\input{labpreamble}
\begin{document}
\subsubsection*{PHSX 101L\hfill Experiment 10  \hfill Name:\underline{\hspace*{2in}}}

\underline{Title:} Electrical Heating

\underline{Object:} To compare two units which are commonly used for 
energy and work, the calorie and the joule, by electrically heating 
some water.

\underline{Theory:} The calorie is a common unit of thermal energy and
the energy of foods (though a food Calorie is 1000 physics calories)
and the joule is the common unit of chemical, mechanical, or
electrical energy.  Of course, because of the relationship between
energy and work, these units are also used for both energy and work.

In this experiment we intend to compare the calorie with the joule by 
comparing a measured decrease in the level of electrical energy (in joules) 
with the equivalent increase in thermal energy (in calories).

When an electric current flows from a high potential to a low potential 
through a circuit, the energy of the flowing electrons is reduced because 
negative work is done on the electrons.  But comparable positive work must 
be done by the electrons on some other part of the system and hence the 
energy of that part of the system increases.

In this experiment the electric current flows through a resistive
Nichrome wire (with resistance $R$).  The current interacts with the
metal atoms of the wire, does work on those atoms, and increases their
average energy and thereby heats it.  Thus, some of the energy of the
electrons now appears as thermal energy of the atoms.  The atoms of
the metal, which are in contact with the water, do work on the water
molecules and thereby increase their energy as well.

The atoms do negative work on the electrons and decrease their energy but 
the electrons do equal positive work on the atoms and increase their energy.
\begin{eqnarray}
\label{joule}({\rm decrease\ in\ electrical\ energy})&=&({\rm increase\ in\ thermal\ energy})\\
({\rm power})({\rm time})&=&({\rm specific\ heat})({\rm mass})({\rm 
temperature\ change})\\
({\rm current})({\rm voltage})({\rm time})&=&({\rm specific\ heat})({\rm mass})({\rm 
temperature\ change})\\
IVt & = & cm\Delta T
\end{eqnarray}
Using Ohm's Law ($V=IR$, where $I$ is the current and $V$ the voltage) 
the equation can be written:
\begin{equation}
I^{2}Rt=cm\Delta T
\end{equation}
\noindent The left hand side of this equation will be in joules (the 
official S.I. unit of energy), while 
the right hand side is in calories; thus the number of joules in one 
calorie can be determined.

\underline{Historical Note:} The discipline of physics began gradually
long before the likes of Aristotle---they called it ``natural
philosophy'' back then.  There were what we would now call furtive
efforts to describe their universe.  By the time of Aristotle the
universe was considered to be composed of the Sun, the Moon, many
twinkly things a few thousand feet in the air (some of which had
strange motions), and the Earth.  The Earth was thought to be composed
of combinations of four substances: earth, water, air, and fire.  Our
understanding has improved since then.  We know about the several
(approximately 100) elements and their combinations, we know that water
is but one of those many combinations, we have studied the mixture of
gasses which make up the air, but we still stumble a lot about
Aristotle's ``fire.''  By the time of the Renaissance, the ``fire''
was thought to be a fluid substance, then called ``caloric,'' the
amount of which was constant/conserved in the universe.  Count Rumford
(n\'ee Benjamin Thompson, 1753-1814), while he was supervising the
boring of canon, realized that this concept was nonsense, but for a
century few would acknowledge the validity his very important
discovery.

Currently most physics textbooks do acknowledge this---they say that
there is no such thing as ``caloric''---and then the authors go right
on talking about ``heat'' as if it were a fluid which moves from one
place or object to another: they teach that ``heat'' moves from hot
things to cold things, that ``heat rises,'' and about the amount of
``heat'' inside an object or system.

There is no such thing as ``heat''---there are hot \emph{things} but no 
object or substance called ``heat.''  Hot \emph{air} rises, 
for example, but heat doesn't rise since it doesn't exist.  (The word 
``heat'' can be correctly used as a verb, but we'll avoid its use as 
a noun.)

The result is that physics has retained much of the old language of 
thermodynamics and made it very difficult to talk or write about.  Thus the 
language of this presentation seems a little strained---it is correct but 
admittedly not the way most physics texts would present it.


\underline{Procedure:} 
\vspace{-.25in}
\begin{enumerate}

\item Measure the room temperature with your thermometer and record it.

\item Determine the resistance of the coil with an ohmmeter.  You can 
also monitor the resistance during the experiment by measuring the current 
and voltage, $R=V/I$.

\item Connect the voltmeter, ammeter, and heating coil into the 
circuit with the power supply.  (Do not turn on the power until 
checked by the instructor.  Also do not heat coil until it is 
immersed.)

\item Fill the calorimeter can 3/4 full of cold water (from the
drinking fountain, about 10 \degree C below room temperature).

\item Adjust the current to 4.5 amps and begin to time the 
experiment.  Heat the water until it is as much above room temperature 
as it started below.  Gently stir occasionally.

\item Record time, current, voltage, mass of water (take the water mass 
equivalent of the calorimeter, 14.7 g, into account), and temperature 
rise.

\item Divide the electrical energy (in joules) by the increase in
thermal energy (in calories) to determine the number of joules in a
calorie.  Compare with the accepted value.

\end{enumerate}


\newpage

\underline{Apparatus:} Make a diagram of the apparatus used; include 
a wiring diagram.
\vspace{2.0in}

\underline{Results:} 
\vspace{-.2in}

\renewcommand{\arraystretch}{2.4}
\begin{tabular}{lr}
	
Resistance of coil: & \underline{\hspace{2.1in}} \\

Voltage: &  \underline{\hspace{2.1in}} \\

Current: &  \underline{\hspace{2.1in}} \\

Heating time: &  \underline{\hspace{2.1in}} \\

Electrical energy consumed (in J):  &  \underline{\hspace{2.1in}} \\

Mass of water (include allowance for can):\ \ \   &  \underline{\hspace{2.1in}} \\

Initial temperature: &  \underline{\hspace{2.1in}} \\

Final temperature:  &  \underline{\hspace{2.1in}} \\

$\Delta T$: &  \underline{\hspace{2.1in}} \\

Thermal energy generated (in cal.): &  \underline{\hspace{2.1in}} \\

Calculate joules per calorie by dividing:  &  \underline{\hspace{2.1in}} \\

Accepted value:  &  \underline{\hspace{2.1in}} \\

\% difference:\ \ \  &  \underline{\hspace{2.1in}}

\end{tabular}

\newpage

\underline{Questions:} 
A typical 50 gallon household hot water heater contains 189 kg of 
water.  Water from the water line is about 15 \degree C, and is 
heated by an electrical heating element having a resistance of 11 
$\Omega$.  The supply line voltage is 220 V.
\vspace{-.2in}
\begin{enumerate}
\item By how many calories must the thermal energy of the 
tank of water be increased to heat it to bath temperature (40 \degree C)?
\vspace{.7in}

\item Calculate the power ($P=IV$) of the electric heating element.
\vspace{.7in}

\item Calculate the time required to heat up the water in the tank.
\vspace{.7in}

\item Calculate the cost of heating the water if the utility company 
charges 8 cents per kW-hr.
\vspace{.7in}

\end{enumerate}

\vspace{-.1in}
\underline{Conclusions:} 

\end{document}
