**I.BRIEF INTRODUCTION**

The motivation for creating a mathematical model to study liquid cooling came from a recent happenstance moment when I was waiting for a cup of piping hot tea to cool down to a reasonable temperature, after having burnt my tongue upon sipping some. Tongue stinging, I was blowing on the tea to cool it, and thought to myself in exasperation “How much time will this take to cool !?”. Et voil*à*, here we are …

On a serious note, this problem is indeed an interesting one from a physics point of view. We all know that the major factor in liquid cooling is the phenomenon of evaporation, but the exact laws governing this process are not familiar to most of us and aren’t explicitly taught in school either. So, it is interesting to scientifically study such an ordinary, yet important phenomenon that we all take for granted. Indeed, from sweating, to drying our clothes, and waiting for a hot beverage to cool down, to transpiration in trees, evaporation is an invisible process that surrounds us. On a bigger scale, it is a major part of the water cycle and hence, of Earth’s biosphere and life itself. Where there is water, there is life, and also evaporation. From the following model, one can see that liquid cooling can indeed be modeled to a very good approximation using little more than high-school physics.

**II. MODEL DESCRIPTION **

- A hot, evaporative liquid is filled up to a height ‘h’ in a cylindrical, insulating, open vessel with cross-sectional area ‘A’.
- Cooling has two components – radiative and evaporative, both of which occur only at the liquid surface.
- We assume that the liquid remains in quasi-thermal equilibrium at all times.
- Due to the open vessel, the liquid vapour from evaporation escapes into the atmosphere and therefore, the reaction of vapour condensing back to liquid is negligible.
- During evaporation, the bulk-mass of the liquid does not change much.
- The following symbols represent the physical quantities involved in our calculations:

ρ = liquid density

r_{liq} = mean liquid particle radius

m, M = liquid particle mass and molar mass respectively

T = liquid temperature

t = time

S = liquid specific heat capacity

H_{evp} = enthalpy of vaporization of the liquid in J.mol^{-1
}m_{liq} = bulk mass of the liquid

p = effective number of liquid molecules on the surface as a fraction of the total number of molecules on the surface, which may contain impurities (any substance particles which do not contribute to evaporation).

a_{B =} Stefan-Boltzmann constant

e = emissivity coefficient of the liquid

N_{A} = Avogadro number

K_{B =} Boltzmann constant

R = universal gas constant

**III. THE PHYSICS OF LIQUID COOLING**

In an infinitesimal time interval ‘dt’, the heat lost by the bulk of the liquid is given by

Where the infinitesimal heat/temperature change can be expressed as a sum of two contributions – one from cooling due to black-body radiation, and another from cooling due to evaporation, both of which occur at the liquid surface

__Radiative cooling contribution:__ –

The heat loss due to black-body radiation at the liquid surface is given by the Stefan-Boltzmann law:

T_{atm} is the temperature of the surroundings/atmosphere in contact with the liquid surface, which exchanges heat with the liquid via black-body radiation. T_{atm }is equal to the room temperature, which here we take to be the standard value of ~25 degrees Celsius.

The bulk-mass of the liquid can be expressed in terms of its density and volume,

Thus, from above, we obtain the contribution to the rate of change of temperature due to radiative cooling:

……… (1)

__Evaporative cooling contribution:__ –

We formulate here the physics of evaporative cooling. Firstly, we assume that the molecules at the liquid surface follow the classical Maxwell-Boltzmann statistics. However, we know that this is not exactly true since every molecule of the liquid experiences several intermolecular forces of attraction. It is quite difficult to theoretically model these forces as they depend on the exact chemical structure and spatial distribution of the liquid molecules. Instead, we can find a clever way around this by using the experimentally known enthalpy of vaporization of the liquid ‘H_{evp}’ (in SI units, the energy in Joules required to vaporize one mole of the liquid), which accounts for both the energy needed to overcome the intermolecular forces and the work done by the vapor to push back the atmosphere. H_{evp} thus effectively encapsulates information about the liquid properties and the physical properties of its surroundings (therefore it is not constant and varies with the temperature and pressure). Hence, if we consider a molecule at the liquid surface which has at some given time, the upward component of its velocity as ‘v_{z}’, then the minimum kinetic energy that it needs to possess (in the upward direction) to overcome all forces of attraction and escape from the liquid surface must be equal to the heat of vaporization per unit molecule of the liquid: –

Thus, we can define an ‘escape velocity’ for a liquid particle at the surface. Any liquid particle which has a vertical component of velocity greater than or equal to the escape velocity will escape into the atmosphere and contribute towards evaporation. Note that the molecular mass times Avogadro number is the molar mass, ‘M’.

Thus, we model the liquid surface like an ideal gas (since we use the Maxwell-Boltzmann statistics, which is the foundation of the Kinetic Theory of Gases). But unlike a true gas, the liquid molecules are not free to escape into the atmosphere unless they have sufficient energy (modeled as the escape velocity defined above) to overcome the various inter-molecular forces of attraction experienced inside a liquid.

We can take a detour here to qualitatively interpret the cooling effect of evaporation based on the above model. An ideal gas with a given temperature has a velocity distribution given by the Maxwell-Boltzmann statistics. This is governed by the random thermal motions and collisions of the gas particles. If one were to ask what would happen if we selectively start removing from the system only those particles which have acquired (because of their random thermal motion/collisions) a velocity greater than a threshold value, then one would find that the result would be a decrease in the net energy, average velocity, and hence the temperature of the gas. This is exactly the cooling effect that occurs in liquids due to evaporation, which we have modeled here thus.

For a surface particle with v_{z} v_{z-esc}, we define ‘evaporation’ to have occurred when the particle travels upwards a distance of at least 2r_{liq }in a short time interval Δt_{evap,} during which a new particle from inside the bulk of the liquid would have filled the void left by the escaping particle. Since the backward reaction of condensation is negligible, the process is not reversible and the particle escapes into the atmosphere, thus contributing towards evaporation.

The number of particles at the liquid surface can be expressed as the total number of particles in the bulk of the liquid times the fraction of the total volume that is the liquid surface:

From the Maxwell-Boltzmann statistics, the fraction of liquid particles at the surface which have an upward component of their velocity between v_{z} ± dv_{z} is given by

Each evaporating liquid particle removes on average, H_{evp}/N_{A} amount of heat from the liquid. Hence, the rate of heat loss due to the evaporating particles (i.e. those particles at the liquid surface that have an upwards velocity greater than or equal to the escape velocity) is

Substituting the expressions for the terms in the integral and carrying out the integration, we obtain the contribution to the rate of change of temperature due to evaporative cooling (we use the relation N_{A}.K_{B }= R, the universal gas constant):

…….. (2)

Note that if the liquid contains some non-evaporative impurities, this reduces the number of liquid particles available at the surface which can potentially contribute towards evaporation. This is represented through the fraction ‘p’. Also, since there is heat loss, the term has an overall negative sign.

It is also interesting to note that the rate of cooling obtained here is inversely proportional to the height of the liquid in the vessel and is independent of the surface area of the liquid. We know about the dependence of the rate of evaporation on the surface area of the liquid, but this tells us that it is the liquid height that matters. Spreading out a liquid increases its area, but also decreases its height, and that is the reason why we observe a higher rate of evaporation in ‘thin films’. In other words, liquid in a vessel with a small cross-sectional area filled up to a lesser height can cool faster than liquid in a vessel with a large cross sectional filled up to a greater height. Also, liquid in a large open container versus a small open container can cool at the same rate if it is filled up to the same height in both.

Finally, another interesting point is that although we use r_{liq} to calculate some intermediate quantities, it does not explicitly appear in the final expression for the rate of cooling.

__Rate of liquid cooling__:

The contribution from (1) and (2) give the net rate of liquid cooling as a function of the liquid properties, temperature, properties of the surroundings, and the height up to which it is filled in the vessel: –

__Rate of mass-loss__:

Although we have assumed that the bulk-mass of the liquid does not change much (and this is a good approximation if we have a large quantity of liquid), it however does not hold good for say, thin film evaporation, where the mass-loss is significant compared to the tiny quantity of liquid in the film. It is not too difficult to derive the rate of mass-loss. The derivation is the same as that of the rate of cooling due to evaporation, except we must replace the term H_{evp}/N_{A} with ‘m’, since each evaporating molecule removes a mass of ‘m’ from the bulk of the liquid. Thus, we have

This gives us the expression for the rate of change of the height of the liquid in the vessel (where the impurity fraction ‘p’ and the overall negative sign are included as done earlier): –

**IV. RESULTS**

Solving the ordinary differential equation for the rate of cooling will give us the variation of the liquid temperature with time, starting from some initial (high) temperature and gradually falling to a final temperature.

If we wish to account for the mass-loss along with the cooling, then both ODEs (cooling rate and mass-loss rate) must be solved simultaneously, because they are coupled.

An analytical solution for either case is difficult, and hence we solve the equations numerically.

H_{evp} is an empirical quantity in our model which also varies with the temperature and pressure. Hence, for a liquid in contact with the atmosphere (i.e. at a given constant pressure), the temperature variation of its H_{evp} needs to be recorded from an empirical data-sheet. Interpolating this data will allow us to calculate the value of H_{evp }at any temperature, which is then fed into the equations to be numerically solved.

Choosing 95% pure water as our evaporative liquid filled up to a height of ~8cm in an average sized drinking cup kept outside (making this system equivalent to say, a regular cup of coffee), and accounting for the mass-loss of the liquid, we obtain the results shown in the plots below.

From the results obtained in the plots we can see that it would take a cup of boiling water around **~10 mins** to cool down to room temperature (provided that air is continuously blown on the water to remove the vapor molecules, minimizing the backward reaction of condensation, as is the requirement for our model of liquid cooling to be applicable). During this time, the height of the water in the cup would fall by around **~1cm** owing to the mass lost by the liquid due to evaporation.

Also, if a hot beverage is served at 80 degrees Celsius, making it too hot to drink, it would take close to **~2 mins** of continuous air blowing to cool it down to 50 degrees Celsius, which is generally regarded as the prefect drinking temperature for a hot beverage.

These results seem to agree reasonably well with observations, but a proper experimental study would be required to verify the accuracy of this model.

**V. DISCUSSIONS AND CONCLUSIONS**

- Experimental realization of this model: For experiments to reasonably agree with the results of this model, the experimental conditions should be such that the assumptions under which this model holds must be maintained. An important requirement is that the reaction of vapor condensing back to liquid should be negligible. This can be achieved by continuously blowing on the liquid surface to remove the vapor molecules.
- Liquid cooling another body: The cooling effect of evaporation is a well-known and widely used phenomenon. Our bodies sweat when it is gets hot so that the water in the sweat can absorb the excess heat and evaporate, thus cooling us down. The rate of cooling of a body in contact with an evaporating liquid can be obtained by solving the heat transfer equation for the body and the liquid, using our model to account for the heat loss from the system due to liquid cooling.
- Evaporative freezing: Under the right conditions, for instance, inside a vacuum chamber, it is possible for evaporation to cool a liquid till it freezes. It is however important to note that in a vacuum chamber, upon reducing the pressure, we quickly move to a different point on the phase diagram for that substance, and hence evaporation can technically become boiling. However, the rate of cooling can still be modeled using the same concepts as done here, with some modifications considering that boiling is a bulk phenomenon whilst evaporation is a surface phenomenon. Also, to apply our model for this case, it is important to use the correct values of H
_{evp}, since it is a function of the temperature and pressure. - While numerically solving for the temperature and height variation using the equations derived above, care must be taken that the initial and final temperature values do not go beyond the boiling and freezing points of the liquid respectively (at a given pressure), where the state of matter changes and the physics of liquid evaporation no longer holds.
- Although our model is a good approximation of real liquid evaporation, it does suffer from some limitations. One of these is the assumption that the liquid remains in quasi-thermal equilibrium at all times. While this assumption may be reasonable for thin films, there is bound to be differential cooling and heat transfer (convection, etc.) inside the liquid if the liquid height is not small. In that sense we are overestimating the of rate of cooling because our assumption implies that heat transfer occurs instantly between different parts of the liquid.
- Another limitation is perhaps the way to model Δt
_{evap}. An alternate, but more complex way would be to calculate the probability for a liquid molecule at the surface to achieve a velocity greater than or equal to the escape velocity due to collisions with other liquid molecules. This can be calculated using the Maxwell-Boltzmann statistics. Using this, we can obtain the average number of collisions required before such a velocity may be achieved. Multiplying this with the mean free time between two collisions would give us the average time for a liquid molecule at the surface to evaporate, i.e. Δt_{evap}. This can improve the accuracy of our model, but it would make the expression for the evaporative cooling rate more complex. The current way to model Δt_{evap }seems to overestimate the cooling rate. Thus, our model provides an upper limit on the liquid cooling rate or a lower limit on the cooling time. - The contribution of the radiative term to liquid cooling is generally small compared to the evaporative term. Hence, the major factor in liquid cooling is evaporation.

**VI. END NOTES**

Following the scientific method, this model now needs to be tested against experiments to check how accurate its predictions are. So, if you are jobless, and are waiting for your 80 degrees Celsius hot cup of tea filled up to a height of 8 cm to cool down to the perfect drinking temperature of 50 degrees, please help test out this model by checking if it takes around 2 minutes of blowing on it continuously to get the job done!

**VII. REFERENCES**

‘Calculating the optimum temperature for serving hot beverages’ (Brown F, Diller KR.)