Introduction
My inspiration for this investigation arose after taking a sip of my favorite drink, coffee. Before this, I have spent extended amounts of time in order in class going into depth on how to create an equation and extract that information in order to calculate an exponential graph. In my confusion, I wanted to be able to apply what I have learned in class to something that was relevant and physical, something that would allow me to easily justify what I have learned was truly real and are a part of the real-world problems. At first, it was difficult to be able to think of anything that was involved in my daily life or apart of anyone else’s daily life that could hold some relationship to the convoluted methods of calculus and algebra, but I then finally settled on something that was a fundamental aid to my math internal assessment and my diploma: coffee.
Sometimes, I make myself a cup of coffee in order to stay awake throughout the all-nighters I pull. It is until a few days ago I realized that an unfinished cup of coffee cools down over time after I used boiling water for it to create it. An inspiration suddenly appeared into my mind as I suddenly started to think how my cup of coffee was able to cool down after being created with boiling water. From my understanding of my higher level in Biology, I was able to use my scientific knowledge and was able to understand the difference of the cooling body and the ambient temperature (room temperature) changes the rate of a body cooling down. The larger differences between temperatures cause a large rate of change while the smaller the differences are between a cooling body and the ambient temperature cause a smaller rate of change. From common sense, I knew that there was no possible way a cup of coffee could cool to below the room temperature at any point which meant that the coffee would approach room temperature with the rate getting smaller and smaller just like the difference between the temperature of the coffee and the ambient temperature. From this, I can understand that this may sound like an exponential curve that could be drawn on a graph. I aimed to create a model of the rate of cooling and produce an equation that would allow me to calculate how much time through my all-nighters I pull it would take me before my cup of coffee’s temperature makes it undrinkable.
My focal goal for this investigation is to simply apply the use of mathematics into the real-world, somewhere that isn’t in a lesson, school-like environment. I want to be able to solve an equation so that I can produce a value that bears a significance in the real world, rather than being able to answer a question extracted from a textbook.
GATHERING DATA
In order to gather the data, I needed to produce the graph, I had to first start off by measuring a cup of coffee cooling down. Using a thermometer and a graphing software, I was able to produce a set of readings that measured the temperature of the coffee every 5 minutes for 2 hours. The graph drawn on the graphing software reflect on the exponential function I needed, with the asymptote of the results approaching the room temperature: 73° F or 22.78° C.
It has come to my attention that the every 10-minute intervals help provide enough accuracy to form a reasonably accurate equation when calculating the temperature of the coffee at any given time. The readings are listed in a table below.
TEMPERATURE OF COFFEE
Time
(mins) 0 10 20 30 40 50 60 70 80
Temp.
(°c) 80.2 67.9 53.7 47.3 43.9 37.4 35.7 32.2 31.6
Time
(mins) 90 100 110 120
Temp.
(°c) 28.3 27.5 27.1 26.9
INITIAL GRAPHING
The next stage logically was to graph the readings listed in the table of the temperature of the coffee in order to produce a graph of my own using a software I’ve seen very often. I used the Microsoft Excel in order to produce a scatter graph with time plotted as X and the temperature plotted as Y.
FROM THIS IT COULD ASSUME:
The power of the exponential is negative; the graph has a negative correlation, so, whatever the power of the function is, it must always be negative
When time (t) = 0, temperature T= 80.2 °C; the initial temperature of the cup of coffee (the y intercept) is 80.2 °C
The ambient/room temperature is 22.78° C; the temperature of this cup of coffee will never go below this temperature
From these observations, I started to create a formula for the exponential curve displayed above. I chose the power e so that I could be able to solve equations later when finding the parameters of the equation. Through this I was able to use T (Temperature) and t (Time) and deduced to:
EXPONENTIAL DECAY
T =e^(-t)
THE ENTIRE GRAPH WAS TRANSLATED UP THE Y AXIS BY A DEGREE OF 22.78, SO THERE WAS A TRANSLATION CONSTANT
T =〖Ta+e〗^(-t)
(In which Ta =22.78 is the representative of the ambient temperature)
Drawing this graph produced this:
This is obviously not quite the correct formula; the axis of the original data was used in order to demonstrate how contrasted between the original data and this was.
Through this section of the investigation, I had a lot of issues. Being able to produce an equation to solve I found myself lost and confused on how to solve an equation such as this one. My initial thinking was that I could use simplistic algebra in order to solve this and retrieve two results from my collected data in order to solve a simultaneous equation.
To start off, I first rearranged the formula in order to make e^(-at) more approachable using the natural logs:
T =〖Ta+e〗^(-at) ln(e) = 1
T -〖 Ta=e〗^(-at)
ln (T -〖 Ta)=ln(Ke〗^(-at))
ln (T – Ta) = (lnK) + (–at) x ln(e)
ln (T – Ta) = lnK – at
Secondly, I let t = 0 and 80 minutes and T = 80.2 and 31.6 degrees Celsius
Using two results from the data in order to create a pair of two variable equations in which where Ta = 22.78
ln (80.2 – 22.78) = lnK – 0a
ln (31.6 – 22.78) = lnK – 80a
With this, I solved it algebraically:
ln (80.2 – 22.78) = lnK Subtract equation A from equation B
ln (31.6 – 22.78) = lnK – 80a
ln (8.82) – ln (57.42) = (lnK – lnK) – 80a
ln (8.82) – ln (57.42) = – 80a Please note, all calculated values
ln (8.82 / 57.42) = – 80a are simplified to 3.s.f that
-1.87 / -80 = a might result in the model
0.0234 = a not matching the results
Solving for K:
ln (80.2 – 22.78) = lnK – (0 x -0.0234)
ln (57.42) = lnK
4.05 = K
Producing the final equation:
T =〖22.78+4.05e〗^(-0.0234t)
This seems to produce a graph that matches the original data.
I started to review the equation more into depth and realized that the exponential form may be more difficult to solve compared to the original form, but straight-line graphs are much easier to solve in my opinion. Simplifying the equation by using logs, I was able to retrieve an equation much similar to y=mx+c.
Initial equation:
T(t) =〖Ta+ke〗^(-at)
T(t) – Ta = 〖ke〗^(-at)
ln (T– Ta) = ln(k) – at(ln)(e)
ln (T– Ta) = –at + ln(k) ln (T– Ta) <– y
= –at <– mx
+ ln(k) <– c
y=mx+c
Extracted from this, I used the experimental data in order to plot the graph of ln (T– Ta) on Microsoft Excel, producing the following straight-lined graph:
GRAPH
This line can be visibly seen to be accounting for the small changes in the line created by real world factors.
Due to the limitations of the Microsoft Excel, the graph was unable to produce enough detail to be able to work out the equation, however, the graphing software produced a line of best fit and displayed the equation of that line shown on the graph. It is shown below.
GRAPH
dT/dt is proportional to (T – Ta)
The coffee is cooling down is the sign of the derivative( it will be negative):
dT/dt=-k(t-Ta) from dT=-ky
I can do the integration here because the above equation can be written in this form:
C(t)=C0e-kt
In this problem:
C(t) =T(t)-Ta = Temperature difference between coffee and temperature in the room at time t.
Co = T(0) -Ta = To-Ta = Initial temperature difference at time t=0
substituting it will let me get:
T(t)-Ta=(To-Ta) e-kt
T(t)=Ta+(To-Ta) e-kt
T =〖Ta+e〗^(-at)
In my research, I tried to solve the cooling equation that was discovered and used by Isaac Newton, Newton’s Law of Cooling. In which it states,
“the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).”
dT/dt is proportional to (T – Ta)
T(t)=Ta+(To-Ta) e-kt
In which: T = Temperature
Ta = Ambient Temperature
To = Initial Temperature
t = time
k = constant
Factoring in the conditions from my investigation, the equation ended up looking like this:
T =〖80.2 – 22.78e〗^(-kt)
So, once more using the experimental data, I let T= 43.9 and t=40:
43.9 =〖22.78+(80.2 – 22.78)e〗^(-k(40))
And solved for k
43.9 – 22.78 / 80.2 – 22.78 = e^(-k(40))
ln (43.9 – 22.78 / 80.2 – 22.78) = -k (40)
ln (43.9 – 22.78 / 80.2 – 22.78) / -40 = k
-0.07591700862 = k
Producing the equation:
T(t) =〖22.78 – 57.42e〗^(-0.0759t)
GRAPH
OUTCOME OF THIS INVESTIGATION
To answer my original question: “How much time through my all-nighters I pull would it take me before my cup of coffee’s temperature makes it undrinkable?”
Using the equation:
T(t) =〖22.78 – 57.42e〗^(-0.0759t)
Let us assume that 30° C is when my cup of coffee is undrinkable:
30 = 〖22.78 – 57.42e〗^(-0.0759t)
30 – 22.78 / 57.42 = e^(-0.0759t)
ln (30 – 22.78 / 57.42) = -0.0759t
ln (30 – 22.78 / 57.42) / -0.0759 = t
t = 80.2 minutes
I can safely study and do school work during my all-nighters for 1.3 hours before my cup of coffee becomes cold and undrinkable.
This value however only represents how long I can leave my cup of coffee in a 22.78° C room before it cools down. The limitations of this equation include the fact that it doesn’t take into account on the insulation of a body cooling down in real life (or what type of mug/cup is the coffee in – perhaps material wise matters too) nor does it take into account on the idea of the ambient temperature moving. For a further and in-depth investigation, it would be intriguing to see if an equation can be produced that takes into account of perhaps the surface area of a body cooling down – a body after death, and the changing ambient temperature. Perhaps by plotting the time (mins) it takes for a specific temperature against the surface area of a cooling body after death, a graph could be produced in order to be solved as the ones above, giving a clear indication of the cooling time. However, in order to investigate the bi-variant data, it is a must to do comparisons of a variety of equations.
However, the equation produced above, despite all the minor difficulties, has produced perhaps an accurate model of my cup of coffee cooling down in my ambient temperature. I have fulfilled my goal to take the more abstracts of the mathematics (laws of logarithms and solving simultaneous equations) allows me to understand its significance to the everyday real-world. Take the natural sciences for example, the use of Newton’s Law of Cooling uses math such as the ones I’ve mentioned earlier that apply it to the real world to suggest for example, the time of death of a human body (forensics sciences). This investigation showed significance to the real world and the relevance to my life.