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Water has a latent heat of vaporization equal to $2,260\ \mathrm{kJ/kg}$. While ethyl alcohol has a latent heat of vaporization equal to $846\ \mathrm{kJ/kg}$.

Even though ethanol's latent heat is almost $1/3$rd as that of water, ethanol has a stronger evaporative cooling effect. In my understanding its because ethanol is more volatile, so even though it absorbs less heat (than water) while evaporation, it does so faster which causes the stronger cooling effect.

Is there any way to quantify this phenomenon, like a latent power of evaporation or something?

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  • $\begingroup$ I would say ethanol faster evaporative cooling effect rather than stronger, $\endgroup$
    – Protein
    Commented Jul 26, 2020 at 10:49
  • $\begingroup$ It will take away far lesser heat than water $\endgroup$
    – Protein
    Commented Jul 26, 2020 at 14:40
  • $\begingroup$ The missing piece in the above discussion is that the effect is faster, but it is smaller on a mole per mole basis. If you have a large reservoir of each substance, it becomes rather tricky to quantify how these will interplay with one another. $\endgroup$ Commented Apr 13 at 13:07
  • $\begingroup$ Evaporation rates can be calculated using the Hertz-Knudson equation: en.wikipedia.org/wiki/Hertz%E2%80%93Knudsen_equation $\endgroup$ Commented Apr 13 at 15:59

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the speed with which a given liquid evaporates depends on its vapor pressure and the amount of heat required for a given amount of it to change phase from liquid to gas is expressed by its heat of vaporization. The best liquid for evaporative refrigeration would be one with high vapor pressure, a large heat of vaporization, and zero cost (since the liquid is lost when it evaporates). Water has a lower vapor pressure than alcohol but its heat of evaporation is large and its cost is extremely low, so it is the only evaporative coolant in large-scale use.

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  • $\begingroup$ It depends on too many factors like area $\endgroup$
    – Protein
    Commented Jul 26, 2020 at 10:52
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The rate of mass transfer is controlled by the diffusion of the evaporating species away from the interface, its partial pressure in the surrounding gas, and the rate at which heat transfer is conducted to the interface through the air. So to model this properly, one needs to include both mass diffusion, conductive heat transfer. and heat of vaporization. See Mass Transfer Operations by Treybel

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Considering the coolant being used, one may have to also consider Saturation Temperature and Pressure of the coolant being used.

When liquids evaporate they basically gain heat from their surroundings equivalent to their Latent heat of Vaporization, change state (gaseous state) and escape the liquid body.

The liquid gaining heat from its surroundings implies that the surroundings lose heat, thereby reducing surrounding temperature.

Phase change is just a part of the energy transaction, for the coolant to be able to evaporate it first has to reach its saturation temperature at the surrounding pressure, which again requires heat to be added to the coolant which is directly proportional to its Heat Capacity and Mass of the coolant.

Also, considering the fact that heat transfer (conduction of heat between coolant and surrounding) is a surface phenomenon, the Area of contact (surface Area) of the coolant and the object being cooled is also necessary.

The Thermal conductivity of the coolant and the object to be cooled also plays an important role in the rate of cooling.

Fourier's Law of Heat Conduction :

$\frac{dQ}{dt}=-KA(\frac{dT}{dx})$

OR:

enter image description here

Solving the above equation will give you heat transfer values between contact surface of coolant and the object to be cooled, rest of the calculations are basically energy conservation.

Considering all these facts, we already have enough terms to satisfactorily and accurately predict the rate and extent of cooling given that we know the necessary data.

And considering the fact that there are many variables to consider, it is difficult though maybe not impossible to quantify this phenomenon with some new constant at given conditions. And we also have the necessary tools to calculate and predict the outcomes already, so I don't think there is really any need to define any new term altogether for the cause.

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Rate of evaporation depends on too many factors. A simple way to quantify the cooling effect (just for sake of comparing two liquids ) would be the ratio

Mass of liq evaporated: fall in temp of body

Of course the liquids will have to be studied under exactly similar conditions .

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