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Introduction: If you are a part of the large portion of our society who enjoys a steaming hot beverage most mornings, you know how quickly your once scalding hot coffee or tea can become a lukewarm liquid. I noticed how quickly this temperature decrease occurred as well as, every morning, I made my too-hot-to-drink tea and rushed off to school. In an attempt to keep up with the class and stay engaged in the moment, by the time I was able to finally take my first sip of tea, it had already begun to go cold. Knowing that the temperature of a substance will cool or heat in order to reach the temperature of the environment it is in, I knew that my tea was doing the same thing; cooling quickly at first, then slowing in temperature decrease in order to reach room temperature.
Through my work with graphs, formulating equations, modeling functions I felt that creating a model for a cooling cup of tea would not only be achievable considering the exponential rate at which it cools, but be useful in determining the time span in which a hot tea will remain at a satisfying temperature. Through this investigation, my aim was to create an equation for the rate at which tea cools through the use of Newton’s Law of Cooling and apply what I have learned in my class to a real life scenario.
Data Collection: To gather data to use for creating an equation, I first made a cup of tea and measured its initial temperature (temperature at 0 minutes). Using a digital thermometer, I recorded the temperature of the tea every 5 minutes for 2 hours. Because the temperatures were measured every 5 minutes instead of measured continuously, the accuracy of my data decreased. To make up for this loss of data, I collected data for a greater span of time.
Processes: After gathering data for the temperature of tea for 2 hours, I created a scatter graph of the data with the time in minutes on the x axis and the temperature in degrees fahrenheit on the y axis.
The graph produced from the data shows that the y intercept is at the point (0, 197. 6). This means that when t=0 (t being the time in minutes), T=197. 6 (T being the temperature in degrees fahrenheit). Knowing that the temperature of the tea could never fall below room temperature and the temperature of the room at the time the data was collected was 72. 6°F, the horizontal asymptote was at T=72. 6. Through the graphed data, I was also able to come to the conclusion that the x value needed to be negative due to the negative slope of the graph.I began solving for a mathematical equation to model the data I collected by using the formula for quantities that decrease at a continuous rate: much like my tea. T(t) = a+e^t In this equation, “a” represents the horizontal asymmptote in which no value can go below this point. In this case, the horizontal asymptote would be the temperature of thr room at the time of the experiment. The ‘t’ in the equation represents the x value which is the time in minutes. I used the value of ‘e’ in the equation to represent the rate of change in the equation because ‘e’ is the base rate of growth in continuous processes.
Since the equation must have a negative x value, I arrived at the new equation: T(t) = a+e^-tUsing 72. 6 as the ‘a’ value in the equation because it was the room temperature during the investigation, I graphed the equation T=72. 6+e^-t
The equation graphed above clearly does not represent the data I collected. In order for this function to better represent the data I collected, a vertical translation upwards would need to be applied to match the T(0) temperature. T=a+ne^-tA horizontal stretch would also need to be incorporated. This lead me to the equation:
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