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\title{\ifnum\TeacherEdition=1 TEACHER VERSION 
\else STUDENT VERSION 
\fi\\How Good is Your Water Bottle?}


\markboth{How Good is Your Water Bottle?}{How Good is Your Water Bottle?}

\author{Eli Goldwyn\\

University of Portland\\
Portland  OR 97203 USA \\
goldwyn@up.edu
}

\makeStitlePDFLaTex


\begin{abstract}
This modeling scenario involves Newton's Law of Cooling and should be appropriate in the first week or even the first day of an introductory differential equations course. The scenario uses an Inquiry Based Learning approach to walk the students through the creation and understanding of a differential equation describing how fluid in a water bottle will change its temperature to approach the ambient temperature in a room. The goal is for the students to understand how the right hand side of the differential equation and the initial condition relate to the physical phenomenon being described. This scenario also hints at the notion of an equilibrium and the speed at which an equilibrium is approached.
\end{abstract}


\teacher{
\section*{INTRODUCTION TO THE TEACHER}
Heat transfer problems appear in a variety of engineering and physics courses and have many important applications in those fields. Broadly speaking, most heat transfer occurs through conduction, convection, and radiation. Newton's Law of Cooling assumes only convection and is popular in differential equations courses since it is formulated as a linear differential equation. Assuming that convection is the only heat transfer method is reasonable if the temperature difference between the fluid and the room is relatively small and if any surface touching the bottle is of similar temperature to the air. \\ \\
While this scenario is designed to be done in one full day of class with students in small groups, it should be flexible enough for instructors to use in a variety of different ways including as homework. To assess the learning outcomes, we recommend that students turn in their completed work, either individually or as a group. The scenario is all building to where the students can write down Newton's Law of Cooling in problem \#6. This is likely to be the most difficult problem and many student groups may require some hints from the instructor. Problems \#4 and \#5 can be done in either order. Problems \#7 and \#8 are additional questions that can be used of as optional or bonus questions. Bringing in two water bottles of different insulations is not required, but is helpful as a visual aid for the students. \textcolor{red}{Answers and commentary to the individual questions appear in red and are only available on the Teacher Version.}
}

\section*{SCENARIO DESCRIPTION}
We will explore how the temperature of cold water (or hot tea) in a water bottle changes when left in a room at constant temperature. Our goal is to build a differential equation describing this rate of change.

\begin{enumerate}
	\item The temperature of our classroom is set to always be exactly $70^\circ F$. The water inside my high-quality water bottle is currently $40^\circ F$. Let us pretend that I accidentally leave my water bottle in this room for a long time.
	\begin{enumerate}
            	\item Sketch a plot of water temperature by time (first quadrant only). Be sure to label your graph. 
                 \item Imagine that instead of my high-quality insulated water bottle, I have a cheap thin plastic water bottle. Sketch the plot for the water temperature in this bottle.  
                 \item Compare the two sketches.
	\end{enumerate}                   
	\teacher{\textcolor{red}{These two graphs should start at $40^\circ F$ and increase asymptotically, approaching the room temperature, $70^\circ F$. The cheap plastic bottle should approach the room temperature faster than the higher quality bottle}}
     	\item What if instead of the $40^\circ F$ cold water, I had filled my high quality bottle with hot tea at a temperature of $130^\circ F$. Sketch the plot for the temperature of the hot tea.\\
	\teacher{	\textcolor{red}{This graph should be similar to the first graph except now it starts at $130^\circ F$ and decreases to the room temperature.}
	}
	
	\item Let $x(t)$ be the temperature of the liquid in the water bottle at time $t$. 
	\begin{enumerate}
		\item Find $x(0)$ in each of the three previous examples (this is called the initial condition). \\	
		\teacher{	\textcolor{red}{$40^\circ F$, $40^\circ F$, $130^\circ F$}	}
		\item Find $\lim\limits_{t\rightarrow \infty}x(t)$ in each of the three previous examples, i.e. what temperature do we expect the water to approach if I leave it the bottle out forever?	\\
		\teacher{	\textcolor{red}{$70^\circ F$. Many students will likely need some help on this one}	}
	\end{enumerate}
	
	\item Give a reason why the following derivatives do NOT describe the rate of change of the water temperature in an arbitrary water bottle, $\frac{dx}{dt}$. Let $T_a$ be the temperature of the room and $k>0$ be a constant that depends on the quality of the water bottle.
	\begin{enumerate}
		\item $\frac{dx}{dt}=-kx$			\hspace{2.8em}\teacher{\textcolor{red}{The temperature decays to 0, not the room temperature.}}
		\item $\frac{dx}{dt}=x-t$			\hspace{2.8em}\teacher{\textcolor{red}{The rate of change of the temperature gets very negative as $t$ gets large. }}
		\item $\frac{dx}{dt}=\sin{t}$		\hspace{3em}\teacher{\textcolor{red}{The temperature oscillates.}}
		\item $\frac{dx}{dt}=\frac{x-T_a}{t}$	\hspace{2em}\teacher{\textcolor{red}{The temperature changes with the wrong sign, if water is colder than the room its temperature will decrease. There is nothing that distinguishes the two water bottles. When $t$ is small the rate of change is very large. }}
	\end{enumerate}
		
	
	\item Let's try to determine what quantities effect the rate of change of the water temperature, i.e. the derivative $\frac{dx}{dt}$.
		\begin{enumerate}
		\item Does the rate at which the water changes temperature depend on the current temperature of the water? Would you expect $50^\circ F$ water in this room to warm up faster, slower, or at the same rate as $68^\circ F$ water?		\\
		\teacher{\textcolor{red}{Yes, the rate at which water changes temperature does depend on its current temperature.}}
		
		\item Does the rate at which the water changes temperature depend on how much time has elapsed? Consider the scenario where the water in a bottle started at $40^\circ F$ and a few hours later has warmed to $50^\circ F$. At that moment we bring in an IDENTICAL water bottle with water at $50^\circ F$. Would we expect the water in these two bottles to change temperature at the same or different rates?			\hspace{4em}\teacher{\textcolor{red}{No.}}
		
		\item What has to be true for the water in the bottle not to be changing temperature, i.e. when $\frac{dx}{dt}=0$?	
		\hspace{4em}\teacher{\textcolor{red}{The water must be the same temperature as the air.}}
		
		\item Using (a) and (b), should we see an $x$ on the right side of the derivative? Should we see a $t$? What other quantities are important? How are the derivatives for my bottle and the cheap plastic bottle similar, yet different? Are there differences between the example with water and the example with hot tea?	\\
		\hspace{4em}\teacher{\textcolor{red}{Yes there should be an $x$. no there should not be a $t$. The room temperature and the quality of the water bottle are both important. The differential equations for the water and hot tea are identical, but the initial conditions will be different.}}
	\end{enumerate}
			
	\item Put everything together, assume the rate of change in temperature is proportional to the difference between the fluid temperature and the room temperature and find $\frac{dx}{dt}$ ($\frac{dx}{dt}$ is a `differential equation' for the temperature of the water or tea and when combined with an initial condition is called an `Initial Value Problem'). \\
		\hspace{4em}\teacher{\textcolor{red}{$\frac{dx}{dt}=-k(x-T_a)$ or $\frac{dx}{dt}=k(T_a-x)$}}

	
	\item The differential equation you found above is known as `Newton's Law of Cooling'. Newton's Law of Cooling assumes that convection is the only method of heat transfer. This problem examines the heat gain from thermal radiation. While the heat gained by convection is linearly proportional to the difference between the water and room temperatures, heat gained by radiation is proportional to the difference between the fourth power of the water temperature and the fourth power of the room temperature.
	\begin{enumerate}
		\item Write down the differential equation describing the change in water temperature in my water bottle coming from both convection and radiation.	\\
		\hspace{4em}\teacher{\textcolor{red}{$\frac{dx}{dt}=k(T_a-x)+k_r(T_a^4-x^4)$}}
		\item Using Newton's Law of Cooling we expect that $40^\circ F$ in a $70^\circ F$ room would heat up at the same rate as $50^\circ F$ water in an $80^\circ F$ room. Would we expect the same to hold true if we include the heat gain from radiation?	\\
		\hspace{4em}\teacher{\textcolor{red}{No! The $50^\circ F$ water will heat up faster. }}
	\end{enumerate}
	
	\item Let's add the effect of a phase change (ice melting) to Newton's Law of Cooling. Let the water inside my water bottle be  $60^\circ F$ and the room be  $70^\circ F$. Now we put several ice cubes into my water bottle. Sketch a graph of the plot of water temperature by time.	\\
		\teacher{\textcolor{red}{The temperature will initially decrease, hit a minimum (above $32^\circ F$), then increases asymptotically to $70^\circ F$.}}
	
	
\end{enumerate}


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