**General comments**

The lecture focused on the meaning and use of the specific heats. I wanted to start with a PRS question regarding the applicability of a particular form of the first law, but since the systyem wasn't up yet, I discussed it instead. It is important that you recognize the various assumptions implicit in the different forms of the equations and that you use the appropriate equations when solving problems. We will frequently deal with systems that have negligible changes in PE and KE, only p-v work, and are quasi-static. The compendium of equations lists most of the common forms along with their applicability.

In order to use the first law to calculate changes in temperature, pressure,
etc. of a system, we need to be able to relate changes in internal energy (or
enthalpy) to other properties. We do this using the specific heats. These are
experimentally measured properties of a material. More details are contained
in the notes. Here are the most important
points: 1) cv and cp tell us how many Joules of heat transfer are required to
change the temperature of a material by 1 Kelvin via a constant volume or constant
pressure process, respectively. 2) These are properties of the material we are
dealing with (that is they are different for different gases, solids, etc.).
3) In general, u is a function of two properties of the system (as is h), however,
if u=u(T) only and h=h(T) only then du=cvdT and dh=cpdT. If this gas also obeys
pv=RT (a thermally perfect gas), then the gas is called an __ideal gas__.
Many of the gases we deal with behave this way. 4) cv and cp allow us to relate
changes in h and u to changes in T. As to which to use when, if you want to
calculate the change in internal energy use cv, if you want to calculate the
change in enthalpy, use cp. 5) Generally, cv and cp are functions of temperature
(that is, the specific heat of a material changes with temperature), but for
most of what we do in this class we will assume they are constant. For air for
example, du=cvdT with cv=716.5 J/kg-K implies that if the temperature of 1 kg
of airs changes by 1 K, then its internal energy has changed by 716.5 J. Similarly,
since cp=1003.5 J/kg-K for air, its enthalpy (u+pv) would change by 1003.5 J.

One subtle point is that in general two thermodynamic properties are needed to specify u and h. However, for many gases (that we will can ideal gases) u and h are only functions of T. To clarify this, think of yourself as the person doing the experiment to measure cp. You designed the experiment up to be a constant pressure process, added heat and measured the change in temperature. Then one day you realized that the piston was stuck and the pressure had been changing -- but you still got the same change in temperature for the same amount of heat addition as when it was a constant pressure process. You conclude that the process doesn't matter for this gas (an ideal gas)--it doesn't have to be a constant pressure process--dh is always equal cpdT (and similarly du = cvdT if you think about the constant volume experiment).

We had a second PRS question regarding the relative magnitudes of cv and cp for gases (and we discussed the case of solids).

We concluded with an experiment and estimation exercise to allow you use specific heat to relate temperature change to a change in internal energy. Note that one item that was not included in the estimations made by many was the energy required to heat the flask. For reference the flask had a mass that was similar to the water (Carl measured it at 0.128 kg) and a specific heat that was about 700 J/kg-K -- so it would absorb a substantial amount of energy also.

**Responses to 'Muddiest Part of the Lecture Cards'**

(53 respondents out of 66 students)

1)* Unclear on Cv, Cp, the derivation, etc.*
(8 students) Please review the section of
the notes as well as my general comments above.

2) * When do you use du=cvdT and when dh=cpdT?*
(2 students) If you want to find the change in u, multiply the change in temperature
by cv. If you want to find the change in h, multiply by cp. So it depends on
what you are looking for.

3) * Why do we associate u with cv and h with cp?*
(2 students) Because du = delq for a constant volume process (for small changes
in KE and PE and quasi-static p-v work), and dh = delq for a constant pressure
process (under the same assumptions).

4) * Is cv approximately equal to cp for liquids
too? * (2 students) Yes, liquids are nearly incompressible.

5) * How do you estimate then efficiency of the hot
plate?* (2 students) I would do it just based on % of area covered
by the flask.

6) * Will we need to go through the derivation ourselves?*
(1 student) No, but you must understand the assumptions in the various equations
and models you use.

7) * What defines an ideal gas and how good of an
approximation is air to an ideal gas?* (1 student) See above for
the first part of your question. For the second part, the answer is "very
good" at most (but not all) conditions we encounter in aerospace applications.
For additional information, please see Section 3.4 of Sonntag, Borgnakke and
Van Wylen "Fundamentals of Thermodynamics" (on reserve in the library).

8) * Why do you multiply cv by mass in the final
problem?* (1 student) Th units on cv are J/kg-K.

9) * If the piston was massless, would you still
have a constant pressure because of the outside atmospheric pressure?*
(1 student) Yes, unless the atmospheric pressure was zero.

10) * How are cp and cv related?* (1
student) We will get to this, cp-cv=R

11) * I am just amazed that u and h are only a function
of T for many gases. Is there some possible intuition for that?*
(1 student) Not that I can think of off-hand, but it sure does make our life
easier when estimating the performance of systems.

12) * When do you use partial derivatives and when
not? * (1 student) If I am interested in the rate of change of
a function with respect to only one variable I use the partial derivative. Whereas
the total rate of change of a function will be comprised of components due to
each variable. Your calculus textbook can provide you additional information
for review.

13) * No mud.* (28 students) I was
very happy to get this many "no mud" cards. In the past I have struggled
to present this material clearly. I think I did better this time.