Liquid boiling and Evaporation

[EEristavi]

In the normal conditions (sea level) water evaporates at 100 C.

In thermodynamics, we say: the amount of energy Q, can raise temperature of the liquid by the formula Q1=cm(t2-t1); when the liquid reaches the boiling point (100 C), we write Q2=Lm.
Q2 is entirely spent on changing liquid state into gaseous state, but as I know it's not necessary for liquid to evaporate (liquid goes into the gaseous state more and more rapidly as we approach the higher temperatures).

The question is: isn't the energy we give, is partially spent on changing state on the <100 C temperatures; and if so - why we don't consider it

 

[russ_watters]

I think I understand your question and it is really a matter of specifying what part of a chain of events you are referring to. So the way I view it is this: when the temperature is <100C, all of the energy put into the water is to change its temperature. But simultaneous to that, some energy is being removed from the water (lowering its temperature) by convection and evaporation (if we're talking about a pot on a stove). You can add the three effects (superposition) to find the net effect (temperature is only rising).

 

[anorlunda]

Your question is a bit muddled.

It sounds like you are saying that Q1 is the heat required to warm the water, and Q2 is the heat of vaporization. But if you put in Q2 additional heat, all the water will be boiled off into steam while the temperature remains constant at 100C. But you don't need to boil any of the water.

You may be confusing evaporation of water from the surface with boiling. It's explained better in this WIKI ariticle.
https://en.wikipedia.org/wiki/Water_vapor#Evaporation

 

[sophiecentaur]

The question is: isn't the energy we give, is partially spent on changing state on the <100 C temperatures; and if so - why we don't consider it

But we DO consider every possible mechanism of Energy transfer when we try for the highest accuracy. Evaporation is very dependent on the experimental details and it is very hard to estimate so heat experiments are often done in sealed containers and, of course, they are well insulated and their thermal capacity is also included in calculations.
Calorimetry is a highly sophisticated branch of measurement but it is not one of the most accurate - not compared with Frequency and Time measurement, for instance.

 

[EEristavi]

Thank you all for answers.

Now I have different kind of question:
If we have some material (for example sand), which has moisture/water 10%, how much energy we must give in order to evaporate water from sand.

Previously, I would have tried the way a posted above (Q=cmΔt;...), but now I'm not sure if its right way

 

[sophiecentaur]

It's very hard to make predictions about that sort of thing. A wet pile of sand or sponge will have a huge surface area compared with a pool of water alone. The 'drying effect' will be helped a lot by that and some of the latent heat of evaporation will be supplied by the surrounding air. As for the total amount of energy needed, I can only think that it will only be different for the one or two layers of water molecules that can be regarded as being in contact with the surface. For those, I would imagine that the energy needed to detach them will depend on whether the material itself is hydrophilic or hydrophobic. It would have to be a small effect except for a nearly dry surface.

 

[anorlunda]

how much energy we must give in order to evaporate water from sand.

Your question is unclear. Water evaporates all the time without adding any heat. Do you mean how much heat to turn wet sand into completely dry sand?

There is also a question of time. As I said, water evaporates all the time, with no added heat, or with only a little heat. The more quickly you want to dry the sand, the more power you need to put in.

 

[EEristavi]

Do you mean how much heat to turn wet sand into completely dry sand?

Yes - I want to turn wet sand into completely dry sand (or Almost dry, let's say moisture is 0.1%), and "Maybe Yes" - if you mean that heating something means giving energy.

There is also a question of time. As I said, water evaporates all the time, with no added heat, or with only a little heat. The more quickly you want to dry the sand, the more power you need to put in.

Yes it is question of time and as you mentioned - The power, Energy per time. So if i want dry sand, I must "put power in it", or as I said previously put/give energy. Of course if I use more power (give energy more rapidly), evaporation will be faster. This is why I asked "how much energy we must give", and how fast I want to evaporate is up to my will. In addition, my assumption is that without interference water evaporation is very slow.

 

[sophiecentaur]

Water evaporates all the time without adding any heat.

Hmm. That can be interpreted wrongly. Natural evaporation doesn't occur for free. Heat is transferred from the surroundings because the temperature of the water drops as molecules leave the surface.

 

[EEristavi]

It would have to be a small effect except for a nearly dry surface.

Good point. This is why I added "almost dry" in my comment above.

 

[sophiecentaur]

Good point. This is why I added "almost dry" in my comment above.

In which case the nature of the substance could be fairly relevant,

 

[EEristavi]

Lets say it's hydrophilic - we were talking about sand and also, it will be more interesting (for me)

 

[zbikraw]

Referring to thermodynamics and standard data is not the best in mentioned situations. Energy circulation in wet sand drying depends on the "way" of process, so it may be drastically dependent on process velocity. It means the entire process is nonequilibrium one. Sand is porous material and pore inlet-outlet trajectories may be different in diameters (water is anisotropic and oriented in silica, so these trajectories might be different). Pores changes their geometry with temperature and pressure, free surface area also changes, so the energy required for absorption, adsorption, and desorption depends on these parameters. Additionally in nearly any solid material there exists "centers" more and less hydrophilic/hydrophobic both on surface and inside. In practice it might be better to use Clausius-Clapeyron relation than direct calorimetric measurement, mainly as thermogravimetric curves. It permits determination of temeperature and pressure dependence of specific heat capacity and specific heat of evaporation.
Sand is not very standarized material, but even pure silica solid has properties strongly dependent on the -"way" it was prepared.
Sub-boiling temperature evaporation is mainly surface process, depending strongly on water vapor removing from surface of liquid and solid. Boiling refers to bubble formation in all the liquid volume.
Regards,
zbikraw

 

[Mister T]

In the normal conditions (sea level) water evaporates at 100 C.

It also evaporates at temperatures lower than that.

In thermodynamics, we say: the amount of energy Q, can raise temperature of the liquid by the formula Q1=cm(t2-t1); when the liquid reaches the boiling point (100 C), we write Q2=Lm.

If a glass of water is sitting on my kitchen counter top at a steady temperature of 22 °C we would still write $Q=mL$ to describe the heat energy transferred to the water to make it evaporate. This is why we call it evaporative cooling, it takes a transfer of heat energy to the water to make it evaporate.

If we have some material (for example sand), which has moisture/water 10%, how much energy we must give in order to evaporate water from sand.

About 334 joules to evaporate each gram of water.