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\centerline{\textbf{Ice Calorimetry Lab}}
\bigskip

In this lab, we will be measuring how much energy it takes to melt ice
and heat water.

\begin{multicols}{3}
\paragraph{Materials:}
\begin{itemize}
\item Styrofoam cup
\item Heating element
\item Scale
\item 2 digital multimeters
\item Temperature guage
\item Ice and water
\end{itemize}
  \includegraphics[width=2in]{eeicelabhandfig1.png}
\end{multicols}

\LARGE \textbf{The setup}

\normalsize You will put some mass of ice (about 50g) and ice-cold water (about
150g) into your styrofoam cup.  Use the scale to record the mass of
the ice and water as you add them to the cup.  Finally, add your
ice-cold heating element and thermometer through the lid of the cup.

%% An easy way to add the ice (or water) is to add the desired mass on
%% one side of the scale, opposite the cup.  Then add ice (or water)
%% gradually to the cup until the scale tips.  At that point, you adjust
%% the balance to determine the actual weight of the filled cup.

\LARGE \textbf{Collect data}

\normalsize We will be measuring the temperature of the water and the power
dissipated in the heating element (which is just a resistor).  Thus we
can find out how much energy was added to the water, and how this
changes the temperature.  In order to keep the temperature measurement
reasonable, we will need to periodically stir the cup and heat it
moderately slowly.

You will be collecting temperature data using the computer, so before
you turn on the heater, you should make sure the computer is taking
data.  Turn on the heater, and write down the time you do so as well
as the current and voltage, from which you can find the power
dissipated in the resistor.  If the current or voltage changes during
the course of the experiment, take note of the new values---and the
time.

\textbf{Question 1.1 Plot your data I} \hs Plot the temperature versus total energy
  added to the system (which you can call $Q$).  To do this, you will
  need to integrate the power.  Discuss this curve and any interesting
  features you notice on it.


\textbf{Question 1.2 Plot your data II} \hs Plot the heat capacity versus
  temperature.  This will be a bit trickier.  You can find the heat
  capacity from the previous plot by looking at the slope.
  $$
    C_{p} = \left(\frac{\partial Q}{\partial T}\right)_{p}
  $$
  This is what is called the \emph{heat capacity}, which is the amount
  of energy needed to change the temperature by a given amount.  The
  $p$ subscript means that your measurement was made at constant
  pressure.  This heat capacity is actually the total heat capacity of
  everything you put in the calorimeter, which includes the resistor
  and thermometer.


\textbf{Question 1.3 Specific heat} \hs From your plot of $C_{p}(T)$, work out the
  heat capacity per unit mass of water.  You may assume the effect of
  the resistor and thermometer are negligible.  How does your answer
  compare with the prediction of the Dulong-Petit law?
 
 \medskip
 
\textbf{Question 1.4 Latent heat of fusion}

  \begin{enumerate}
  \item What did the temperature do while the ice was melting? How much
    energy was required to melt the ice in your calorimeter? How much
    energy was required per unit mass? per molecule?
  \item The change in \emph{entropy} is easy to measure for a
    reversible isothermal process (such as the slow melting of ice),
    it is just
    $$
      \Delta S = \frac{Q}{T}
    $$
    where $Q$ is the energy thermally added to the system and $T$ is
    the temperature in Kelvin.  What is was change in the entropy of
    the ice you melted? What was the change in entropy \emph{per
      molecule}? What was the change in entropy per molecule divided
    by Boltzmann's constant?
  \end{enumerate}

\textbf{Question 1.5 Entropy for a temperature change} \hs Choose two temperatures
  that your water reached (after the ice melted), and find the change
  in the entropy of your water.  This change is given by
  $$
    \Delta S = \int \frac{\delta Q}{T} 
  $$
  $$
    = \int_{t_i}^{t_f} \frac{P(t)}{T(t)}dt
  $$
  where $\delta$ represents an inexact differential, $P(t)$ is the heater power as a function of time, and $T(t)$ is
  the temperature, also as a function of time.
  
    
\vfill
\leftline{\textit{by David Roundy}}
\leftline{\copyright DATE David Roundy}

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