Cooling Water

By: Laura Lowe

Problem: Use the following exploration to generate a function to predict observed data.

a.    Take a cup of hot water and measure its initial temperature (time = 0) and then record temperature readings each minute for 30 minutes. Make note of the room temperature . . .

b.   Enter the data on a spread sheet and construct a function that will model the data.

c.    Using the function predict the temperature after 45 minutes, 60 minutes, or 300 minutes.

d.   Calculate a measure of the error between your model and the observed data by taking the square of the difference for each time, sum the squares, and divide by the number of data points. You can use this statistic to guide refinement of your function to model the data.

I began by boiling water and then pouring it into a coffee mug.  The room temperature was 72¡F.  I used my probe thermometer to measure the water temperature every minute for 30 minutes.  My data was as follows:

After I collected the data I used Microsoft Excel to plot the points on a scatterplot.

This data looks relatively linear, but it also has a leveling off as time increases, so it could be exponential decay.

I used Excel to find a linear regression equation relating time and temperature.  (Figure 1)

Figure 1

This looks like a pretty good regression line.  R2 is very close to 1, which means r, the coefficient of correlation, is also very close to 1.  (r = 0.9695)  In fact when I used Excel to average the squared differences of the observed temperature and the predicted temperature, I got 21.64.  This is a rather large error.  It also predicts a temperature of 86.2¡F in 45 minutes, 55.3¡F in 60 minutes, and -439.9¡F in 300 minutes.  This just doesnÕt make sense when the room temperature is 72¡F.  I think we can do better.

Next I used Excel to find an exponential regression equations relating time and temperature.  (Figure 2)

Figure 2

This function does not look any better than the linear equation.  R2 is slightly better (r = 0.9830), but when I used Excel to average the squared differences of the observed temperature and the predicted temperature, I got 14.49.  Again, this is better than the linear function.  However, this equation predicts a temperature of 97.5¡F in 45 minutes, 79.4¡F in 60 minutes, and 2.98¡F in 300 minutes.  This is the same problem I ran into with the linear function.  There is no way the water will reach 2.98¡F when the room is 72¡F.  Now what?

When I took a second look at the regression equation Excel gave me, I can see the problem.  Logically an exponential function of the water temperature should have a horizontal asymptote y = 72, but the regression function in Excel assumes a horizontal asymptote y = 0.  I adjusted for this by subtracting 72 from each data point, and then having Excel find the exponential regression.  (Figure 3)

Figure 3

This looks better.  Now I adjusted the equation by adding 72.  So my regression equation became y = 110.33e-0.0267x + 72.  This time when I used Excel to average the squared differences of the observed temperature and the predicted temperature, I got 9.23.  This equation predicts a temperature of 105.2¡F in 45 minutes, 94.2¡F in 60 minutes, and 72.0¡F in 300 minutes.  This makes a lot more sense.

After all this I concluded that a good equation to predict the temperature of my water is:

(I also concluded that I need to replace my thermometer.)

NewtonÕs Law of Cooling

NewtonÕs Law of Cooling describes the cooling of a warmer object to the cooler temperature of the environment.  The formula is:

T(t) is the temperature of the object at a time t

Te is the constant temperature of the environment

T0 is the initial temperature of the object

k is a constant depending on the properties of the object.

When I applied this formula to my water, using (0, 195) and (30, 123) to find k, I found my equation should have been:

When I averaged the squared differences of the observed temperature and the predicted temperature, I got 48.5.  This is worse than the linear function!  However, the equation predicts a temperature of 104.9¡F in 45 minutes, 93.2¡F in 60 minutes, and 72.0¡F in 300 minutes, it is better than linear in the long term.  Why is the error so large?

This graph shows me that the equation overestimates the temperature for the majority of the time.  So NewtonÕs Law isnÕt perfect, but it does a pretty good job if you only know 2 data points, and the room temperature.

NewtonÕs Law of Cooling is often used to estimate the time of death when a body is found.  Where else would you use it?