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\begin{document}

\subsubsection*{PHSX 101L\hfill Experiment 9\hfill Name:\underline{\hspace*{2in}}}

\underline{Title:} Specific Heat

\underline{Object:} To determine the specific heat of various 
substances.

\underline{Introduction:} Specific heat is an indication of how much 
the temperature of an object will change when thermal energy is added 
or subtracted.  Specific heat varies from substance to substance 
which means that two objects of the same mass but of different 
substances will change temperatures by different amounts when the 
same amount of energy is added or subtracted.  For example, have you 
ever held a hot piece of pizza by its crust only to have the moister 
parts burn your mouth when you take a bite?  The meats and cheeses 
have high specific heat, while the crust has a low specific heat 
capacity.  The specific heat of a substance $c$ would be the slope of a 
straight line of energy ($E$ in general, or $Q$ specifically for 
thermal energy) vs.\ the mass of the object times the change in 
temperature.
\begin{equation}\label{c}
	E= cm\Delta T
\end{equation}
Therefore the specific heat of a substance can be calculated by measuring the 
change in temperature of an object when a known amount of energy is 
added (such as by doing mechanical work on the object).

Specific heats of different substances can be compared by measuring 
temperature changes as thermal energy flows from one object to 
another.  For example, if a sample of hot metal is dropped into a 
quantity of cold water then they both arrive at a common temperature 
somewhere between the extremes.  By using a good calorimeter to 
minimize losses or gains to or from the room we can assume that the 
thermal energy $Q$ lost by the metal is gained by the water.
\begin{equation}
	Q_{lost}=Q_{gained}
\end{equation}
\begin{equation}\label{compare}
	c_{\rm m}m_{\rm m}\Delta T_{\rm m}=c_{\rm w}m_{\rm w}\Delta T_{\rm w}
\end{equation}

\noindent
We'll first see the equivalence of mechanical work and thermal energy by doing 
work on some lead, assuming the work is transformed into thermal energy, and 
using that assumption to deduce the specific heat of lead.  Then we'll 
use our new knowledge (and equation \ref{compare}) to measure the 
specific heat of water; and then use the specific heat of water (and equation 
\ref{compare}) to measure the specific heat of a different metal.

\underline{Apparatus:} Draw the apparatus. 


\pagebreak
\underline{Procedure:} 
% \vspace{-.2in}

\underline{Using equation \ref{c}}:
\vspace{-.2in}
\begin{enumerate}
	\item Put some lead shot in a plastic bottle and find the mass of the lead. 
	\item Measure the initial temperature (in \degree C) of the lead shot 
	with a sensitive thermometer.
	\item Drop the bottle $n$ times (about 20 or 30) in quick succession 
	from the lab bench to the floor.
	\item Immediately measure the final temperature of your lead shot (the rise 
	won't be much).
	\item Calculate the mechanical work done (energy added) from $W=n\,Fd$ where $F$ is the 
	weight (in newtons) of the shot and $d$ is the distance (in meters) from the floor 
	to the lab bench.  The result will be in units of Nm; and one Nm is 
	equal to one joule.  Now convert the energy from joules to calories.
	\item Make a plot of energy (or work) (in calories) vs. mass times the 
	change in temperature (in g~\degree C).  Your lead shot sample will 
	provide one data point; also acquire the data from the other groups 
	in the class on this procedure for other data points to plot.  
	Use a ruler and eyeball a best-fit straight line through the origin.
	\item The slope of your straight line is your experimental value for 
	the specific heat of lead (in cal/g~\degree C); compare it to the accepted value.
\end{enumerate}

\underline{Using equation \ref{compare}}:
\vspace{-.2in}
\begin{enumerate}
	\item Measure the mass of a solid piece of lead or a quantity of 
	lead shot.
	\item Suspend the solid lead by a string or the shot in a cup in the steam 
	generator to raise its temperature to the temperature of boiling water 
	(not necessarily 100 \degree C at our altitude).
	\item Put a quantity (about 1/2 the mass of your lead sample) of 
	cold water in the calorimeter and measure its mass and initial temperature.  
	Include 12.2 grams to account for the effect of the aluminum calorimeter cup.
	\item Record the initial hot temperature of the lead.
	\item Quickly dry the lead (if you used the solid) and put it in the 
	cold water in the calorimeter.
	\item As they equilibrate (gently use your stirring rod 
	occasionally), record the highest water temperature as 
	the final temperature for both the lead and the water.
	\item Use these data along with the specific heat for lead (use the 
	accepted value) to compute the specific heat for water from equation 
	\ref{compare}.
\end{enumerate}

Now that you have found the specific heat for water, use the 
procedure above (for equation \ref{compare}) to find the specific 
heat of a different solid metal sample, such as aluminum or steel or 
brass.

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

\underline{Using equation \ref{c}}:
% \vspace{-.2in}

Mass of shot (first in g, then in kg): \hfill $m_{\rm Pb}=$\ \underline{\hspace{1.2in}} 
\hfill $m_{\rm Pb}=$\ \underline{\hspace{1.2in}}

Weight of shot (in newtons): \hfill $(mg)_{\rm Pb}=$\ \underline{\hspace{1.2in}}

Initial Temp.:  $T_{{\rm Pb}_{i}}$ = \underline{\hspace{.8in}}  \hfill
Final Temp.: $T_{{\rm Pb}_{f}}$ = \underline{\hspace{.8in}}
\hfill $\Delta T_{\rm Pb}=$\ \underline{\hspace{1.2in}}

Energy added (first in J, then in cal):
\hfill  $E=$\ \underline{\hspace{1.2in}}\hfill  $E=$\ \underline{\hspace{1.2in}}

Experimental value for specific heat of lead (slope of line): 
\hfill $c_{\rm Pb}=$\ \underline{\hspace{1.2in}}

Compare with accepted value:\hfill  $c_{\rm Pb}=$\ 
\underline{\hspace{1.2in}}\hfill \% error $=$\ \underline{\hspace{1.2in}}

\vspace{.2in}

% \vspace{.3in}
\underline{Using equation \ref{compare}}:
% \vspace{-.2in}

Mass of lead (in grams): \hfill $m_{\rm Pb}=$\ \underline{\hspace{1.2in}}

Initial Temp.:  $T_{{\rm Pb}_{i}}$ = \underline{\hspace{.8in}}  \hfill
Final Temp.: $T_{{\rm Pb}_{f}}$ = \underline{\hspace{.8in}}
\hfill $\Delta T_{\rm Pb}=$\ \underline{\hspace{1.2in}}


Mass of water (add 12.2 g to account for the calorimeter cup): 
\hfill $m_{\rm w}=$\ \underline{\hspace{1.2in}}

Initial Temp.:  $T_{{\rm w}_{i}}$ = \underline{\hspace{.8in}}  \hfill
Final Temp.: $T_{{\rm w}_{f}}$ = \underline{\hspace{.8in}}
\hfill $\Delta T_{\rm w}=$\ \underline{\hspace{1.2in}}

Specific heat of water (solve equation \ref{compare} for $c_{\rm w}$): 
\hfill $c_{\rm w}=$\ \underline{\hspace{1.2in}}

Compare with accepted value:\hfill $c_{\rm w}=$\ 
\underline{\hspace{1.2in}} \hfill \% error $=$\ \underline{\hspace{1.2in}}

\vspace{.2in}
\underline{Using equation \ref{compare} for a different metal} \ \ (metal is \underline{\hspace{1.2in}}) :
% \vspace{-.2in}

Mass of water (add 12.2 g to account for the calorimeter cup): 
\hfill $m_{\rm w}=$\ \underline{\hspace{1.2in}}

Initial Temp.:  $T_{{\rm w}_{i}}$ = \underline{\hspace{.8in}}  \hfill
Final Temp.: $T_{{\rm w}_{f}}$ = \underline{\hspace{.8in}}
\hfill $\Delta T_{\rm w}=$\ \underline{\hspace{1.2in}}

Mass of metal (in grams): \hfill $m_{\rm m}=$\ \underline{\hspace{1.2in}}

Initial Temp.:  $T_{{\rm m}_{i}}$ = \underline{\hspace{.8in}}  \hfill
Final Temp.: $T_{{\rm m}_{f}}$ = \underline{\hspace{.8in}}
\hfill $\Delta T_{\rm m}=$\ \underline{\hspace{1.2in}}

Specific heat of metal (solve equation \ref{compare} for $c_{\rm m}$): 
\hfill $c_{\rm m}=$\ \underline{\hspace{1.2in}}

Compare with accepted value:\hfill  $c_{\rm m}=$\ \underline{\hspace{1.2in}} 
\hfill \% error $=$\ \underline{\hspace{1.2in}}


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

\begin{enumerate}
	\item Why might the results in the first procedure not 
	be very accurate?
	\vspace{.6in}
	\item How could one improve the results of this whole experiment?
	\vspace{.6in}
	\item Water is often used as a heat storage medium.  Is there a 
	physics reason for this, or is it just cheap and plentiful?  What 
	property of water determines the answer to this question?  How unique 
	is water with respect to this property?
	\vspace{.6in}
% (The next lab is similar to the first procedure, but with a 
% different goal.)  
\end{enumerate}

\vspace{-.3in}
\underline{Conclusions:} 

\end{document}
