Investigating the Optimum Time and Temperature to Drink Coffee

Introduction

What inspired this investigation emerged after my in depth search, whilst looking for the perfect

Internal Assessment topic. What I had concluded from my research was that I wanted to include

what I had learned in the classroom to the real world, in hopes of justifying the relevance of it. It

was a challenge in the beginning to come up with a concept that had some sort of link to calculus

and differential equations. However, I was able to decide on something that I really enjoyed, as

well as it being a crucial component in my everyday life. Coffee.

It has become obvious to me that an odd pattern could be detected, every time I make coffee.

After numerous sessions of me being too preoccupied with studying, once I finally get around to

drinking my coffee I find that it has become too cold for me to drink. Hence, this lead me into

questioning how exactly did coffee cool down? From my prior knowledge that I have gained

from me being a chemistry and physics student, I was aware that “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 (i.e. the temperature of its

surroundings).”1 In other words meaning that the difference in temperature overtime would

never be able to reach a temperature that is below the room temperature. Therefore, this

statement lead me to believe that the curve modeled will be an exponential decay curve, because

the graph will only continue to exponentially decrease due to the cooling of the temperature.

The main aim of my investigation was to figure out what the optimum temperature and time is to

drink coffee, by applying the math in an environment other than the classroom. Additionally, I

wanted to be able to find a solution to an equation so that I can produce a value that holds

significance in the real world, rather than it simply being just an answer in a textbook.

My Experiment

For my internal assessment it required of me to preform an experiment to gather my data. To

measure the cooling down of a cup of coffee, it required coffee, a mug, a thermometer, and a

stopwatch. Using those materials I was able to produce the data required for my investigation.

Fist, I made the coffee and poured it into a mug, I then went ahead and measure the initial

temperature when time was zero. After that I started the stopwatch and I noted down the

temperature and time difference every five minutes. Hence, the results of my experiment can be

seen down below displayed in table 1.

1″Other Differential Equations.” Newton’s Law of Cooling, www.ugrad.math.ubc.ca/

coursedoc/math100/notes/diffeqs/cool.html.

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Table 1: Data Gathered From My Experiment:

Time/min

Temperature °C

0

90.0

5

82.2

10

74.2

15

69.3

20

64.1

25

59.4

30

55.2

35

53.6

40

50.2

45

47.9

50

45.5

55

43.8

60

41.6

65

40.2

70

38.4

75

36.7

80

35.1

85

33.9

90

32.5

95

31.3

100

29.9

105

28.9

110

27.2

115

26.2

120

25.2

125

24.0

130

22.8

135

21.7

140

20.3

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Graph 1- Original Data Collected:

Data for the temperature a cooling cup of coffee over time

Equation of the graph: x

y = 78.1357 ? (0.990347)

Using a TI-inspire Calculator I used the data I have collected to graph the exponential function

displayed above. The graph does not look extremely accurate and that may be due to some

limitations such as the heat being used in the surrounding.

The formula of Newton’s Law of Cooling states:

dT ?(T?Tr)

In which:

dT : The rates of degrees of temperature per unit time

T: The initial temperature of the coffee at a certain time

Tr: The ambient temperature (room temperature)

K: Is the constant of proportionality

This can then be written as:

dT= ?K(T?Tr)

dt

dt

dt

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If y=(T?Tr)

T= 90°C Tr= 20°C

Now the previous formula can be converted into a first order differential equation, which would

look like this:

Solve for y

dT= ?Ky

dt

dY= ?Kydt

dt

By separating the variables, we can rewrite the equation as follows:

1×dY= ?Kdt

y

By Integrating both sides:

? 1 × dY = ??Kdt

y

lny= ?Kt+c

here, c is the constant of integration

l n70=?K(t)+c

ln70= ?K(0)+c

ln70 = c

lny= ?Kt+ln70

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t is the Initial Time, which is 0

lny?ln70= ?Kt

ln(y)= ?Kt

70

By the changing the logarithmic equation in to an exponential equation we get

e ?Kt = y

Make y the subject,

70

y = 70 e?Kt

As stated earlier, y = (T ? T r), so;

As the room temperature was 20°C; Substitute Tr by 20

(T ? Tr) = 70 e?Kt

T ? 20 = 70 e?Kt

Now, choose any given time; I decided on 20 minutes, and its respective temperature

64.1C°,from the temperature drop chart I made for this IA.Integrate this information into the

equation.

64.1 = 20 + 70e?K(20)

64.1 ? 20 = 70e?K(20)

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Hence,

44.1 = 70e?20K

Now we want to find out what e?10K is, so we will divide 44.1 by 70

44.1 =e?20K

70

0 .63=e?20K

To find out what K is, we have to multiply by ln on both sides,

ln0.63 = lne?20K

?0.462 = ? 20K

?0.301 =K

?20

0.0231 = K

Now that we have the answer for our constant term K, we can use it to find out the temperature

of the coffee at any given time.

The official formula to do so looks like this,

T = 20 + 70e?0.023t

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Graph 2: Official Formula Calculated Graph:

90.00

67.50

45.00

22.50

Temperature Of Coffee Over Time

0.00

0 35 70 105 140

Time (minutes)

The curve that is modeled in the graph above is from the equation that I found, and it looks more

accurate because the exponential curve looks consistent unlike in the first graph.

The equation I just found is going to be able to help me determine when is the maximum time

before coffee becomes undrinkable.

Hence,

T = 20 + 70e?0.023t

25°C is the temperature that I am assuming that coffee become undrinkable, therefore

substituting T with 25 to find t.

25 = 20 + 70e?0.023t

25 ? 20 = 70e?0.023t

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Temperature (°C)

25?20 =e?0.023t

70

ln(25 ? 20) = ln(e?0.023t)

70

l n(25?20)= ?0.023t

70

ln 25?20

( 70 )

?0.023 =t

t = 114.7 minutes

t = 114.7

60

t = 1.9 hours

t ? 2 hours

This means that my coffee will still remain drinkable when the time is 1.9 hours (approximately

2 hours)

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Conclusion and Evaluation

Overall, the equation that I have found helped me in answering my initial aim, which I have set

at the very beginning of this investigation which was using math outside of the classroom to

discover the optimum time and temperature to drink coffee. I have successfully achieved my

aim, because by using integration and differential equations I was abled to find an equations that

will give me the temperature of coffee at any given time. Additionally, this investigation has

helped me think outside of the box and learn something new as this was something I was

extremely interested in. I was very happy with my results, I even told a lot of my friends and

family members about it because coffee is considered to be a universal drink that everyone could

relate to it by one way or another.

However, it can be noted that some limitations might have occurred. Such as, the value found in

the end specifically represents the time and temperature of coffee when the ambient temperature

is 20°C. Additionally, the equation doesn’t include the fact that heat may be lost to the

surroundings or the insulation of the mug used in my experiment. If I was to have the chance of

doing my experiment again, I would want to use a water bath so the temperature wont be lost to

the surroundings, and I would have preferred using a data logger to get accurate readings.

Nonetheless, in a controlled environment where the said variables would be constant; my

equation is bound to work because as can be seen above in my second graph produced the

exponential decay curve is a very precise model of the cooling down of a cup of coffee.

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