This stands where Paper 429 stands. There, the same air crossed the same height and came out warmer, and latent heat held the answer—it holds the answer here too. Phase change carries heat that does not show in a temperature.
Observation 4.5 It overlaps Paper 460 as well. There, humidity decided the room for evaporation—here it decides whether sprinkling works. The same mechanism acts on a person and on the ground.
Observation 4.6 And on Paper 463. There, the conspicuous quantity (torque) and the acting one (clamp force) differed—here the conspicuous one (water temperature) and the acting one (room to evaporate) differ.
Observation 4.7 To be honest—humidity is treated crudely. The evaporable fraction was taken as $1-\mathrm{RH}$, whereas in reality it follows the saturation deficit and air movement—the value $14.2206$ depends on that model; the conclusion that it is zero at saturation does not.
5. Roll Call and Closing
- Sensible is $4.1860\times10^{4}$ J (Computation 2.1): the temperature route.
- Latent is $2.4500\times10^{6}$ J (Computation 2.1): $58.53$ times.
- Sensible is $1.68$ % of the total (Observation 2.2): almost all evaporation.
- Twenty degrees buy $3.3882$ % (Observation 2.4): the core.
- $100$ m$^3$ by $20.3151$ degrees (Computation 3.1): from one bucket.
- $14.2206$ degrees at $30$ % humidity (Computation 3.3): dry air works.
- $0.000000000000$ at $100$ % (Observation 3.4): identically zero.
- Linear in humidity (Observation 3.6): $4/7$ exactly.
- The separator is sensible or latent (Observation 4.3): distinct roots by Paper 300.
- Humidity is treated crudely (Observation 4.7): the zero survives.
Closing: Sprinkling water is thought to work better cold—put something cold down and things cool, a natural thought. But water evaporates, and in doing so carries off a great deal of heat without changing temperature. There are two routes. The two routes differ enormously in size—one litre removes $\mathbf{4.1860\times10^{4}}$ J by temperature difference and $\mathbf{2.4500\times10^{6}}$ J by evaporation, a factor of $\mathbf{58.53}$. The temperature difference contributes $1.68$ per cent of the total. So a $20$-degree change in water temperature, from $5$ to $25$, moves the total from $2.5546\times10^{6}$ to $2.4709\times10^{6}$ J, a difference of $\mathbf{3.3882}$ per cent. Iced water beats tap water by three per cent, which is not worth the trouble. Latent heat is large because it is the work of breaking hydrogen bonds: raising a temperature and changing a phase differ by orders of magnitude, the same word “heat” with different destinations. The effect is large, provided evaporation can happen—one litre evaporating cools $100$ m$^3$ of air by $\mathbf{20.3151}$ degrees. One bucket moves the air around a house that far, which is why sprinkling is felt at all. But taking the evaporable fraction as $1-\mathrm{RH}$, the cooling is $\mathbf{14.2206}$ degrees at $30$ per cent humidity, $8.1260$ at $60$, $2.0315$ at $90$ and $\mathbf{0.000000000000}$ at $100$. At saturation it is not approximately but identically zero: with no evaporation, the latent route closes, and what remains is the sensible $1.68$ per cent. The cooling is linear in humidity, with $60$ per cent giving exactly $4/7$ of $30$. So whether sprinkling works is decided by humidity and not water temperature—$3.3882$ per cent against everything from zero to a hundred. One thing separates them—is the cooling sensible, or latent. Sensible, and the temperature difference acts: colder is better and humidity is irrelevant. Latent, and whether evaporation is possible acts: the water temperature hardly matters and humidity is decisive. They are distinct roots by the criterion of Paper 300, one a quantity of temperature difference and one of phase change. This stands where Paper 429 stands, where the same air crossed the same height and came out warmer with latent heat holding the answer. Phase change carries heat that does not show in a temperature. It overlaps Paper 460 too, where humidity decided the room for evaporation: the same mechanism acts on a person and on the ground. And Paper 463, where the conspicuous quantity and the acting one differed: here it is water temperature against room to evaporate. To be honest—humidity is treated crudely. The evaporable fraction was taken as $1-\mathrm{RH}$, whereas in reality it follows the saturation deficit and air movement. The value $14.2206$ depends on that model; the conclusion that it is zero at saturation does not.
References
- Black, J. (1803). Lectures on the Elements of Chemistry. Longman & Rees.
- Watt, J. (1769). A New Invented Method of Lessening the Consumption of Steam and Fuel in Fire Engines. British Patent 913.
- Oke, T. R. (1987). Boundary Layer Climates (2nd ed.). Routledge.
- Yamagishi, Y. (2026). The Three Hundredth Census: Whether Two Things Share a Root Is Decidable [Paper 300]. Zenodo. 10.5281/zenodo.22171914
- Yamagishi, Y. (2026). The Same Air Crosses the Same Height and Comes Out Warmer [Paper 429]. Zenodo. 10.5281/zenodo.22700352
- Yamagishi, Y. (2026). The Limit of Heat Is Set by Temperature and Humidity Together [Paper 460]. Zenodo. 10.5281/zenodo.22733136
- Yamagishi, Y. (2026). Of the Seven Hundredfold Geometry Promises, One Tenth Arrives [Paper 463]. Zenodo. 10.5281/zenodo.22733205
Reference
Yamagishi, Y. (2026). Chilling the Water Changes the Effect by Four Per Cent [Paper 464]. YOI: https://www.yamagishiyuuki.com/yoi/464
DOI: 10.5281/zenodo.22733251
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 464. This paper is a preprint and has not been peer reviewed.