Boiling Water
By
Janet Shiver
EMAT 6680
Investigation: Take a cup of boiling water and measure its initial temperature. Then record the temperature of the water each minute for thirty minutes. Graph the data and then construct a function to model the data. Calculate a measure of the error between the model and the observed data. Finally, using the function, predict the temperature after 45 minutes, 60 minutes, or 300 minutes.
Collect the Data
After boiling a cup of water, I recorded its temperature every minute for thirty minutes. The data is graphed below, with the independent variable as time in minutes and the dependent variable as temperature.
Time | Temperature |
0 | 210 |
1 | 200 |
2 | 189 |
3 | 180 |
4 | 174 |
5 | 167 |
6 | 161 |
7 | 156 |
8 | 151 |
9 | 148 |
10 | 145 |
11 | 142 |
12 | 139 |
13 | 136 |
14 | 134 |
15 | 131 |
16 | 128 |
17 | 126 |
18 | 125 |
19 | 123 |
20 | 121 |
21 | 119 |
22 | 117 |
23 | 116 |
24 | 114 |
25 | 113 |
26 | 112 |
27 | 110 |
28 | 109 |
29 | 108 |
30 | 107 |
Finding a model
Our next task will be to develop a model for this data. We must attempt to find a cooling function that will “best” fit our data. As we do this, we must keep in mind that since our data is “real-world” it is highly unlikely that we will be able to find a curve that matches it exactly. We need to focus instead on finding a function that will best match our data.
We will begin by first examining the general shape of the graph and its behavior. Several basic functions can be eliminated based on the appearance of the graph. The graph of the data does not lie on line so the linear function can be eliminated as a possibility. It is also does not resemble a parabola so the quadratic function would not be a good choice. We can also eliminate the cubic function and higher order polynomials for the same reasons of appearance. So we can safely say that the graph does not resemble any polynomial functions. It is also safe to eliminate the trigonometric functions since the graph has no periodicity.
The function that the graph most closely resembles is the exponentional decay function. As with all exponential decay models, our graph shows a rapid decrease at the beginning of the time. As time progresses, the graph continues to decrease in temperature but at a much slower rate. We could conjecture that if we continued to take readings that the water would ultimately reach room temperature which in this case was 77 degrees.
Exponential Decay Model
The general form of any exponential model is
Time | Temperature | y=210e^(-.028*t) | Square of the difference |
0 | 210 | 210.00 | 0.00 |
1 | 200 | 204.20 | 17.65 |
2 | 189 | 198.56 | 91.46 |
3 | 180 | 193.08 | 171.10 |
4 | 174 | 187.75 | 189.04 |
5 | 167 | 182.57 | 242.28 |
6 | 161 | 177.52 | 273.05 |
7 | 156 | 172.62 | 276.31 |
8 | 151 | 167.86 | 284.13 |
9 | 148 | 163.22 | 231.69 |
10 | 145 | 158.71 | 188.09 |
11 | 142 | 154.33 | 152.08 |
12 | 139 | 150.07 | 122.56 |
13 | 136 | 145.93 | 98.55 |
14 | 134 | 141.90 | 62.38 |
15 | 131 | 137.98 | 48.72 |
16 | 128 | 134.17 | 38.07 |
17 | 126 | 130.47 | 19.94 |
18 | 125 | 126.86 | 3.47 |
19 | 123 | 123.36 | 0.13 |
20 | 121 | 119.95 | 1.09 |
21 | 119 | 116.64 | 5.56 |
22 | 117 | 113.42 | 12.81 |
23 | 116 | 110.29 | 32.61 |
24 | 114 | 107.24 | 45.64 |
25 | 113 | 104.28 | 75.99 |
26 | 112 | 101.40 | 112.29 |
27 | 110 | 98.60 | 129.88 |
28 | 109 | 95.88 | 172.11 |
29 | 108 | 93.23 | 218.05 |
30 | 107 | 90.66 | 267.02 |
Sum of the squares | 3583.74918 |
119.458306 |
By calculating the difference between my model and the actual data using the least squares method we can see that our model needs to be improved on. 119.46 is much higher than we would like; we would prefer that the difference between the actual data and the model be much closer to zero.
Now lets look at it graphically.
We can see that our model does not fit the curve of the actual data well. Although our model is decreasing it is decreasing too slowly for the first 20 minutes and then too quickly during the last 10.
How can we improve this model? We must now consider that most decay models approach the value of zero but this is not the case for our model. This experiment was conducted in a room that was 77 degrees so once our water cools completely it should maintain a temperature of the air in the room or 77 degrees.
We must now consider Newton’s Law of Cooling. Newton’s law of cooling states that the final temperature of an object that is warmer than the air around it will be the room temperature. This temperature can be determined by the formula
The model we will use is
Time | Temperature | y =77+133e^(-.0586t) | Square of the Difference |
0 | 210 | 210.00 | 0.00 |
1 | 200 | 202.43 | 5.91 |
2 | 189 | 195.29 | 39.58 |
3 | 180 | 188.56 | 73.25 |
4 | 174 | 182.21 | 67.39 |
5 | 167 | 176.22 | 85.03 |
6 | 161 | 170.57 | 91.66 |
7 | 156 | 165.25 | 85.52 |
8 | 151 | 160.23 | 85.10 |
9 | 148 | 155.49 | 56.07 |
10 | 145 | 151.02 | 36.25 |
11 | 142 | 146.81 | 23.12 |
12 | 139 | 142.83 | 14.71 |
13 | 136 | 139.09 | 9.53 |
14 | 134 | 135.55 | 2.41 |
15 | 131 | 132.22 | 1.49 |
16 | 128 | 129.08 | 1.16 |
17 | 126 | 126.11 | 0.01 |
18 | 125 | 123.32 | 2.83 |
19 | 123 | 120.68 | 5.37 |
20 | 121 | 118.20 | 7.86 |
21 | 119 | 115.85 | 9.91 |
22 | 117 | 113.64 | 11.29 |
23 | 116 | 111.55 | 19.76 |
24 | 114 | 109.59 | 19.46 |
25 | 113 | 107.73 | 27.74 |
26 | 112 | 105.98 | 36.19 |
27 | 110 | 104.33 | 32.10 |
28 | 109 | 102.78 | 38.70 |
29 | 108 | 101.31 | 44.74 |
30 | 107 | 99.93 | 50.02 |
Sum of the squares 1041.616
34.72
Lets use our model to predict the temperature at 45, 60 and 300 minutes. Click here to see the model. Move point b to determine the temperature at any given time. Pay attention to the temperature as the time increases and record your results for 45 minutes, 60 minutes and 300 minutes.
Scroll Down For The Solutions
Solutions: 45 minutes - 86 degrees, 60 minutes - 81 degrees, and 300 minutes - 77 degree
Did you notice that as the time increased the temperature went to room temperature, 77 degrees. For further investigation you might want to determine when the model first reached 77 degrees and what this means. You might also want to investigate whether the model will drop below 77 degrees and when this occurs.
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