You are about to erase your work on this activity. Are you sure you want to do this?

Updated Version Available

There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed?

Mathematical Expression Editor

We study Newton’s Law of Cooling as an application of a first order separable
differential equation.

Newton’s Law of Cooling

Newton’s law of cooling states that if an object with temperature at time is in a
medium with temperature , the rate of change of at time is proportional to ; thus,
satisfies a differential equation of the form

Here , since the temperature of the object must decrease if , or increase if . We’ll call
the temperature decay constant of the medium.

For simplicity, in this section we’ll assume that the medium is maintained
at a constant temperature . This is another example of building a simple
mathematical model for a physical phenomenon. Like most mathematical
models it has its limitations. For example, it’s reasonable to assume that
the temperature of a room remains approximately constant if the cooling
object is a cup of coffee, but perhaps not if it’s a huge cauldron of molten
metal.

To solve (eq:4.2.1), we rewrite it as
Since is a solution of the complementary equation, the solutions of this equation are
of the form , where , so . Hence,
so
If , setting here yields , so

Note that decays exponentially, with decay constant .

A ceramic insulator is baked at C and cooled in a room in which the temperature is
C. After 4 minutes the temperature of the insulator is C. What is its temperature
after 8 minutes?

Here and , so (eq:4.2.2) becomes
We determine from the stated condition that ; that is,
hence,
Taking logarithms and solving for yields
Substituting this into (eq:4.2.3) yields
Therefore the temperature of the insulator after 8 minutes is

An object with temperature F is placed outside, where the temperature is F. At
11:05 the temperature of the object is F and at 11:07 its temperature is F. At what
time was the object placed outside?

Let be the temperature of the object at time . For convenience, we choose the origin
of the time scale to be 11:05 so that . We must determine the time when .
Substituting and into (eq:4.2.2) yields
or
We obtain from the stated condition that the temperature of the object is F at 11:07.
Since 11:07 is on our time scale, we can determine by substituting and into (eq:4.2.4) to
obtain
hence,
Taking logarithms and solving for yields

Recall that .

Substituting this into (eq:4.2.4) yields
and the condition implies that
hence,
Taking logarithms and solving for yields
Therefore the object was placed outside about minutes and seconds before 11:05;
that is, at .

Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored
and Edited Books & CDs. 8. (CC-BY-NC-SA)