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\newpage




\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\SUBLIMATION OF CARBON DIOXIDE}



\markboth{Sublimation of Carbon Dioxide}{Sublimation of Carbon Dioxide}








\author{Brian Winkel,
Director SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
}


    \makeStitlePDFLaTex


\begin{abstract}
We offer data on the sublimation of dry ice (carbon dioxide) collected in a classroom setting so that students can model the rate of change in the mass of a small solid carbon dioxide block with a differential equation model, solve the differential equation, estimate the parameters in the model using the data sets, and validate the model.
\end{abstract}



\section*{SCENARIO DESCRIPTION}


Matter exists in several states: plasma, gas, liquid, and solid. When a solid becomes liquid the object melts. When a liquid becomes a gas the object vaporizes. When a solid becomes gas directly the object  {\it sublimates\/}. A readily available example of sublimation occurs when the solid, dry ice (solid carbon dioxide - ${\rm CO}_2$), becomes gaseous.

Now if we had a solid block of dry ice we would notice it ``disappear'' as it sublimates into its gaseous state. Indeed, the mass of the block of dry ice would decrease. A reasonable question to ask is, ``At what rate does the mass decrease as the dry ice sublimates?" A related question is, ``What does the rate of change of the mass depend upon?''

Some years ago in a course we were teaching we too wondered whether we could model the sublimation of a piece of carbon dioxide. At the time (over 30 years  ago) dry ice was used to pack ice cream shipments.  One day, because we had received some things in a shipment containing dry ice, we thought of collecting data on the sublimation of dry ice. While we had colleagues who were true engineers and scientists on our teaching team, we decided, as a mathematician who was  pretty much an inept lab fellow, to go this one alone.  

We set up a modest device consisting of a tray, in which we placed our dry ice on a scale which could render mass in grams to the thousandth of a gram. We zeroed out the scale (i.e. we noted the mass of the tray  and then noted the mass with the initial dry ice amount on it so that we could tell from total mass data exactly what our dry ice mass was as time progressed) and began to collect data. We should say we really lived dangerously, for we did this all for the first time in class with  two students in the first row recording the mass every 30 sec.  We have since lost that data, but there was a very unexpected thing that happened. The mass of the dry ice, rather we should say the mass of the zeroed  dry ice and tray, went up(!) in the first few minutes. We continued to collect data on that run and we all went home to ponder what was going on. I had the advantage of holding the equipment and some more dry ice. After class I tried the experiment again and found the problem. The dry ice was so cold, below sublimation temperature of dry ice, -78.5$^{o}$ C, that it was causing water to condense on the tray, and upon inspection, I could see the ice film on the tray. That was why the mass went up on this configuration for a while.




\begin{figure}[h]
\centering
\includegraphics[width=8.9cm]{1-12-SubApparatus.png}



\parbox{12 true cm}{\caption{\label{fig:1-12-SubApparatus} Sketch of apparatus for data collection consisting of a scale on which is mounted  a ring stand to hold the solid block of $CO_2$ out from  the base. This configuration will not  allow condensed moisture  realized by the falling cold sublimated $CO_2$ to collect on any surface involved in the scale apparatus.}}

\end{figure}



Most of the students were able to explain exactly what we saw, only without seeing it - they were sharpies! We all realized that we needed to somehow do the experiment by, perhaps, suspending the dry ice so that when the sublimated (but still very cold) ${\rm CO}_2$, which was heavier than air, fell it did not fall on any part of our scale device. The next day the teacher arrived with a proper set of equipment (see Figure~\ref{fig:1-12-SubApparatus}) and we began to collect data again. 


\begin{table}[t]
\begin{center}
\small
\begin{tabular}{||r|r|c|r|r||} \hline
Time (s) & Mass (g) &&Time (sec) & Mass (g)\\ \hline 
0&	25.525&&        1800&	13.553\\ \hline  
120&	24.512&&	1930&	12.910\\ \hline 
240&	23.524&&	2040&	12.331\\ \hline
360&	22.639&&	2170&	11.689\\ \hline 
480&	21.765&&	2280&	11.188\\ \hline 
600&	20.890&&	2410&	10.566\\ \hline  
720&	20.043&&	2530&	10.043\\ \hline
840&	19.221&&	2690&	9.377\\ \hline 
960& 	18.431&&	2780&	9.011\\ \hline
1080&	17.677&&	2880&	8.616\\ \hline  
1200&	16.936&&	3060&	7.945\\ \hline
1320&	16.220&&	3220&	7.404\\ \hline	 
1440&	15.548&&	3380&	6.877\\ \hline 
1570&	14.828&&	3480&	6.593\\ \hline 
1680&	14.213&&        3600&	6.244\\ \hline 

\end{tabular}

\parbox{12.4 true cm}{\caption{\label{table:MakkarWernerData} Data collected by students Masood Makkar and Paul Werner  (3 December 1992) on successful run for mass of dry ice (g) as a function of time (s).}}

\end{center}
\vspace*{-.8 true cm}
\end{table}


We collected the data in a room of temperature 19.3$^{o}$ C using a small box shape piece of dry ice we had carved (wearing protective gloves) to the size of two sugar cubes, 1 cm $\times$  1 cm $\times$ 2 cm. This time we did not experience any unusual phenomena. We present the collected data in Table~\ref{table:MakkarWernerData}.

 
\begin{table}[h!]
\begin{center}
\small
\begin{tabular}{||r|r|c|r|r||} \hline
Time (s) & Mass (g) &&Time (sec) & Mass (g)\\ \hline 
0&	7.570&&	480&	5.871\\ \hline  
30&	7.457&&	510&	5.776\\ \hline 
60&	7.338&&	540&	5.682\\ \hline 
90&	7.220&&	570&	5.589\\ \hline 
120&	7.110&&	600&	5.497\\ \hline 
150&	6.995&&	630&	5.405\\ \hline 
180&	6.885&&	660&	5.313\\ \hline 
210&	6.778&&	690&	5.224\\ \hline 
240& 	6.673&&	720&	5.136\\ \hline 
270&	6.571&&	750&	5.048\\ \hline 
300&	6.464&&	780&	4.960\\ \hline 
330&	6.363&&	810&	4.878\\ \hline 
360&	6.265&&	840&	4.790\\ \hline 
390&	6.163&&	870&	4.707\\ \hline 
420&	6.067&&	900&	4.625\\ \hline 
450&	5.969&&	&\\ \hline 
\end{tabular}

{\parbox{7.9 true cm}{\caption{\label{table:MottelData} Data collected by Dr.~Ed Mottel  on  successful run for mass of dry ice (g) as a function of time (s).}}}

\end{center}
\vspace*{-.8 true cm}
\end{table}


Incidentally, a few days later Dr.~Ed Mottel, a chemist in our teaching group, collected additional data in his lab using the same set up. We show this in Table~\ref{table:MottelData}. We do not have the room temperature data for this second collection, though.
 
So where are we?  Well, we are going to ask you to  model the amount of mass of dry ice as a function of time by producing a differential equation which models the {\it rate of change in mass\/}. That is, if $m(t)$ is the mass in grams (g) at time $t$ in seconds (s) we seek to find, solve, and use the differential equation: 
\vspace{-.1 true cm}
\begin{equation} \label{eq:sublimate}
m'(t) = \rule{3 true cm}{.01 true cm} \quad  {\rm where } \quad   m(0) = m_0\  \quad   {\rm is \,\,\, the \,\,\, initial \,\,\, condition }, 
\end{equation}
\noindent to determine $m(t)$ and then use one of the two data sets (Table~\ref{table:MakkarWernerData} or Table~\ref{table:MottelData}) to estimate any parameters introduced into your model.            
 
You should be asking yourself what the rate of sublimation depends upon.

\vspace*{.4 true cm}

Produce a model differential equation with initial condition using reasonable and defended assumptions; solve it; estimate the parameters in your model for one of the two data sets (Table~\ref{table:MakkarWernerData} or Table~\ref{table:MottelData}); and validate that your model is reasonable. Offer up a discussion on your modeling process. Try your approach on the second data set to confirm your modeling skills.


\teacher{

\section*{NOTES FOR TEACHERS}

We have used this experiment and this data in a number of settings over the years. It is usually a 2-3 day assignment with time for class development (say 15-30 minutes) and part of a follow-on class  for clarifying questions before a final write-up is due. Below we offer up the complete text, with some improvements as suggested by a careful referee,  of the article \cite{Winkel2012} from which this modeling scenario was extracted. 

\section*{APPENDIX -- 
Modeling Sublimation of Carbon Dioxide}



\section*{Introduction\label{intro}}

In high school science we learn about change of state for various elements, e.g., melt, evaporate, and sublimate.  Some years ago we wondered if we could model the sublimation of a piece of solid carbon dioxide. At the time (over 30 years  ago) frozen $CO_2$ or as it is called ``dry ice'' was used to pack ice cream shipments.  Left alone the dry ice would disappear without a trace. This disappearance is  called {\it sublimation\/}, the transition from a solid to a gas without any apparent transition through the liquid state. 

We taught a course \cite{WinkelFroyd1992} in which all first-year college science, engineering, and mathematics courses were integrated into one 12 credit course.  As such, we were free to mix it up and do science with mathematics, engineering with science, and mathematics with engineering. One day, because we had received some things in a shipment containing dry ice we thought of collecting data on the sublimation of dry ice. While we had colleagues who were true engineers and scientists on our teaching team,  we decided, as a mathematician who was  pretty much an inept lab fellow, to go this one alone. 

We set up a modest apparatus consisting of a tray in which we placed our dry ice and then placed the tray on a scale which could render mass in grams to the thousandth of a gram. We zeroed out the scale (i.e. we noted the mass of the tray  and then noted the mass with the initial dry ice amount on it so that we could tell from total mass data exactly what our dry ice mass was as time progressed) and began to collect data. We  really lived dangerously, for we did this all for the first time in class with  students in the first row recording the mass every 30 sec.  A very unexpected thing  happened. The mass of the dry ice, rather we should say the mass of the zeroed  ice and tray, went up(!) in the first few minutes. We continued to collect data on that run, but  we all went home to ponder what was going on. After class I tried the experiment again and discovered the problem. The dry ice was so cold, below sublimation temperature of dry ice, -78.5$^{o}$ C, that it was causing water to condense on the tray. Indeed, upon inspection I could see the ice film on the tray. That was why the mass went up on this configuration for a while.



Most of the students were able to explain exactly what I saw, only without seeing it. We all realized that we needed to somehow do the experiment by, perhaps, suspending the dry ice so that when the sublimated (but still very cold) $CO_2$, which was heavier than air, fell it did not fall on any part of our scale device. The next day I arrived with a proper set of equipment (Figure~\ref{fig:1-12-SubApparatus}) which did not produce the collected moisture problem found in the previous  equipment and we began to collect data again. 


We collected the data in a room of temperature 19.3$^{o}$ C. using a small, box-shaped piece of dry ice we had carved (wearing protective gloves) to the size of two sugar cubes, 1 cm $\times$  1 cm $\times$ 2 cm with an initial mass of approximately 7.5 g. This time we did not experience any unusual phenomena. We present the collected data in Table~\ref{table:EdData} with a plot rendered in  Figure~\ref{fig:1-12-SubEdDataPlot}.

\begin{table}[t]
\label{table:EdData}
\begin{center}
\small
\begin{tabular}{||r|r|c|r|r||} \hline
Time (s) & Mass (g) &&Time (sec) & Mass (g)\\ \hline 
0&	7.570&&	480&	5.871\\ \hline  
30&	7.457&&	510&	5.776\\ \hline 
60&	7.338&&	540&	5.682\\ \hline 
90&	7.220&&	570&	5.589\\ \hline 
120&	7.110&&	600&	5.497\\ \hline 
150&	6.995&&	630&	5.405\\ \hline 
180&	6.885&&	660&	5.313\\ \hline 
210&	6.778&&	690&	5.224\\ \hline 
240& 	6.673&&	720&	5.136\\ \hline 
270&	6.571&&	750&	5.048\\ \hline 
300&	6.464&&	780&	4.960\\ \hline 
330&	6.363&&	810&	4.878\\ \hline 
360&	6.265&&	840&	4.790\\ \hline 
390&	6.163&&	870&	4.707\\ \hline 
420&	6.067&&	900&	4.625\\ \hline 
450&	5.969&&	&\\ \hline 
\end{tabular}

\parbox{8 true cm}{\caption{\label{table:EdData}  Data collected by Dr.~Ed Mottel   on successful run for mass of dry ice (g) as a function of time (s).}}

\end{center}
\vspace*{-.3 true cm}
\end{table}



We were studying chemical reaction kinetics  models \cite{Winkel2011} in our integrated course. Thus, it seemed that perhaps these models might work for sublimation.  In reaction kinetics the rate of reaction is determined by the concentrations. Thus the amount of dry ice present in our block at time $t$ seconds can be called $m(t)$ (in grams) and the rate of change of mass might  be modeled by the differential equation, (\ref{eq:order}), where the exponent, $r$, is referred to as the {\it order of the reaction\/}. This is an example  of {\it The Law of Mass Action\/}.  

\begin{equation}\label{eq:order}
m'(t) = - k\, m(t)^r , \quad m(0) = 7.57\, ,
\end{equation}
 

The three reaction orders of interest in introductory chemistry texts are $r =  0$, $r = 1$, and  $r=2$, so-called zeroth, first, and second orders, respectively.  Thus we sought  to determine whether the reaction was one of these three types. We had done an experiment with the evaporation of acetone in a beaker and found it to be 0$^{\rm th}$ ($r=0$) order, as evaporation was only proportional to the constant cross sectional surface area in the beaker. 

We can determine the optimal values of best fitting  $k$ for $r = 0, 1, 2$ by solving the respective differential equation (\ref{eq:order}),  forming a sum of square errors between observed data and model data, and  determining the value of the parameter $k$ which minimizes the  sum of square errors. For example, if we were attempting to estimate $k$ in the case $r=1$ then our model from solving (\ref{eq:order}) would be
$$m_1(t) = 7.57 e^{-k\, t} \, .$$

\noindent When we form our sum of square errors between the theoretical model, $m_1(t)$, and our actual data values we get $SSE_1(k)$ in (\ref{eq:ss}).

\begin{equation}\label{eq:ss}
SSE_1(k) = \sum_{i=1}^{i=n} (m_1(t_i) - m_i)^2 = \sum_{i=1}^{i=n} (7.57 e^{-k\, t_i} - m_i)^2
\end{equation}

\noindent Here $n$ is the number of data points we have and $m_i$ is the value of the mass at time $t_i$, $i=1, \ldots n$. In this case we find our best value of $k=0.000536791$. 






We do a similar least sum of square errors analysis for $r=0$ to obtain $m_0(t) = .57 - 0.00340764\, t$ and for $r=2$ to obtain $m_2(t) = \frac{1}{0.1321+k\, t}$ to determine the optimal values of the respective $k$'s for these two reaction models. In Figure~\ref{fig:1-12-SubrPlots} we show all three plots over the data and note the respective $r$ values.

In light of the closeness for the model when $r=1$,  we examine just the $r=1$ best fitting model when compared to the data in Figure~\ref{fig:1-12-Sub1Plot}. This looks terrific. We are prepared to suggest that  sublimation of $CO_2$ is a first order reaction.  We make note of the fact that the minimum sum of square errors  is 0.0119927 when $r = 1$, and the resulting best $k=  0.000536791$.

\begin{figure}[t]
\centering
\includegraphics[width=8cm]{1-12-SubEdDataPlot.png}



\parbox{9.5 true cm}{\caption{\label{fig:1-12-SubEdDataPlot} Plot of data from Table~\ref{table:EdData}.}}

\end{figure}



\begin{figure}[h]
\centering
\includegraphics[width=8cm]{1-12-SubrPlots.png}



\parbox{13 true cm}{\caption{\label{fig:1-12-SubrPlots} Plots of best fitting 0$^{\rm th}$ ($r=0$), 1$^{\rm st}$ ($r=1$), and 2$^{\rm nd}$ ($r=2$)  order fits to data in  Table~\ref{table:EdData}. Here  when $r = 0$, $k = 0.00340764$, when $r = 1$, $k = 0.000536791$, and when $r = 2$,  $k = 0.0000841212$.  }}

\end{figure}


\begin{figure}[!h]
\centering
\includegraphics[width=8cm]{1-12-Sub1Plot.png}



\parbox{13.5 true cm}{\caption{\label{fig:1-12-Sub1Plot} Plot of best fitting  1$^{\rm st}$ ($r=1$), order fit to data in  Table~\ref{table:EdData}. Here  when  $r = 1$, $k = 0.000536791$. This is a very convincing plot of a model which fits the data.  }}
\vspace*{-.3 true cm}
\end{figure}





Since that time, over 30 years  ago, when we first did this experiment, using the same data, we found something very interesting. Students have suggested that the order of the reaction in (\ref{eq:order}) does not necessarily have to be $r = 0, 1, 2$. Indeed, maybe we need to consider what is really happening, i.e. examine the physics of the situations. Most descriptions of sublimation of dry ice say something like, ``When exposed to air, dry ice changes directly to vapor.''\cite{World2011} It is for this reason that students argue that sublimation should probably be greatest (and maybe exclusively) at the surface of the cube of $CO_2$. This leads them to conjecture that the rate of sublimation is proportional to surface area, i.e. to  $$ {\rm cm}^2 = \left({\rm cm}^3\right)^{\frac{2}{3}} \, .$$
\noindent Now mass is directly proportional to volume when the mass is uniform which,  no doubt, it is in our cube of $CO_2$, so just maybe, the students suggested,   (\ref{eq:order}) should look more like (\ref{eq:twothird}),  


\begin{equation}\label{eq:twothird}
m'(t) = - k\, m(t)^{\frac{2}{3}} , \quad 
m(0) = 7.57 \, .
\end{equation}
 
If we solve (\ref{eq:twothird}) and do a least sum of square errors analysis we find that we have not really improved on our $r = 1$, first order model as measured by our sum of square errors measure (see Table~\ref{table:summary}), even though this two-thirds order model renders a nice model plot (Figure~\ref{fig:1-12-SubrPlot}) when compared to our data. We summarize our results in Table~\ref{table:summary}.

\begin{table}[h]\label{table:summary}
\begin{center}\small
\begin{tabular}{|c|l|l|} \hline
Order $(r)$ & Parameter $k$ & SSE \\ \hline
0 & 0.00340764& 0.12126\\ \hline
1 & 0.000536791& 0.0119927\\ \hline
2 & 0.0000841212& 0.251683\\ \hline
$\frac{2}{3}$ & 0.00184154& 0.0361216\\ \hline
\end{tabular}

\parbox{11 true cm}{\caption{\label{table:summary} Summary of  data analysis for various orders $r$ with best estimate of parameter $k$ and respective minimum sum of square errors.}}

\end{center}
\vspace*{-.3 true cm}
\end{table}



\begin{figure}[h]
\centering
\includegraphics[width=8cm]{1-12-SubrPlot.png}



\parbox{13.5 true cm}{\caption{\label{fig:1-12-SubrPlot} Plot of best fitting  ``$\left(\frac{2}{3}\right)^{\rm rds}$'' ($r=\frac{2}{3}$), order fit to data in  Table~\ref{table:EdData}. Here  when  $r = \frac{2}{3}, \quad k = 0.00184154$. This is not as   convincing a plot (Figure~\ref{fig:1-12-Sub1Plot}) of a model  fit of the data as we have when $r = 1, \quad k = 0.000536791$.  }}
\end{figure}






\section*{General $r^{\rm th}$ order model and parameters}


In an effort to offer a most complete order analysis we turn to the  general  model in (\ref{eq:massrate}) and see whether  we can determine simultaneously both $r$ and $k$ so that the solution of (\ref{eq:massrate}) has the least sum of square errors as a function of the two variables $r$ and $k$.

\begin{equation}\label{eq:massrate}
m'(t) = - k\, m(t)^r, \quad m(0) = 7.57 \, .
\end{equation}

\noindent The solution for (\ref{eq:massrate}) is

\begin{equation}\label{eq:solm}
m(t) = \left((r-1) \left(k\, t-\frac{7.57}{7.57^r (1- 
   r)}\right)\right)^{\frac{1}{1-r}}\, .
\end{equation}

If we use (\ref{eq:solm}) and form a sum of square errors with our data as a function of the two variables, $r$ and $k$, we can find the true optimal order, $r$,  as well as a value for $k$.  Here we get a minimum sum of squares errors in parameters $r$ and $k$ of $0.000529786$ for our best fitting values of $r = 0.751209$ and $k = 0.0008506$. We note in Table~\ref{table:finalsummary} that this sum of square errors is much less than the values when we found the best fit for the  $r=0, 1, 2$ or $r = \frac{2}{3}$ values. Thus we have an optimal order model when $r = 0.751209$. 



\begin{table}[h]\label{table:finalsummary}
\begin{center}\small
\begin{tabular}{|c|l|l|} \hline
Order $(r)$ & Parameter $k$ & SSE \\ \hline
0 & 0.00340764& 0.12126\\ \hline
1 & 0.000536791& 0.0119927\\ \hline
2 & 0.0000841212& 0.251683\\ \hline
$\frac{2}{3}$ & 0.00184154& 0.0361216\\ \hline
0.751209 & 0.0008506 & 0.000529786\\ \hline
\end{tabular}

\parbox{13.0 true cm}{\caption{\label{table:finalsummary} Summary of  data analyses for various orders $r$ with best estimate of parameter $k$ and  respective minimum sum of square errors. We see that none of $r = 0, 1, 2$ or $r = \frac{2}{3}$ are best. Through our optimization analysis of values of $r$ rather than presuming set values of $r$ we obtain the best order model  when $r = 0.751209$. We show the plot of the model over the data in Figure~\ref{fig:1-12-SubOptPlot}.}}

\end{center}
\vspace*{-.3 true cm}
\end{table}

Moreover, this final model makes  sense, since it gives a better representation of the physical situation. For, as students conjectured, the sublimation or conversion from solid to gas occurs at the surface, largely. But perhaps some sublimation occurs internal to the solid -- intramolecular, so that instead of the rate of sublimation being proportional to $r = \frac{2}{3}$, it is a bit more than that, say $r = 0.751209$, but nowhere near $r = 1$ or $r = 2$ and not down  near $r = 0$. See plot in Figure~\ref{fig:1-12-SubOptPlot}.


\begin{figure}[h]
\centering
\includegraphics[width=8cm]{1-12-SubOptPlot.png}



\parbox{13.8 true cm}{\caption{\label{fig:1-12-SubOptPlot} Plot of optimal  order fit to data over the data in  Table~\ref{table:EdData}. Here    $r = 0.751209$ and $k = 0.0008506$. This is a very convincing plot of our solution   for (\ref{eq:solm}) with these parameters.  }}
\vspace*{-.4 true cm}
\end{figure}

\section*{Analysis of the Results}

The data used in this paper was not from our first successful collection, but  rather from a collection experiment conducted by Dr.~Ed Mottel, our chemistry colleague in our integrated curriculum. (See Table~\ref{table:EdData}.) It was collected in the same manner as the students did, but you have to believe it was done with more professional skill. Besides, Dr.~Mottel did this in his laboratory without the ``noise'' of class going on while collecting the data. However, the data in Table~\ref{table:MakkarWernerData} is the real thing, collected by students in our classroom after redesigning our device so as not to permit water vapor condensing on the tray holding the dry ice. 



\begin{table}\label{table:StudentData1}
\begin{center}
\begin{tabular}{||r|r|c|r|r||} \hline
Time (s) & Mass (g) &&Time (sec) & Mass (g)\\ \hline 
0&	25.525&&        1800&	13.553\\ \hline  
120&	24.512&&	1930&	12.910\\ \hline 
240&	23.524&&	2040&	12.331\\ \hline
360&	22.639&&	2170&	11.689\\ \hline 
480&	21.765&&	2280&	11.188\\ \hline 
600&	20.890&&	2410&	10.566\\ \hline  
720&	20.043&&	2530&	10.043\\ \hline
840&	19.221&&	2690&	9.377\\ \hline 
960& 	18.431&&	2780&	9.011\\ \hline
1080&	17.677&&	2880&	8.616\\ \hline  
1200&	16.936&&	3060&	7.945\\ \hline
1320&	16.220&&	3220&	7.404\\ \hline	 
1440&	15.548&&	3380&	6.877\\ \hline 
1570&	14.828&&	3480&	6.593\\ \hline 
1680&	14.213&&        3600&	6.244\\ \hline 

\end{tabular}

\parbox{12.4 true cm}{\caption{\label{table:StudentData1} Data collected by students Masood Makkar and Paul Werner  (3 December 1992) on successful run for mass of dry ice (g) as a function of time (s).}}

\end{center}
\vspace*{-.3 true cm}
\end{table}


We leave it to the reader (and the reader's students) to determine the best fitting model in the manner we have described above for this data. Hint:  When we applied our techniques to this data set we got an order of $r = 0.708891$ which was reasonable when compare to  our $r = 0.751209$ with the first data set.

One might ask,  ``Why the difference?'' There may be a number of reasons, each of which could merit further investigation. We list some here.

\begin{enumerate}
\item Density of dry ice pack may differ depending on how it is formed.
\item What shape is the piece of dry ice? Should we pay attention to surface area to volume ratio which can vary from shape to shape?  E.g., a cube of unit volume has a ratio of 6 while a sphere of unit volume has a ratio of 4.84 and a tetrahedron  of unit volume has a ratio of 7.21)\cite{Wiki2011}. Presumably, the higher the ratio is  the more sublimation per unit time there is due to greater surface area for fixed volume.
\item Internal sublimation within the solid may be  occurring.
\item What is the temperature of the environment in which the sublimation takes place? NB: We did not record the temperature for the student data, so while we presume it was close to what Dr.~Mottel's sample experienced, it could be different.
\item Impurities can conduct heat and thus cause small, local changes in the rate of sublimation.
\end{enumerate}








\section*{Conclusion}

We have offered the reader two sets of data and several approaches to determine the rate of sublimation of a piece of solid dry ice. We presented several models and their analyses for one data set and discovered that a  good model for the rate at which the change of mass in dry ice occurs is proportional to the mass to some power, $r$, (called the order)  which is approximately $r = 0.75$. Additionally, we have some first principle physics notions with respect to surface area assumptions that make our model reasonable.  

Students enjoy collecting data and then using it in mathematics class. It is not the norm in such classes. Moreover, in a modeling context data is the confirming or validating opportunity for our model that we can provide our students.  Certainly collecting additional data on varying shapes and initial conditions might permit a finer resolution of the order of the reaction, and hence the rate of sublimation.
 
\section*{Acknowledgments}

We thank our colleague, Dr.~Ed Mottel of the Department of Chemistry, Rose-Hulman Institute of Technology, for collecting this data and for many helpful conversations on things chemical and, more broadly, scientific. We acknowledge the willingness of students, who every time they are asked to step up and do something extra in class, in this case collect data and listen attentively, say ``Yes." In this case we appreciate the efforts of former students Masood Makkar and Paul Werner.



There is an exceptional web site with lots of information about dry ice at \cite{DryIceInfo2011} as well as pointers to ideas for experiments involving dry ice.



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\bibitem{Wiki2011} Wikipedia.  2011. Surface-area-to-volume ratio. \href{http://en.wikipedia.org/wiki/Surface-area-to-volume\_ratio}{http://en.wikipedia.org/wiki/Surface-area-to-volume\_ratio}.  Accessed 23 July 2011.

\bibitem{WinkelFroyd1992} Winkel, B. J. and J. E. Froyd. 1992. An Integrated, First-Year Curriculum in Science, Engineering, and Mathematics: Progress, Pitfalls, and Promise. {\it Proceedings of the Frontiers in Education 1992 Conference. 11-15 November 1992\/}. Vanderbilt University, Nashville TN. 557-562. 

\bibitem{Winkel2011} Winkel, B. J. 2011. Parameter Estimates in Differential Equation Models for Chemical Kinetics. {\it International Journal of Mathematical Education in Science and Technology\/}. 42(1): 37-51.

\bibitem{Winkel2012} Winkel, B. J. 2012. Modeling Sublimation of Carbon Dioxide.  {\it International Journal of Mathematical Education in Science and Technology\/}. 43(8): 1077-1085. 

\bibitem{World2011}  World of Forensic Science. 2011. Sublimation. \href{http://www.enotes.com/forensic-science/sublimation}{http://www.enotes.com/forensic-science/sublimation}. Accessed 7 October 2011.
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}

\end{document}
