Cooling Project

Initial Thoughts:

1. How will the temperature of the beverage change the next hour?

The temperature will decrease exponentially over the course of the next hour, though the outside

factors such as material of cup, various of “room temperature”, and initial temperature of the actual

liquid.

2. Assume we record the temperature of the beverage every minute during the hour. Will the changes

in temperature be the same throughout the hour? If not, how will the changes in the first few minutes

differ from the last few minutes?

No, I don’t believe that they will be. Since it is anticipated to be an exponential decay, then it will start

decreasing in temperature faster in the beginning and slow down as time goes on.

3. Speculate on how he changes in temperature from minute to minute may be related to the tempera-

ture itself and the temperature of the room.

The initial temperature of the liquid and the ambient temperature of the room will impact the cooling of

the liquid. I think that the temperature of the room will act on the liquid by allowing the heat to dissipated

until it reaches a temperature approximately equal to that of the room. Then I presume it won’t play too

much of a role after that. The initial temperature will probably impact the rate at which the temperature

decreases, because theoretically, without interference, an exponential decay will drop faster the higher

the initial temperature.

4. If you take two temperature measurements, h seconds apart, and they differ by d degrees, what are

the units of dh and what does this quantity represent?

The units of d/h would be °(degrees)/seconds. The physical meaning of this quantity is a rate of change

of degrees per unit time, which shows how temperature will be changing over time.

Experiment:

Procedure:

Once the temperature sensor, resistor were connected correctly and then connected to the Raspberry

Pi (Figure 1), the code was written and tested for a shorter interval of time. After verification of function,

water was boiled and placed inside of a ceramic mug (Figure 2) and the probe was inserted and the

program was started. Using the following code, the temperature of the cup was measure over an hour

and 20 minutes.

t={}

RunScheduledTask[(deg=temp;AppendTo[t,deg]),{60,60}]

It is unclear why the program ran for as many repititions as it did. The code was written using the count

variation of the RunScheduledTask command, and should have taken a reading every 60 seconds, for

60 repetitions. It said in the unique ID box that the program would run for 60 repitions, but output 120

data points, which can be seen in the graph (Figure 3).

Data:

{86.625,87.5,87.375,86.125,85.687,83.687,83.375,81.812,81.5,80.187,79.937,78.562,78.25,77.062,76.7

5,75.625,75.312,74.25,74.,73.,72.75,71.75,71.5,70.687,70.375,69.437,69.312,68.375,68.187,67.375,67.

187,66.375,66.187,65.437,65.25,64.437,64.312,63.687,63.437,62.75,62.562,61.937,61.75,61.125,61.,6

0.375,60.25,59.625,59.5,58.875,58.75,58.187,58.062,57.562,57.375,56.875,56.75,56.187,56.062,55.5,5

5.375,54.875,54.812,54.375,54.25,53.812,53.75,53.25,53.187,52.75,52.625,52.25,52.125,51.75,51.625,

51.25,51.125,50.75,50.687,50.312,50.187,49.812,49.75,49.375,49.312,48.937,48.875,48.5,48.437,48.0

62,48.,47.687,47.625,47.25,47.187,46.875,46.812,46.5,46.437,46.125,45.75,45.375,45.062,44.625,44.3

12,44.,43.687,43.312,43.,42.687}

Figure 1:

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Figure 2:

Cooling Project Write Up.nb 3

Figure 3:

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20 40 60 80

20

40

60

80

Post-Analysis:

Questions:

1. Does the data that you collected match what you anticipated before collecting the data? If not, how

was it different?

It definitely resembles what I thought it would look like, however there is a spike on the second reading,

and I didn’t anticipate that. My guess is that it took until the second reading to get the actual starting

temperature, but I’m not entirely sure. It also didn’t decrease as quickly as I thought it would. Based off

of experience with hot beverages, I thought it would cool faster.

2. What further observations can you make about the data you collected? Are any of the lists of data

you computed related to one another in some way? Keep in mind that this is experimental data, so

there will be some “noise” that you’ll have to ignore.

The data could indicate cooling rates within different material vessels. I did my experiment in ceramic,

which disperses heat more evenly and thus, the contents cool more slowly and evenly. Depending on

the material of the vessel, the graphs could have some variation..I also notice that it would take much

longer for the liquid to reach room temperature than just an hour.

3. Of all the calculations you’ve done, report on the relationship that seems to best describe what’s

going on in the physical situation.

Of the calculations that I did, the best relationship to describe the physical situation present in this

Cooling Project Write Up.nb 5

experiment was a proportional relationship. The equation I ended up crafting was dT/dt=k(T(t)-A).

However, this equation does not take into account the initial temperature of the liquid, and the material

of the container. I’m not sure how to include a rate of change for different materials.

Model:

Equation:

dT

dt= k(T(t) - A)

Solution:

DSolve[y'[t] k * y[t] - k * A, y[t], t]

y[t] A + k t C[1]

Model:

data = {86.625, 87.5, 87.375, 86.125, 85.687, 83.687, 83.375, 81.812, 81.5, 80.187,

79.937, 78.562, 78.25, 77.062, 76.75, 75.625, 75.312, 74.25, 74., 73.,

72.75, 71.75, 71.5, 70.687, 70.375, 69.437, 69.312, 68.375, 68.187, 67.375,

67.187, 66.375, 66.187, 65.437, 65.25, 64.437, 64.312, 63.687, 63.437, 62.75,

62.562, 61.937, 61.75, 61.125, 61., 60.375, 60.25, 59.625, 59.5, 58.875,

58.75, 58.187, 58.062, 57.562, 57.375, 56.875, 56.75, 56.187, 56.062, 55.5,

55.375, 54.875, 54.812, 54.375, 54.25, 53.812, 53.75, 53.25, 53.187, 52.75,

52.625, 52.25, 52.125, 51.75, 51.625, 51.25, 51.125, 50.75, 50.687, 50.312,

50.187, 49.812, 49.75, 49.375, 49.312, 48.937, 48.875, 48.5, 48.437, 48.062,

48., 47.687, 47.625, 47.25, 47.187, 46.875, 46.812, 46.5, 46.437, 46.125,

45.75, 45.375, 45.062, 44.625, 44.312, 44., 43.687, 43.312, 43., 42.687};

NonlinearModelFit[data, A + c * Exp[-k * t], {A, c, k}, t]

FittedModel 35.7855 + 52.5991 -0.0164141t

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