# Exponential growth and decay (Part 12): Newton’s Law of Cooling

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss Newton’s Law of Cooling, which describes how quickly a hot object cools in a room at constant temperature. (This law is not to be confused with Newton’s Three Laws of Motion.)

While Newton’s Law of Cooling is easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that the rate at which a hot object cools is proportional to the difference between its current temperature and the surrounding temperature. In other words, in a room at a temperature of $70^\circ$F, a very hot object at $270^\circ$F will cool twice as fast than when its temperature drops to $170^\circ$F.

The above paragraph can be rewritten as a differential equation. Let $T(t)$ be the temperature of the object at time $t$, and let $S$ be the (constant) surrounding temperature that surrounds the object. Since the rate at which the substance decays is $dT/dt$, we find that

$\displaystyle \frac{dT}{dt} = - k (T - S)$,

where $k$ is a constant that depends on the object. The negative sign on the right-hand side isn’t strictly necessary, but it’s a reminder that amount present decreases as time increases.

This differential equation can be solved in several ways, including separation of variables (below, I’ll be sloppy with the constant of integration for the sake of simplicity):

$\displaystyle \frac{dT}{T-S} = -k$

$\displaystyle \int \frac{dT}{T-S} = - \displaystyle \int k \, dt$

$ln |T-S| = -k t + C$

$|T-S| = e^{-kt+C}$

$|T-S| = e^{-kt} e^C$

$T-S = \pm e^C e^{-kt}$

$T-S = C e^{-kt}$

$T = S + C e^{-kt}$

To solve for the constant, we usually use the initial condition $T(0) = T_0$, a number that must be given in the problem:

$T(0) = S + C e^{-k \cdot 0}$

$T_0 =S + C \cdot 1$

$T_0 - S = C$

Plugging back in, we obtain the final answer

$T = S + (T_0 - S) e^{-kt}$

Of course, students in Algebra II or Precalculus (or high school physics) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

The careful reader will notice that this derivation essentially parallels the previous derivations for continuous compound interest and for radioactive decay. In other words, these phenomena from apparently different realms of life have similar solutions because the governing differential equations are very similar.

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