Cool!
Version 3.14
Background
In this activity we will create a mathematical model from a simple scientific principle –
Newton’s Law of Cooling. In the early 1700s, Sir Isaac Newton stated a very basic
assumption that concisely captures some of the basics of heat transfer. Imagine a cup of
hot coffee sitting on a desk top. Further, imagine how the coffee cools over time.
1. What physical characteristics affect how the coffee cools?
2. Does the coffee keep cooling indefinitely? Explain.
3. Think about the coffee as it is first set out and after it has been sitting out about 20
minutes. How might the rate of cooling be different at these two different periods
of time?
4. Produce a possible graph of the temperature of the coffee over time. Is it linear?
Why or why not?
Newton
Newton said (probably in Latin): the rate of heat loss of a body is proportional to the
difference in temperatures between the body and its surroundings. [When Newton used
the word “body”, he was probably referring to any generic heated object. However, you
might have seen various crime shows in which the time of death of a murder victim is
estimated by the temperature of the actual “body”.] While this statement is very
powerful, it is also pretty much just common sense. A hot object in a hot room will not
cool very quickly. A hot object in a cool room (say a refrigerator) will cool very quickly
(at least initially). Notice that the rate of change of the temperature of the cooling object
does not depend explicitly on either the temperature of the object nor the temperature of
the room, but rather the difference between these two temperatures. Let’s see how we
can use this simple assumption to develop a mathematical description (formula) of a hot
object cooling in a room.
The Model
Let’s say we are modeling the temperature of a cup of coffee sitting in a classroom.
There are three different temperatures that we need to keep track of, so a little notation
needs to be developed. Let
 A represent the room temperature (often referred to the ambient
temperature). We will assume that this value remains constant.
 T represent the temperature of the coffee. This will change over
time so we can use a subscript to denote the coffee temperature at
different times: T0 , T1 , T2 ,
 D represent the difference between the temperature of the coffee
and the classroom. This will also change over time and we will
have formulas like: D0  T0  A and Dn  Tn  A .
Now we need to figure out how to use Newton’s law of cooling to help us calculate how
the coffee temperature changes over time. Newton claimed that the rate of heat loss (i.e.
the amount the coffee temperature changes) is proportional to the difference in
temperatures (for us, that is D). Thus, we can interpret this mathematically to mean:
Change in coffee temperature = kD
where k is a proportionality constant. The value of this constant depends on the
insulatory properties of the coffee cup (stainless steel, Styrofoam, paper, etc.). Let’s see
how this might work is a typical scenario.
Let’s Calculate
Let’s start with a 178°F cup of coffee sitting in a 70°F room. Using our notation from
above we have:
 A = 70

T0  178

D0  T0  A  108
All we are missing is a value for k.
5. Should k be positive or negative? Explain.
Let’s use k = -0.025 and assume that each time step refers to one minute. So after one
minute our model would predict that the new temperature of the coffee would be:
T1  T0  (temperature change)
So,
T1  178  (0.025)(108)  175.3
6. Calculate the new temperature difference, D1 .
7. Calculate the temperature of the coffee over the next 10 minutes and graph your
results.
8. Does this graph resemble the graph you created in question 4 above. Point out
any similarities or differences.
9. Compare the change in coffee temperature during the first one minute interval and
the last one minute interval. Does this agree with your intuition from question 3
above? Explain.
10. How easy would it be to calculate the temperature of the coffee after one hour?
Finalizing the Model
Our current understanding of this model requires us to know the temperature of the coffee
(and hence the difference between the coffee temperature and the room temperature) at a
particular time in order to calculate the temperature one minute later. So we would have
to be able to calculate the temperature of the coffee at each of the previous 59 minutes if
we wanted to know what the temperature of the coffee would be after an hour of sitting in
the room. Even worse, if we wanted to know how long it took for the coffee to reach a
particular temperature, say 77°F, we would just have to plug and chug until we reached
that temperature. In other words, the current form of our model is not very powerful.
However, a little thought and some algebra can help us turn this into a more useful
calculating tool.
Our first step in this process will be to focus our attention solely on the values D n , the
temperature differences. We’ll worry about the actual coffee temperatures later.
11. Using the starting values above (initial coffee temperature of 178°F, room
temperature of 70°F, and k=-0.025), what is D0 ?
12. What is D1 ? Hint: D changes by the same amount as the coffee.
13. Develop a simple formula for D1 of the form D1 = r D0 . How are r and k related?
14. Using your answer from above, develop a simple formula for D2 that depends
only on r and D0 .
15. Find a general formula for D n that depends only on r and D0 .
16. Use this formula for D n to help you determine a simple formula for the
temperature of the coffee at any time, Tn .
17. Use this formula to calculate the temperature of the coffee every ten minutes for
two hours and produce a graph showing the cooling of the coffee over time.
Reality Check
How does our mathematical model compare to a real life situation. If you have access to
accurate temperature probes you can collect your own data and see how this real data
compares to our mathematical model. You will be able to easily calculate the initial
temperature of the coffee, the ambient temperature (room temperature), and the initial
difference, D0 . The only hard part will be in determining the value of k that best
represents the physical properties of the particular object that is cooling. Remember that
different objects will have different k-values, depending on how well the object conducts
heat. The following questions will guide you though a process that will show you how to
fit Newton’s Cooling Model to a set of real data.
A cup of coffee is placed in a room and its temperature is recorded at various times. Here
is the resulting data:
Time
(minutes)
0
6
12
18
24
30
40
51
140
165
177
Coffee
Temperature
178
156
141
132
124
113
107
100
75
72
71
After nearly three hours, the temperature of the coffee is 71°F, so we can probably safely
assume that this represents the temperature of the room. Thus, we have A=71.
18. What are T0 and D0 in this example?
19. Now we need to determine k. Using your formula from question 15 above, show
D
how the ratio 2 can be used to determine k.
D1
D
20. What if you used the ratio 4 instead?
D3
Basically, any successive ratios of the Ds can be used to calculate k. Let’s try this
technique on our temperature data. Before we can quickly do this there are some
obstacles:
 The temperature of the coffee was not taken every minute and
 We are not given the values for D n .
In our previous work, there was nothing crucial in having n represent the number of
minutes that the coffee had been cooling. It could have represented any equal increment
of time. In the coffee temperature dataset, notice that initially the temperature of the
coffee was taken every six minutes. So, let’s just assume that each “time step” represents
six minutes. That is

T0 = initial coffee temperature

T1 = temperature after six minutes

T2 = temperature after twelve minutes
 Etc.
Since we are assuming that room temperature is 71°F, it should be easy to calculate the
first several differences.
21. Add a third column to the data table representing the temperature difference, D
and fill in the first six values (representing D0 through D5 ). We can’t calculate
D7 or higher since the time interval has changed again.
Dn 1
.
Dn
23. Use this information to deduce a plausible value for k? Explain your reasoning.
24. Use this value of k in your mathematical model and use your formula from
question 16 to calculate the predicted value of the coffee temperature for three
hours. Careful! Each time step represents six minutes of real time!.
25. Graphically compare the actual coffee temperature with the temperature predicted
by the model. Comment on the validity of the model.
22. Add a fourth column to the data table representing successive ratios of
Extensions
There was nothing special about cups of coffee in the mathematics above. You can use
Newton’s Law of Cooling to model the rate of cooling of any object placed in a room of
constant temperature. To make predictions, one just needs the values of k for different
objects. These values can be estimated experimentally as we have seen. For instance, the
value of k you found above can always be used when you are using that particular cup of
coffee, even if the room temperature is different or the coffee starts out at a different
temperature. Once you know k you know it all!
If a crime scene investigator wanted to use body temperature to estimate the time of death
of an individual, they would just need to have a value of k to use. Most likely such
investigators have lots of past data to estimate the k value associated with a human body.
Of course, if the body was outside then the ambient temperature would not remain
constant. You may try to repeat some of the above calculations in such a scenario. For
instance you might get some temperature data for a particular night at a specific location.
Now, whenever you calculate the change in temperature just remember that the ambient
temperature is different at each “time step”.