Cooling Data

Today, we'll be investigating the rate at which boiling water cools in a cup of water

First we'll need to examine Newton's Law of Cooling

Newton's Law of Cooling says that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient
temperature (i.e. the temperature of its surroundings).

Newton's Law makes a statement about an instantaneous rate of change of the temperature. The solution to this equation will then be a function that tracks the complete record of the temperature over time.

Newton's Law of Cooling is given by the following:

After some calculus, this yields the following equation:

Where:

Table 1. Shows the temperature of boiling water as it cools in a cup of water. The initial temperature is the boiling state at time=0. Each minute (from 0 to 30) the temperature of the water was measured. We have different 31 measurements. Ambient temperature = 71 deg. Initial Temperature of the water =212 deg

 Time Temp 0 212 1 207 2 202 3 198 4 194 5 189 6 185 7 180 8 176 9 173 10 169 11 166 12 162 13 159 14 156 15 153 16 151 17 150 18 148 19 146 20 144 21 141 22 140 23 139 24 137 25 135 26 133 27 132 28 131 29 130 30 129

Now we use Newton's Law of Cooling to find an equation to model our data. I took time=12 and time = 24 to find k in Newton's Equation. I averaged the two values for k.

Table 2 gives us the model temperature, the squared differences between the observed data and the predicted values, and the sum of squares divided by the number of observations

 Time Temp Model Temp Model - Obs squared diff 0 212 212 0 0 1 207 207.279767995136 -0.279767995136126 0.0782701311024874 2 202 202.717554360341 -0.717554360341353 0.514884260044888 3 198 198.308069142653 -0.30806914265284 0.0949065966548561 4 194 194.046199479925 -0.0461994799253773 0.00213439194537534 5 189 189.927003672393 -0.927003672393425 0.859335808630897 6 185 185.945705452698 -0.945705452698405 0.894358803263496 7 180 182.097688447738 -2.09768844773754 4.40029682377153 8 176 178.378490825912 -2.37849082591197 5.65721860894739 9 173 174.783800123568 -1.78380012356837 3.18194288084253 10 169 171.309448244635 -2.30944824463458 5.33355119464575 11 166 167.951406627652 -1.95140662765189 3.80798782644373 12 162 164.705781574599 -2.70578157459931 7.32125392944114 13 159 161.568809736094 -2.56880973609429 6.59878346025284 14 156 158.536853747734 -2.53685374773445 6.43562693739435 15 153 155.606398012521 -2.60639801252069 6.79331059967178 16 151 152.774044624471 -1.77404462447137 3.14723432961577 17 150 150.036509428701 -0.03650942870118 0.00133293838408655 18 148 147.390618213396 0.609381786604359 0.371346161845121 19 146 144.833303029267 1.16669697073348 1.36118182151869 20 144 142.36159863222 1.63840136777998 2.6843590419433 21 141 139.972639045113 1.02736095488675 1.05547053162581 22 140 137.663654234612 2.33634576538793 5.45851153544612 23 139 135.431966899297 3.56803310070279 12.7308602077108 24 137 133.274989365294 3.72501063470554 13.8757042286694 25 135 131.190220585829 3.80977941417095 14.5144191846407 26 133 129.175243241226 3.82475675877413 14.6287642637884 27 132 127.227720935992 4.77227906400798 22.7746474647688 28 131 125.345395489732 5.65460451026772 31.97455216754 29 130 123.526084318756 6.47391568124374 41.9115842478536 30 129 121.767677905341 7.23232209465914 52.3064828808948 270.770313259298 sum of squares 8.73452623417091 sum of squares/31

Using our function we predict the temperature after 45 minutes, 60 minutes, or 300 minutes:

 45 101.463 60 89.2791 300 71.0052

We see that at time = 300 minutes our water has cooled to the initial room temperature! Sweet!

To refine our model we could calculate new k values use different times. Say when t=10 and t=20 we can compute another Newton's Cooling Equation with the new k value. By comparing the sum of squares (divided by the number of data points) for both our models, we can see which one is a better fit! A better fit is determined by a lower sum of squares.

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