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?
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
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)