Cooling Project Experiment

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
d
h
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:
2 Cooling Project Write Up.nb

Figure 2:
Cooling Project Write Up.nb 3

Figure 3:

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 

Cooling Project Experiment

More Related Content

What's hot (18)

Viewers also liked (20)

Similar to Cooling Project Experiment (20)

Cooling Project Experiment