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

% \underline{Title:} Electrical Heating


% \underline{Object:} To determine the electrical equivalent of the 
% calorie in joules by using an electric current to heat a known mass of 
% water.
% 
% \underline{Theory:} In this experiment we intend to find the joule 
% equivalent of the calorie by transforming a measured quantity of 
% electrical energy into thermal energy.  When an electric current flows 
% from a high potential to a low potential through a circuit, electrical 
% energy is consumed and is transformed into mechanical work, heat, or 
% some other form of energy.  In this experiment the electric current 
% flows through a resistive wire (with resistance $R$) and heats it, 
% producing no work.  Thus, the electrical energy consumed is converted 
% entirely into thermal energy.

\underline{Object:} To compare two units which are used for energy and
work: the calorie and the joule.

\underline{Theory:} The calorie is a common unit of thermal energy and
the energy of foods, 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
the degree of either energy or work.

We'll compare the calorie with the joule by comparing a measured
decrease in the level of electrical energy (in joules) with the
equivalent increase in the level of the thermal energy (in calories). 
When an electric current flows from a high potential to a low
potential through a wire with resistance $R$, 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.

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 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{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.  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 consumed (in joules) by the 
thermal energy generated (in calories) to determine the number of 
joules in a calorie. Compare with the accepted value.

\end{enumerate}


\marginparwidth=1.5in \marginparsep=-1.5in \marginpar{\underline{Apparatus:} Make a diagram of the apparatus used; include 
a wiring diagram.}

\vspace{-.5in}

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

\renewcommand{\arraystretch}{2}
\begin{tabular}{lr}
	
Resistance of coil: & \underline{\hspace{1.8in}} \\

Voltage: &  \underline{\hspace{1.8in}} \\

Current: &  \underline{\hspace{1.8in}} \\

Heating time: &  \underline{\hspace{1.8in}} \\

Electrical energy consumed (in J):  &  \underline{\hspace{1.8in}} \\

Mass of water (include allowance for can):  &  \underline{\hspace{1.8in}} \\

Initial temperature: &  \underline{\hspace{1.8in}} \\

Final temperature:  &  \underline{\hspace{1.8in}} \\

$\Delta T$: &  \underline{\hspace{1.8in}} \\

Thermal energy generated (in cal.): &  \underline{\hspace{1.8in}} \\

Calculate joules per calorie by dividing:  &  \underline{\hspace{1.8in}} \\

Accepted value:\ \underline{\hspace{.9in}}\ \hfill \% difference: &  \underline{\hspace{1.8in}}

\end{tabular}



\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 How many calories of thermal energy must be delivered to the 
tank of water to heat it to bath temperature (40 \degree C)?
\vspace{.3in}

\item Calculate the power ($P=IV$) of the electric heating element.
\vspace{.3in}

\item Calculate the time required to heat up the water in the tank.
\vspace{.3in}

\item Calculate the cost of heating the water if the utility company 
charges 8 cents per kW-hr.
\vspace{.3in}

\end{enumerate}

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

\end{document}
