Modelling up to a cup of tea next

Modelling a cup of cooling tea My inspiration to investigate this came from my indisputable favoritism of calculus over other topics, however I had to connect this to a real life situation. I had to find something physical and relevant to justify that what I was learning was real. It was difficult at first to find something related to the perplexing methods of calculus that were applicable to real life, but I finally settled on something I consume routinely and have a fervent admiration for. TeaThere are times where I have woken up to a cup of tea next to my bed and fallen asleep again before drinking it, or making a cup of tea and simply forgetting to drink it, and then having to consume a cold, significantly more undesirable beverage. Naturally I started thinking about how tea cooled, or anything cooled for that matter, and from prior knowledge I knew that the difference between the cooling body and the ambient temperature (room temperature) changed the rate at which the body cooled. Large differences between temperatures cause a large rate of change and smaller differences between the cooling body and the ambient temperature cause a smaller rate. From common sense I knew that tea couldn’t go below room temperature, which means the tea would approach room temperature as time goes on, eventually staying at a constant temperature which would only change if the room temperature changed. This sounds like an exponential decay pattern, which could be graphed. I aimed to model this rate of exponential decay and produce an equation which would allow me to see how much time I had before my tea became cold. To gather the data required to produce this graph I had to measure a cup of tea cooling down. This was easy to do by making a cup of tea and then using a thermometer, checking the temperature of the tea every 5 minutes for 2 hours. I started recording the temperature after the tea was completely made, so after the sugar was dissolved and the tea bag was taken out. The next thing to do was to graph the results. I used microsoft excel to plot a scatter graph with Temperature as the Y axis and Time as the X axis.It can be determined that:The power of the exponent is negative- Since there is exponential decay, the graph has a negative correlation, so the exponent must always be negativeWhen time (t)= 0, the temperature is 71.7°- The initial temperature of the tea before any cooling has taken place is 71.7°The room temperature is 20°- The temperature of the tea can never go below 20°From my observations I needed to construct a formula which could model this. Choosing e meant that I could find the constants needed to solve the equations later. Using temperature (T) and time (t), I made the equation for exponential decay:There is a translational constant because the graph was shifted up by 20°, making the equationThis produced the graphThis is not the correct formula, as the the graph of this formula doesn’t represent the original graph. To represent the original graph, it requires a translation on the x-axis (k), and a gradient smaller than 1 (a). This gives the equationTo make the formula easier to work with, I rearranged and used natural logs.Then I plugged in two data points from my results to create a set of two variable equations.I used t=0 and t=50Subtracting equation A from equation B:However since there is exponential decay, a should be negativeSolving for k:Plugging these values into the original equation gives Graphing this givesThe graph seems to roughly represent the original data, although it can be seen that approximately the first 50 seconds seem to represent a faster decay, however it can be seen on the original graph that it is slower.To find the exact differences, I decided to check the y point at every 5 second time interval on the graph, and also list the differences between the data produced by this formula and the original data.From the table it can be seen that the only two times where the temperature is equivalent is at t=0s and t=50s, meaning the formula does not accurately depict the original data. This inaccuracy could be attributed to the fact that the formula does not take into account the errors that took place in the experiment, such as perfectly stopping and starting the stopwatch without extra time after recording the temperature, and something such as the surface which the cup of tea was placed on, which was a cold surface, could have affected the rate of decay in temperature as well. I figured there must be some mathematical concept to do with temperatures, and after watching some murder mystery series, I was interested to find out how they were estimating times of death, and after researching a bit I found it had to do with body temperatures. After a bit more research I came across Newton’s Law of Cooling, perfect for my tea cooling experiment. Newton’s law of cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (“Other Differential Equations.” Newton’s Law of Cooling). The equation for Newton’s law of cooling is Is the temperature at time tIs the temperature of the surrounding environment Is the initial temperature of the objectIs the constant, which differs for each objectUsing the same numbers as I used in my previous formula, it produces the equation”Other Differential Equations.” Newton’s Law of Cooling, www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html.