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During evaporation process water molecules on the surface absorb energy from the water inside and evaporates after getting certain energy to escape from the surface. As a result remaining water cools in this process. No work is done on the water still energy has flown from the water. How is this possible?

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  • $\begingroup$ Can you explain in more detail in your question why you think work would necessarily be required to cool an object? $\endgroup$ Commented May 22, 2018 at 15:15

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I presume that your question is based on the query of whether the Kelvin-Planck statement of Second Law of Thermodynamics is violated under any scenario.

The simplest answer would be NO. The very first thing that we need to realize is the concept of high grade energy and low grade energy. Low grade energy are those in which the motion of molecules is random and disorderly, and are available in the nature abundantly, but cannot be fully exploited. On the other hand, high grade energy are those forms of energy which are properly exploited and suited for one specific purpose. A conventional example would be the difference between Heat Energy and Mechanical Energy.

Coming back to your question, while it may seem that the water molecules are transforming into vapour phase and no work is being done, it is not so. The molecules are indeed absorbing energy from the environment, so as to move from a state of lower temperature to a state of higher temperature. Also, regarding violation of Kelvin-Planck statement, the statement is particularly suited for a device or a machine which can be used to obtain some useful output or serve a purpose (eg, a Refrigerator). In the situation stated above, although the movement from lower temp. to higher temp. is taking place, it does not provide any useful output as such. In order to have specific output from the process of converting water to steam, large amount of heat is required, which is obtained only by doing work on the system.

To summarize, in many physical phenomena, the transfer of a system from lower temperature to a higher temperature may be possible without any external work-performing agency, but it would not be beneficial as such. In order to obtain beneficial output, a work performing agency is mandatory.

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Basically, evaporation is an endothermic process. This means that it will absorb heat from the surroundings. This energy es equal to the enthalpy of vaporization of the water.

So where is the energy taken to evaporate that water? It comes from the increase of entropy due the increase of vapour concentration in the air. We could say that energy is flowing from a wet ambience to a dry ambience.

This works only until a certain concentration of water is reached in the air.

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The liquid water is colliding with other liquid water. Heat is constantly being exchanged between individual water molecules. Once a water molecule has a specific level of energy, it escapes its brothers, leaves the clan and joins the air forces. Its brothers are sad to see it go but it's gone and he took with it a decent amount of energy.

You stated "as a result remaining water cools": this is explained above however the thing is the water doesn't cool after the molecule has gone, it cools at the time of the brothers departure, not after.

you talk of boiling water however keep in mind that water dosnt need to boil to evaporate keeping this in mind also water at room tempature on the floor will cool the floor and surronding area by absorbing heat energy and using it to evaporate so yes cool things can heat hotter things on an idividual molecular base but in whole like what i mean is that we can not add ice to fire to heat up the fire however on a idividual molecular base this is true

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    $\begingroup$ energy is what is being exchanged by molecules. heat is a thermodynamic variable en.wikipedia.org/wiki/Heat ( i am not the one downvoting) $\endgroup$
    – anna v
    Commented May 23, 2018 at 8:46

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