Janet Kaplan

Cooling Water Function

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In this investigation we are given data involving the temperature of boiling water as it cools, ultimately, to room temperature. Starting at 212 degrees Fahrenheit, the temperature of the water is taken every minute for 30 minutes.

The data is graphed below, with the independent variable as time in minutes and the dependent variable as temperature.

Given this data, how can we predict what the water's temperature will be in 45 minutes, or 60 minutes, or even 300 minutes?

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Our first task will be to determine specifically what the cooling function is. This is real life, however, and chances are the data will not fit neatly into a perfectly smooth line or curve. We must, therefore, determine a best-fit curve for our data.

How do we begin to do this?

Take a look at the shape of the curve. What type of function best models our data? It doesn't look linear, as it is not a straight line. It's not a quadratic function, for it has no resemblance to a parabola. Nor is it any higher level polynomial, again because it has no similarity to a cubic or higher degree equation.

It's clearly not a trigonometric function, for there is no periodicity. There are a few other families of functions that also do not apply. But what our data does resemble is an exponential function. More specifically, a decreasing exponential function, often known as a decay curve.

Notice that it starts off at 212 degrees and decreases fairly rapidly at first. As time moves on, the water temperature is still decreasing, but at a somewhat slower rate. Although it's hard to see on the graph, with each minute the reduction in temperature progresses at a slower rate. At 30 minutes it is down to 130 degrees. We can surmise that the water will continue to cool, and that ultimately it will come to rest at room temperature (let's say 70 degrees).

So we will begin by trying to find the best-fit exponential function for our data.

There's a catch, however. Exponential functions increase or decrease without bounds (to infinity). But experience tells us that the water will maintain the temperature of the air around it indefinitely, not decrease to absolute zero. So we will need to modify our model to incorporate this fact.

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At this point we can approach the problem from one of two ways. We can use a spreadsheet application, like Excel, to show us the results of otherwise time-consuming calculations quickly and easily. Or we can graph a general exponential model using Graphing Calculator software. Often it is helpful to use both avenues simultaneously.

The general form of a decreasing exponential function is y = ae-bx, where a is the starting value, e is the natural base, (-b) is the rate of decrease, and x is the time in minutes. In our example, a will be the starting temperature, and (-b) will be the cooling rate.

Exponential functions increase or decrease by the same multiplicative factor each time period. That is, we multiply or divide by b for each unit change in x. My first model is saying that starting at 212 degrees, the temperature will decrease from the previous temperature by e -0.02 every minute.

Here I have used Excel to list the actual data along with temperature calculations using My First Model, y=212e^-0.02x.

 Actual y=212e^-0.02x Actual My First Minute Temp diff Model sq of diff 0 212 0 212.00 0.00 1 205 7 207.80 7.85 2 201 4 203.69 7.22 3 193 8 199.65 44.28 4 189 4 195.70 44.90 5 184 5 191.83 61.24 6 181 3 188.03 49.38 7 178 3 184.30 39.74 8 172 6 180.65 74.90 9 170 2 177.08 50.09 10 167 3 173.57 43.18 11 163 4 170.13 50.89 12 161 2 166.77 33.24 13 159 2 163.46 19.92 14 155 4 160.23 27.31 15 153 2 157.05 16.43 16 152 1 153.94 3.78 17 150 2 150.90 0.80 18 149 1 147.91 1.19 19 147 2 144.98 4.09 20 145 2 142.11 8.36 21 143 2 139.29 13.73 22 141 2 136.54 19.93 23 140 1 133.83 38.04 24 139 1 131.18 61.12 25 137 2 128.58 70.82 26 135 2 126.04 80.31 27 133 2 123.54 89.44 28 132 1 121.10 118.89 29 131 1 118.70 151.33 30 130 1 116.35 186.37 avg diff 2.73 47.29

I have also calculated the deviation between my model and the actual data using the least squares method. 47.29 is higher than we would like; we want this value to be as close to zero as possible.

Here is a graph of the data, comparing the actual data to my first model. How close did I come?

Notice that my model decreases too gradually at first, and then too quickly later on. I’ll need to do some refinement.

This is also a good time to think about including the limiting temperature in my model. How do I build this in?

Newton’s Law of Cooling states that the final temperature of an object that is warmer than the air around it can be determined by the formula Tf = Tr + (T0Tr)e-rt, where Tf is the final temperature after t minutes, Tr is the temperature of the surrounding air, T0 is the original temperature of the object, and r is the rate at which the object is cooling. Newton’s law of cooling is an example of a modified decay curve.

We can state the same thing this way: y = c + ae-bx , where c is the minimum temperature (70), a represents the range of temperatures in the function (212-70 or 142), (-b) is the cooling rate, and x is the time in minutes.

We’ll use Graphing Calculator now to graph this function. CLICK HERE to manipulate the function yourself.

y = 70 + 142e-0.325x

You may wonder how I knew what value of b to use. Nothing beats trial and error, and nothing beats technology’s capacity to help in this matter. Starting with b = -.1, I just kept using different values in Graphing Calculator until I came as close as comfortable to the curve I was looking for. I then entered my figures into Excel and calculated the deviation again using the least squares method.

 Actual y=70 + 142e^-.0325x Actual GCF Minute Temp (b=-.0325) sq of diff 0 212 212.00 0.00 1 205 2 201 3 193 4 189 5 184 190.70 44.89 6 181 7 178 8 172 9 170 10 167 172.56 30.91 11 163 12 161 13 159 14 155 15 153 157.21 17.72 16 152 17 150 18 149 19 147 20 145 144.13 0.76 21 143 22 141 23 140 24 139 25 137 133.01 15.92 26 135 27 133 28 132 29 131 30 130 123.56 41.47 avg diff 25.28

At a deviation value of 25.28, I came very close to a function that models the actual cooling data.

I then use this model, y = 70 + 142e-0.325x, to predict the water temperature at 45, 60 and 300 minutes. The values are, respectively,

45 minutes            102.90 degrees

60 minutes              90.20 degrees

300 minutes            70.01 degrees