
Thermodynamics Notes
I. Types of Energy
Energy is the ability to do work… that is to apply a force to cause an object to move. Energy comes
in many forms. Primarily in chemistry we are concerned primarily with chemical potential energy
which is stored in chemical bonds. We will also be concerned with heat, the transfer of thermal
energy associated with the kinetic energy of molecules, atoms, and ions. Kinetic energy is also
important for gas laws and the speed of electrons ejected after the photoelectric effect. Photons
carry energy that is generally associated with electrons moving from level to level in the atom.
Nuclear energy is energy stored by the binding energy in the atom.
We will also be concerned with work. Generally, work to a chemist is P-V work: the work done by
a gas as it moves a piston. W=(PV). However, since this is usually done at a constant pressure
the pressure term is able to be separated from the differential. W  PV . Technically, work is the
integral (area under the curve) of a P-V graph in which the volume is x-axis and pressure is the yaxis.

Energy can be measured in a variety of units. Long before I was in school, it was measured in
ergs (a good thing to know if you do crossword puzzles). The two commonly used units these
kg* m 2
days are calories and joules. The SI unit for energy is the joule (which is really
). In
s2
chemical reactions, the numbers are usually so large that we use kJ for convenience. Calories are
convenient when finding heat by calorimetry (mixing or heating) because the specific heat, ‘c’, of
cal
water is 1.00
. 1cal  4.184Joule

g *o C
II. Heat transfer


When heating a substance that does not undergo a phase change, the heat (q) obeys the
equation: q  mcT [ ‘m’ is the mass of the object being heated. ‘c’ is the specific heat constant of
the substance. ‘T’ is the change in temperature caused by the addition or removal of heat
energy.]

For example, to raise the temperature of 125 grams of water from 21.0 oC to 55.5 oC.
q  mcT
q  125(4.184)(55.5  21.0)
Notice that the value is positive because heat energy was
q  18043.5J
absorbed. When q is positive, the process is described as endothermic. If q is negative, then the
process is exothermic.
Since energy cannot be created or destroyed according to the Law of Conservation of energy (the
first law of thermodynamics), whatever heat energy one object gains, another object must lose the
same amount of energy. This leads to the calorimetry equation (we called it the mixing equation in
first year chemistry).


qgaincold  qlosthot
m1c1T1  m2c 2T2
Since heat always flows from high concentration (high temperature)
m1c1(Tf ,1  Ti,1)  m2c 2 (Tf ,2  Ti,2 )
m1c1(Tf  Ti,1 )  m2c 2 (Tf  Ti,2 )
to low concentration (low temperature), the q term for the cold object is always positive
(endothermic) and the q term for the hot object is always negative (exothermic). They must be
equal in magnitude but opposite in sign as seen in the derivation above. Moreover, when the two
objects reach thermal equilibrium, they must have the same final temperature. So if you mix 125
grams of water at 25.0 oC with 265 g of water at 75.0 oC, we know the temperature must end
higher than 25 degrees but lower than 75 degrees. Also, since both substances are the same
(have the same ‘c’ value), the final temperature must be closer to the more massive sample, in this
case the 75 degree sample. So the final temperature can be estimated to be above 50 degrees
(the
simple
average)
and
the
maximum
value
of
75
degrees.
qgaincold  qlosthot
m1c1T1  m2c 2T2
125(4.184)(Tf  25)  265(4.184)(Tf  75)
523(T  25)  1108.76(T  75)
523T 13075  1108.76T  83157
This ‘q’ value can eventually be related to the H of a
1631.76T  96232
T  58.974
reaction.
III. Phase energy Diagrams
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.


Pictured above is a phase-energy diagram. The diagonal sections with a positive slope represent
heating a substance without changing its state of matter (no phase change). They all obey
q  mcT . However, the specific heat, ‘c’, for the solid (farthest left line), the liquid (middle line),
and gas (farthest right line) are usually different. Consequently, the slopes of these steps (we call
them Q1, Q3, and Q5) are different. The two horizontal stages (Q2 and Q4) represent melting and
vaporizing as labeled above. They obey the equation q  mH . Again, step two has a different
constant (heat of fusion) than step four (heat of vaporization).
The total heat energy involved is simply the sum of all the individual steps that are involved.

5
qtotal   qi Not all problems involve moving across all five steps. For example, if you heated liquid
i1
water at 25 oC to steam at 145 oC, you would only need to calculate steps 3,4 and 5.
Additionally, the sign of the steps follows the usual rules. If the process is endothermic (moving to
the right on the graph), then make all the individual q values positive. On the other hand, if the
process is exothermic (moving to the left), then make all the q values negative.
This diagram can be used for any substance. The only differences would be that the constant
values (specific heats, melting point, boiling point, heat of fusion, and heat of vaporization) would
have different numerical values.
It is important to not that the heat of vaporization (Hvap) is ALWAYS larger than the heat of fusion,
(Hfus). This is because it requires substantially more energy to break all the intermolecular
attractions in going from a liquid to a gas than to only break some of them (lattice energy, for
example) in converting a solid to a liquid.