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 , where a is the initial value at time 0, k is the decay constant, and t is time in minutes.  We will use an excel spread sheet to try to find an appropriate model. The model that I first tried was .  I let a = 210 since this was the boiling point of my water.  I found k using an excel spreadsheet and good old trial and error.  You can see the y values that my model produced in the chart below along with the squares of the differences between the actual data and the values produced by my model.

 

 

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 , where  is the final temperature,  is the initial temperature and is the room temperature.

 

 

The model we will use is .  You can see from the sum of the squares of the difference that, at 34.72, we were much closer to zero this time.  You can also see from the graph that our model appears to have the same general shape as our original data.

 

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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