What is boiling water?

What is boiling water?

Water is said to be boiling when its temperature is such that the ambient pressure surrounding the water and the vapor pressure of the water are the same.

We call this temperature the “boiling point for this ambient pressure”.

The vapor pressure of water for a given ambient (air) pressure and for a given temperature, is the pressure of the vapor formed over the water in a closed container when evaporation and condensation are at equilibrium in the container. The vapor pressure of water for a given ambient pressure increases with temperature. For a given ambient pressure, we can do experiments and measure the vapor pressure for different temperatures thereby constructing a temperature vs. vapor pressure curve for water at that ambient pressure. By the definition in the first paragraph, the temperature at which this vapor pressure curve reaches the ambient pressure is the boiling point of water for that ambient pressure.

When water reaches its boiling point it changes from the liquid phase (“liquid water”) to the gaseous phase (“steam” or “(water) vapor”). At the boiling point, any water molecule is equally likely to be in the liquid or gaseous phase. If the temperature is kept at the boiling point, the rate of molecules changing phase in either direction will be the same (in a closed system where no molecules can escape).

This is not to be confused with evaporation where water molecules on the surface having sufficient kinetic energy escape from the liquid to the gaseous phase, or electrolysis where water molecules are chemically changed into hydrogen and oxygen diatomic gas molecules. With boiling and evaporation the water molecules remain intact as water molecules; they just change phase. The main difference between boiling and evaporation is that evaporation is a surface only phase change phenomenon whereas boiling takes place throughout the liquid.

The boiling point temperature is the maximum temperature for liquid water. As more heat is absorbed by the liquid more of it changes to vapor. As heat is absorbed by the vapor, the liquid loses the corresponding amount of heat so heat must be supplied to keep the liquid at the boiling point. Thus all the heat absorbed by the water at the boiling point temperature is used to convert liquid to steam - and not to raise the temperature of the liquid water. 

The amount of heat needed to raise the temperature of a liquid or to convert a liquid to the vapor phase are characteristics of the substance having to do with its molecular and intermolecular structure. The  specific heat of a substance is the amount of energy ( the number of calories or joules) needed to raise one gram of the substance by one degree celsius (cal/g-°C or joules/g-°C). The heat of vaporization of a substance is the amount of energy (number of calories or joules) needed to convert one mole of the substance from the liquid phase at the boiling point to the vapor phase at the boiling point (cal/mole). That is, the heat of vaporization is the difference in the heat content between a mole of the liquid at the boiling point and heat content of the mole of steam it forms. The heat of vaporization of water is 44.0 kJ/mol (44,000 joules per mole). Of course the corresponding heat of condensation has the same magnitude in the opposite direction (i.e. a release of 44 kJ/mol).

Note on units: A mole is a certian number of things, just like a dozen. Whereas a dozen is 12 things (whatever they might be), a mole is  6.02 hundred billion trillion things (whatever they might be). In this case the things are molecules of the substance at hand, i.e., water molecules. Also a calorie is a measure of energy as is a joule. Sort of like both a meter and a foot are measures of length. Just as we can convert feet to meters (1 ft = 0.3048 m) we can convert calories to joules (1 cal = 4.18 j). Heat is a form of energy. For example heat can be absorbed or released as a gain or loss of energy in a system. The calorie is defined as the amount of heat one must add to a gram of liquid water to raise its temperature by 1°C. So the specific heat of water is, by definition, 1.0 cal/g-°C = 4.18 j/g-°C.

Example: Suppose 1 liter of steam at 100°C (373.15 K) and 1 atm of pressure was collected and put into a cooler to form water. What is the heat loss? 

Solution: Use the ideal gas law PV=nRT to get the number of moles of water(steam).

[P=pressure,V=volume, n=number of moles, R=constant=0.082 L-atm/mol-K), T=absolute temperature (K)].

So n= (1.0 atm)(1.0 L)/(0.082 L-atm/mol-K)(373.15K)=0.033 mol. Since 44kJ/mol is released in condensation and we have .033 mol, the total heat released is the product (44)(0.033)=1.45 kJ. 

When water is boiling we get a rapid transformation of liquid to vapor which is characterized by bubbling.

The bubbles formed as the water boils are filled with vapor at the corresponding boiling vapor pressure (equal to the ambient pressure). For this reason the bubbles can form without being crushed instantaneously by the outside pressure as they would be at lower temperatures when the corresponding vapor pressure is less than the ambient pressure. Because the vapor filled bubbles are less dense than the surrounding liquid, they rise to the surface and release the vapor into the air. As we noted above, these bubbles might arise anywhere in the liquid because at the boiling point temperature any molecule is equally likely to be in the liquid or gaseous phase.

Note: The reason we see bubbles first form on the bottom is because the heat is usually applied there so it is the first water to reach the boiling point.

Note: Before the boiling point temperature is reached we do see tiny bubbles throughout the liquid but these are a different sort of bubbles. In the water we are usually boiling there is a certain amount of air (a gas). As the temperature of a liquid is raised, the saturation level of gases dissolved in it is decreased. That is, it can hold less gas in the solution as the temperature rises. So as the temperature is raised, some of the dissolved air comes out of solution and we see these air bubbles rising before the water starts boiling. We are seeing air bubbles as opposed to the vapor bubbles that exist during boiling.

Relearning Chemistry

I had sort of a strange start in Chemistry. Traditionally, everyone in my high school was compelled to study Latin for at least two years. In 1965, a new school rule allowed you to discontinue Latin after one year if you took Chemistry during your sophomore year instead. Only two of us took up this offer which meant we sat in a class made up of Juniors who of course wanted nothing to do with us. Essentially I took the class in complete isolation, unable to ever talk to anyone about anything. Needless to say it wasn't the best way to learn. Because of that experience, there was always a wrinkle in my mind regarding Chemistry. So recently I went back and relearned Chemistry. First from the Khan Academy Chemistry class which is a great way to start and then from a Chemistry textbook…. all 31 chapters, every problem, chapter test, unit test. Of course in this day and age whenever a topic is not quite clear or you want a more in-depth explanation there is the infinite abyss of the internet to explore until you are content. A few topics definitely got me waylayed in a fun exploratory sort of way. I think I understand things better than I did before or at least they are a lot fresher in my mind. To celebrate I am going to include a few interesting chemistry topics on this blog.

Chemistry Lesson Part 2

continuation…

Now P(A) is in direct proportion to [A] and the same for B. So we can conclude that the probability of getting m-molecules of A and n-molecules of B in close enough proximity and with high enough energy to cause a reaction is directly related to the product [A]^m*[B]^n and thus the rate of the reaction, a measure of how many times those individual reactions are taking place is also directly related to this product. This very long discussion leads us to a very nice result.

   Suppose mA + nB <--> xC + yD is a balanced equation of a reaction in equilibrium. We have shown that the forward rate, F=k [A]^m*[B]^n, for some constant of proportionality k, and through the same reasoning, the reverse rate, R=c [C]^x*[D]^y, for some constant c. Now, equilibrium was defined as the state when F=R so (drum roll please) we can define a constant called Keq, the so called equilibrium constant for this equation at this temperature, defined as:
Keq = k/c = [C]^x*[D]^y/[A]^m*[B]^n.    This constant, Keq, is singular to this equation at this temperature. If we add various amounts of one or more of the substances A,B,C,D to the mixture in equilibrium, it will reestablish equilibrium in such a way that the relative concentrations still satisfy Keq.  They might all be different but the ratio will hold.

Example: In water at 25 degrees Celsius (room temperature), the water molecules H2O have a certain kinetic energy distribution in which some of the molecules will have sufficiently high energy so that their collisions could free a Hydrogen nucleus from the electron it is sharing with a greedier Oxygen nucleus. (When two H atoms share their respective electrons with the O atom to fill the outer shell of the O, the bond between the H's and O is a polar covalent bond, meaning that there is a slight negative charge on one side of the water molecule and two slight positive charges on the opposite side due to the  greedier pull of the O on the shared electrons.) This freed H-nucleus, H+, essentially a proton, has a positive charge and quickly aligns it self with another water molecule, drawn by the negatively charged O-side of the H2O molecule. The H2O molecule that lost the H+, but held its electron, is now a hydroxide ion OH- and there is a corresponding hydronium ion H3O+. That is we have an ongoing reaction which we can write in two ways (equivalent):
H2O <--> H+  +  OH-    or       2H2O <--> H3O+  +  OH-

Again, once H+ is freed up it is captured by another H2O to become a H3O+. So [H+]=[H3O]+ and they are just two ways of describing the hydronium ion.

In this case one of our reactants, H2O, is a solvent which means its molecules are everywhere, it has a concentration of 1. Thus we get Keq= [H+][OH-] which we can show is Keq=10^-14, and since the concentrations of hydronium ions and hydroxide ions are the same in pure water auto ionization, they are both 10^-7. Defining the function p=-log, we get pH = p([H+]) =7 for pure water.  :) :)

Hydronium and Hydroxide Ions Part 1

   When a chemical reaction reaches equilibrium it is not that the reactions terminate, it is that the rates at which the forward and reverse reactions continue are equal. This means that relative amounts of reactants and products stay constant at the equilibrium concentrations - both reactions continue but the concentrations are stable.
   If you add one or more of the reactants or products to a mixture in equilibrium, thus changing the relative concentrations, actions will occur until a new equilibrium is reestablished (i.e. the rates will gradually go back to equal after being changed to unequal by the addition.)
   The rate of a reaction is correlated directly with the number of individual reactions that take place. In order for an individual reaction to take place, the required molecules, in the right ratios, must be in close enough proximity and have high enough energy for the reaction (i.e. the redistribution of electrons, changing bonds holding electrons to their nuclei) to occur.
   The  temperature is a measure of the average kinetic energy (KE) of the molecules in our mixture. The actual kinetic energy of the individual molecules is probabilistically distributed about this average - some having greater (KE) and some less (KE) than the average. This distribution of KE for the molecules in our reactions is determined by the temperature of the experiment so the probability of our
molecules having high enough energy to react is a constant determined by the temperature.
   This means that the rate of a reaction, at a given temperature, is proportional to the probability of getting the right molecules in close enough proximity.
   Suppose our reactants are A and B and we need m-molecules of A and n-molecules of B, in close enough proximity and with high enough energy for a reaction to occur. Suppose we have a small volume we call dV that satisfies the requirement for "close enough proximity". The probability that a reaction will occur in dV is the probability of having m-molecules of A and n-molecules of B in dV times the probability that a reaction will occur given they are there (i.e. the probability they have high enough energy). The probability of having the required molecules in dV depends, by definition, on the concentration of A and B molecules in our experiment. Assume a uniform distribution of A (and B) in our mixture. The concentration of A, written [A], is given in moles per liter.  The number of A molecules, N(A), in dV is:
    N(A) = ([A] mol/Liter)  (6.022x10^23 molecules/mole) (k Liter/dV)
where k is a constant (1/k=how many dV's in a Liter). Suppose k was small enough to cancel Avogadro's number. Then we would have N(A) = [A] molecules/dV. That is, we would expect this concentration, a number between zero and one, to be the number of A molecules in dV - verses the required m such molecules. This means we would not have a concentration dense enough for the reaction to occur.
     If P(A) is the probability of getting one A-molecule in a given dV, then the probability of getting two would be P(A)*P(A) and the probability of getting m-molecules of A in dV would be P(A)^m. Similarly, the probability of getting n-molecules of B in dV is P(B)^n and the probability of getting them all in dV is the product.
    To Be Continued…..

How much does an infinite stack of pennies weigh?

Have you ever wondered how much an infinite stack of pennies might weigh? Today you can find out. First, we get a stack of pennies 1 foot high and find it weighs 22 oz = 1.375 lb. That's all we need (along with a little physics and a little math). Let the radius of the Earth be R and the mass of the Earth be M.

Physics part: An object of mass m at a distance D from the center of the Earth weighs GMm/(D^2), where G is the gravitational constant and "D^2" means "D squared". A couple of things to notice in this formula are: the weight of something on the surface of the Earth is GMm/R^2 and that as the object of mass m moves further from the center of the Earth, that is farther from the Earth's surface, its weight decreases by the square of the distance. Since our 1 ft stack of pennies weighs 1.375 lb, its mass is given by m = R^2 (1.375) / GM where we have just plugged the weight into the formula above. That is, GMm/(R^2)=1.375, and solved for m.

Calculus idea: If we had an infinite stack of pennies we could consider the stack as a uniform rod and we could take a tiny piece of it, say dx high, at the distance x from the surface of the earth. Now the infinite stack is made up of such pieces where x ranges from zero to infinity. The total weight of the infinite stack is just the weight of all these little pieces. That is, the total weight W is the limiting value of the sum of the weights of all these dx high pieces as the value of dx gets smaller and smaller (goes to zero).

Calculation: the weight of the dx high piece at distance x from the Earth's surface depends on the mass of the little piece. Since mass is an inherent property of matter it remains constant no matter where the object is in the universe. In the physics part above we saw the mass of the 1 ft-stack, so we know the mass, dm,  of a dx ft-stack will be dx/1 = dx  of the original mass. That is dm= R^2(1.375)/(GM) *dx.
Then, the weight of this little piece at distance x from the surface of the Earth is
dw=GM(dm)/(R+x)^2 = R^2(1.375)/(R+x)^2 * dx.

Calculus part: Adding up all these little weights as x goes from zero to infinity means taking the integral of this expression. We can bring the constant R^2(1.375) outside the integral and simply compute the definite integral of 1/(R+x)^2 from zero to k and then let k go to infinity. From this we get
W= R^2(1.375) [ 1/R - 1/(R+k) ] which goes to R(1.375) as k goes to infinity.

Thus the weight of an infinite stack of pennies is W= (4000 x 5280/2000)(1.375) = 14520 tons!!

Special note: when k=R, that is, when the stack is as high as one Earth radius, R, W= 1/2(R*1.375).
This means half the weight of the infinite stack is found in the first 4000 miles of pennies!!

And now you know!!

Open for business

On this blog I will be writing about things that are not related to life in Chili except that they come from a guy who comes from Chili. I hope they bring some fun and maybe some enlightenment to anyone who visits. The writings will tend to arise after I've been working on something for awhile and hopefully will add some sense of clarity to things that took me some time to work through. I have found that oftentimes it takes some work to get to an understanding of things. I hope to make the path easier or open new paths to anyone who is interested.