Water and heat capacity
Burning fuels releases heat generated by the reaction of the fuel
with air. When we talk about the caloric content of a food, we refer
to the heat energy available by reacting the molecules with oxygen
in living cells, producing water and carbon dioxide.
Every substance can be heated by adding energy to it. The heat
capacity of a substance is the amount of heat needed to raise the
temperature of one gram one degree Centigrade. It takes one
calorie of heat to raise the temperature of 1 gram of water by 1
degree Centigrade.
Question: you and your sister pig out at McDs and eat a 1500 kilo
calorie meal each, and when you weigh yourselves afterwards she
estimates that most of her weight is in the form of 50 kg of water.
If her supersized double cheese quarterpounder meal were
converted entirely to heat and used to heat up this water, how
many degrees would its temperature rise?

Answer: T = 1.5x106 calories/5x104 g = 30 degrees!
Question: Air weighs about 1.25 g per liter and has a heat capacity
of about .25 cal/g/degree C, so it takes about .3 cal ( 1,25 x .25) to
heat a liter or air by 1 degree. If you are standing around and once
in a while walk back and forth a bit, you might breath around 20
half liter breaths per minute and so breathe around 10 liters of air
every minute. Working fairly hard would make you breathe much
harder, so that 50 liters of air per minute might be typical.
Now suppose that you are in Minnesota and it’s -13C outside,
also typical. Your internal temperature is around 37C, so your need
to warm up that air by 50C as you breath it. How many 1500 kcal
supersized double cheese quarterpounder meas do you need to heat
up the air if you just stand around outside all day (for 500
minutes). How about if you work pretty hard?
Answer: Warming up liter of air by 50C costs .3x50 or 15 cal.
Standing around all day you would breathe in about 10x500 or
5000 liters of air, costing you 75 kcal or a bit over .05 of those
jumbo meals. If you are working hard in those conditions, you
need 375 kcal just to warm up the air, which is about what an
average hamburger costs you (about .25 of our putative jumbo
artery clogger). Lumberjacks working very hard in cold weather
can pretty easily need 1000 kcal per day just to warm the air they
breath.
What every 2107 student should know about pH
A + B  C + D simple reaction where A and B are reactants, C
and D are products
Brackets denote concentration in molar units
Keq= ([C] [D])/([A] [B]) where concentrations are at equilibrium
and Keq is the equilibrium constant
H2O  H+ + OH- dissociation of water into protons and
hydroxyl ions
K = [H+] [OH-] = 10-14 equilibrium constant for water – the ion
product (its really dimensionless because its in terms of activities,
but we usually treat it as if the dimensions are M-2; the product of
proton and hydroxyl concentrations is 10-14.
Definition: pH = -log10 [H+]
So [H+]= 10-pH
Example: at pH 8 the proton concentration is 10-8 M, so [OH-] is
10-6 M.
A buffer is a substance which can exist in at least two protonation
states in water. We can consider a simple buffer A which can exist
in solution as A- or as AH. Example- acetic acid and acetate ion
can coexist in solution as CH3COOH and CH3COO-. If both forms
of the buffer are present, adding acid or base causes the pH to
change slowly because acid has to protonate the basic component
of the buffer, and base would have to deprotonate the acid
component of the buffer.
The pK (sometimes refered to as pKa) of actetate is around 4.75;
this is molecular property like molecular mass, while pH is a
solution property. The pK describes the equilibrium for
dissociation of the acid form in solution. Strong acids have low
pKs (<2.5) and release almost all their protons, while weak acids
like acetic acid have high pKs and don’t fully dissociate. The pK is
the pH at which the buffer is half in the protonated (acidic) form
and half in the unprotonated (conjugate base) form, Acetate is a
good buffer near its pK in the range pH 3.75 to pH 5.75.
Problems
What are the concentrations of H+ and OH- ions in (aqueous)
solution at pH 7? Notice they are equal- charge is balanced without
any other ions. Either there are no other ions (pure water) or the
number of other + and – charges is equal ( like if you added NaCl).
What are the concentrations of H+ and OH- ions in (aqueous)
solution at pH 5? How can you account for charge balance?
What are the pH and H+ concentration when the OH- ion
concentration is 10-11 M?
What are the pH and OH- ion concentration when the proton
concentration is 3x10-4 (this is the same as ~10-3.5) ?
Assume HCl fully dissociates so that you get one free proton
(unless something else takes it up) for each HCl you add. How
much HCl do you need to change the pH of pure water from 7 to
4? (it ain’t pure after you add the acid, just to start)
Now assume that you have 200 mM of the deprotonated (conjugate
base) form of a buffer with a pK of 4 in solution at high pH. About
how much HCl will it take to get to pH 4?
You have a 100mM solution of a weak acid with pK~6 at pH 3 and
add 50 mM of NaOH, a strong base. What’s the pH?
Answers
10-7 each (ion product)
10-5 (protons) and 10-9 (hydroxyls) (ion product) there must be
other negative charges (anions) like chloride
3 and 10-3 (ion product and definition of pH)
3.5 and 10-10.5 (ion product and definition of pH)
(10-4 M -10-7 M) ~ 10-4 M
About 100 mM, because to get to the pK you need to neutralize
half of the base.
By adding 50 mM you neutralized half the acid, so pH~pK~6.