Presentation

Mathematical communication

Personal engagement

Reflection

Use of mathematics

As an SL IB math student, I have to spend a lot of time learning different mathematical topics, one of them being calculus. This topic, in particular, leads to me spending many late nights doing homework, studying for tests, and making notes of what we learned in class. The one thing that fuels these late-night study sessions is coffee. I drink both instant coffee, which is cheaper, and freshly brewed coffee, which is more expensive. As coffee is not a necessity for me, I have to pay for it myself, so for my limited budget as a high school student, I want to get the most out of my coffee. I don’t care for the taste, so the most significant factor for me and my coffee purchases is if I’m getting my money’s worth. I wanted to compare the two and find out which cools quicker because if they both cool at the same rate, why should I pay more for the brewed coffee? According to Newton’s Law of Cooling, if the air surrounding an object is more relaxed than the object itself, the object’s temperature will decrease exponentially, leveling off as it approaches the ambient air temperature (History and Applications, n.d).​ ​Logically, it is impossible for the temperature to be below room temperature. Instead, both cups of coffee would approach room temperature, which would be a horizontal asymptote of the graphed function. According to these parameters, the graph of these functions would be an exponential or logarithmic function, which is graphable. Using the equations, I can find out if black coffee or brewed coffee will stay hot longer and exactly how long I have before my coffee becomes cold. I also hope this exploration will allow me to find a practical use for calculus and help me apply these concepts beyond math class. This will also allow me to better understand the reasoning behind these concepts, as well as how I can better apply math to my life. To gather my data, I used a probe thermometer to record the temperature of each cup of coffee every five minutes for 130 minutes. This method did not allow me to produce the most accurate readings, but it was the only precise method accessible to me. I, however, believe that the amount of data I was able to collect is sufficient enough to produce an accurate model of the cooling coffee, with which I can create an equation and use it to predict when each cup of coffee will reach room temperature. Before beginning my experiment, I took the temperature of my bedroom, where I study, and where the investigation would take place. The temperature of my room was 79.1 °F at the time of my experiment. This meant that the coffee would cease cooling once it reached this temperature. Both cups of coffee had different initial temperatures due to their other preparation methods. The brewed coffee was made in a coffee maker, and the instant coffee was made by mixing water that I boiled on the stove with instant coffee powder. Since I only had one thermometer, I collected the data for each cup of coffee separately and then put all my data on an Excel spreadsheet. This most likely affected the data collected, as the ambient temperature changed throughout the day. I put a selection of my data points in the table below. I decided to round all my data and calculations to 4 significant figures. Table 1: Temperature of black instant coffee and black brewed coffee over time

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After collecting my data, I made a scatter plot of both sets using Microsoft Excel, with room temperature as the horizontal asymptote. Both sets of data begin to plateau at room temperature. Figure 1:Graph of the temperature of instant coffee and brewed coffee over time

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According to this graph, it is evident that both functions are exponential functions. Both have a negative correlation. Since the initial temperature or the y-value of the instant x = 0 and brewed coffee are 174.3 °F and 162.2 °F, respectively, these values are also the y-intercepts. I used these values to build my equations. There is a horizontal asymptote at room temperature, 79.10°F, which can be my vertical translation, c. Let “I” equal the temperature of the instant coffee, “B” equal the temperature of the brewed coffee, and “t” is the time in minutes. I decided to make a natural logarithm function, using e. Using these variables, I created my general equations. \(B(f)\ =\ ae^{kt}+\ c\)​​​​​​​ \(I(t)\ =\ ae^{kt}\ +c\)​​​​​​​ Since the horizontal asymptote is 79.10, I added that value as my vertical translation. \(B(t)\ =\ ae^{kt}+79.10\) \(I(t)\ =\ ae^{kt}+79.10\)​​​​​​​ To solve for a, I substituted a value, t=0 and I(t)=162.2, from the original points into my equation. \(162.2\ =\ ae^{k(0)}+79.10\) \(162.2\ =\ ae^{(0)}+79.10\)​​​​​​​ I know that an exponent to the power of 0 is 1, from which I can isolate a. a ≈ 83.10 I used this new ‘a value and a different point from the original values, t =130.0, I(t) = 80.50, to isolate k. \(80.50=83.10e^{k(130.0)}+79.10\)​​​​​​​ \(1.400=83.10e^{130.0k}\) ​​​​​​​ \(0.0168471721=e^{130.0k}\)​​​​​​​ To get rid of e, I used the property ln e = 1 \(In\ 0.0168471721=In\ e^{130.0k}\) \(-4.083572465=130.0k\) \(k\approx-0.03141\)​​​​​​​ I repeated this method with my other equation to give me my final equations: \(B(t)=95.2e^{-0.02643t}+79.10\) \(I(t)=83.10e^{-0.0314lt}+79.10\)​​​​​​​​​​​​​​ I graphed my equations with my original points to compare them and see how accurate my equations were. Figure 2: Graph of final equations and original data points

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The green dots are my original data points for instant coffee, and the black dots are the actual data points for brewed coffee. My equations are very similar to the actual data points, but there are some anomalies in the original data. These points may have occurred due to changes in the ambient temperature or accidental movement of the coffee mug, so I was not too worried as to why my equations didn’t pass through them. I also decided to use Excel to create equations based on my original data points, which I could compare to my equations to see how accurate I was. Using technology, I was able to produce these equations: \(B(t)=-19.32In(t)+175.9\) \(I(t)=-22.98In(t)+193.3\)​​​​​​​ Figure 3: Excel equations

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While researching Newton’s law of cooling, I found a general formula that I could use to find my equations. ​According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings, expressed by this relationship: \(\frac{dT}{dt}\) is directly proportional to (T – Ts)(Newton's Law of Cooling, 2019). Using integration, the website was able to produce a general equation for Newton’s law of cooling, which I can use to find an equation: \(T(t)=T_s+(T_o\ -\ T_s)\ e^{-kt}\)​​​​​​​ Where t is time, T(t) is the temperature based on time, is the ambient temperature, \(T_s\) is the initial temperature, \(T_0\)and k is a constant. Since I had to find 2 equations, I decided to make T(t) I(t), and B(t), to represent instant coffee and brewed coffee, respectively. Using these conditions, I was able to create my equation: \(B(t)=\ 79.10+(174.3-79.10)\ e^{-kt} \)​​​​​​ To solve for k, I substituted in t= ​130 and B(t) = 81.20 \(81.20=79.10+(1743-79.10)\ e^{-k(130)} \) \(2.1=95.2\ e^{-k(130)}\) \(0.0220588235=e^{-k(130)} \)​​​​​​​ I got rid of e; I used ln e = 1 \(In\ 0.220588235\ =-k(130) \) \(-0.0293387892=-k \) \(k\approx0.02934\)​​​​​​​ I repeated this process with my second equation to give me these final equations: \(B(t)=79.10+(174.3-79.10)\ e^{-0.02934t}\) \(I(t)=79.10+(162.2-79.10)\ e^{-0.04567t} \)​​​​​​​ I graphed my final equations and the equations created by Excel and my original data points to see which equations were more accurate. To make things easier, I decided to call \(B(t)=95.2e^{-0.02643t}+79.10 \) final equation 1, \(I(t)=83.10e^{-0.0314It}+79.10 \)final equation 2. \(B(t)=-19.32ln(t)\ +\ 175.9\ excel\ equation\ 1,\ I(t)=-22.98In(t)\ +\ 193.3\ excel\ equation\ 2.\) \(B(t)=79.10\ +\ (174.3-79.10)e^{-0.02934t}Newton\ equation\ 1.\ and\) \(I(t)=79.10\ +\ (162.2-79.10)e^{-0.03347t}Newton\ equation\ 2.\)​​​​​​​ Figure 3: Graph of all equations and original points All equations seem relatively accurate, so I could not decide which was better based on observation alone. I decided to instead compare them by creating a table of values comparing the original data points with a selection of deals from my equations, the technologically developed equations, and the Newton equations. I also decided to find the standard deviations between my original data points and my final equations and between my actual data points and the technologically derived equations. I decided to compare the first 21 points as I felt that would give me an excellent range to compare each equation. Table 2: Original temperature of brewed coffee compared to final equation 1, Excel equation 1, and Newton equation 1

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Table 3: Original temperature of instant coffee compared to final equation 2 and Excel equation 2

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I decided to find the standard deviation between my original points and the final equations, my original points, and the Excel equations, and my original points and the Newton equations using Excel: Figure 7: Calculation of standard deviation using Excel

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Table 4: Mean and standard deviation between original brewed coffee points and equations.

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Table 5: Mean and standard deviation between original instant coffee points and equations

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According to these standard deviations, the most accurate equations are the ones created by Excel. This makes sense because I was working with rounded values, whereas Excel had the exact ones. Additionally, Excel was able to take into account all the data I collected, as well as anomalies. In contrast, I had no way of considering every value as I did my calculations. Final equation 1 was the least accurate compared to the other brewed coffee equations, as well overall. Still, Newton equation 2 was the least accurate compared to the other instant coffee equations. Based on the standard deviations from the brewed coffee equations, I expected final equation 2 to be the least accurate, but it was pretty accurate. The formula created by Newton’s Law of Cooling is restricted by the fact that it doesn’t consider external factors, such as the ambient temperature changing, and instead assumes that everything remains constant. Additionally, using one of the anomaly values would produce an inaccurate model, whether I solved the equation algebraically or with Newton’s Law of Cooling formula. One of the primary reasons I believe that final equation 1 was so inaccurate may have been my randomly choosing values to substitute into my equation. To answer my question of which type of coffee cools faster, I decided to use the Excel equations, as they are the most accurate and will give me the most accurate prediction of how long each coffee will take to cool. This is, however, just a rough estimate, as in real life, the rate of cooling would be affected by factors such as the thermal retention properties of the mug the coffee is in, as well as the size and shape of the mug, changes in ambient temperature, and if other things such as milk, sugar or cream are added, among other things. To take all of these factors into account, I would need to collect more data with these variables and somehow produce an equation that takes these factors into account, creating bi-variant data, which would be more problematic for me.   To solve for t, I substituted in 79.1, which is the room temperature, as B(t): \(79.1=-19.32In(t)+175.9 \)​​​​​​​ \(-96.80=-19.32In(t)\) \(-96.80=-19.32In(t)\) \(5.010351967\ =\ In(t) \)​​​​​​​  \(Since\ log_ab=c,\ \rightarrow b=a^c\ \) \(t=e^{5.010351967}\ \) \(t=\ 149.957507\ minutes \) \(t\approx2.500\ hours \) I repeated this method with my instant coffee excel equation, I(t) = −2.98ln(t) + 193.3 , using 79.10 as I(t) to solve for t, giving me this final value for t: \(t=143.9604673\ minutes\ or\ 2.399\ hours \)​​​​​​​ Therefore, instant coffee cools faster than brewed coffee, as instant coffee reaches room temperature 5.997 minutes before brewed coffee does. Again, this value is only relevant to a particular situation where the room is precisely and consistently 79.10°F, and the mugs are not disturbed, among other factors. I could fulfill my goal of finding a way to apply the concepts I learned in calculus class, such as the laws of logarithms and statistical analysis, to answer a question relevant to my life beyond math class; what type of coffee cools faster? I now know that brewed coffee will stay warmer longer during my study sessions, but since the difference between the two is roughly 6 minutes, I’ll most likely stick to instant coffee, as it is cheaper.

Mathematics AI SL's Sample Internal Assessment

Comparing the cooling rates of instant coffee and black coffee

Table of content

Comparing instant coffee to brewed coffee