## Use Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

$T\left(t\right)=a{e}^{kt}+{T}_{s}$

This formula is derived as follows:

$\begin{cases}T\left(t\right)=A{b}^{ct}+{T}_{s}\hfill & \hfill \\ T\left(t\right)=A{e}^{\mathrm{ln}\left({b}^{ct}\right)}+{T}_{s}\hfill & \text{Laws of logarithms}.\hfill \\ T\left(t\right)=A{e}^{ct\mathrm{ln}b}+{T}_{s}\hfill & \text{Laws of logarithms}.\hfill \\ T\left(t\right)=A{e}^{kt}+{T}_{s}\hfill & \text{Rename the constant }c \mathrm{ln} b,\text{ calling it }k.\hfill \end{cases}$

### A General Note: Newton’s Law of Cooling

The temperature of an object, T, in surrounding air with temperature ${T}_{s}$ will behave according to the formula

$T\left(t\right)=A{e}^{kt}+{T}_{s}$

where

• t is time
• A is the difference between the initial temperature of the object and the surroundings
• k is a constant, the continuous rate of cooling of the object

### How To: Given a set of conditions, apply Newton’s Law of Cooling.

1. Set ${T}_{s}$ equal to the y-coordinate of the horizontal asymptote (usually the ambient temperature).
2. Substitute the given values into the continuous growth formula $T\left(t\right)=A{e}^{k}{}^{t}+{T}_{s}$ to find the parameters A and k.
3. Substitute in the desired time to find the temperature or the desired temperature to find the time.

### Example 4: Using Newton’s Law of Cooling

A cheesecake is taken out of the oven with an ideal internal temperature of $165^\circ\text{F}$, and is placed into a $35^\circ\text{F}$ refrigerator. After 10 minutes, the cheesecake has cooled to $150^\circ\text{F}$. If we must wait until the cheesecake has cooled to $70^\circ\text{F}$ before we eat it, how long will we have to wait?

### Solution

Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation

$T\left(t\right)=A{e}^{kt}+35$

We know the initial temperature was 165, so $T\left(0\right)=165$.

$\begin{cases}165=A{e}^{k0}+35\hfill & \text{Substitute }\left(0,165\right).\hfill \\ A=130\hfill & \text{Solve for }A.\hfill \end{cases}$

We were given another data point, $T\left(10\right)=150$, which we can use to solve for k.

$\begin{cases}\text{ }150=130{e}^{k10}+35\hfill & \text{Substitute (10, 150)}.\hfill \\ \text{ }115=130{e}^{k10}\hfill & \text{Subtract 35}.\hfill \\ \text{ }\frac{115}{130}={e}^{10k}\hfill & \text{Divide by 130}.\hfill \\ \text{ }\mathrm{ln}\left(\frac{115}{130}\right)=10k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{ }k=\frac{\mathrm{ln}\left(\frac{115}{130}\right)}{10}=-0.0123\hfill & \text{Divide by the coefficient of }k.\hfill \end{cases}$

This gives us the equation for the cooling of the cheesecake: $T\left(t\right)=130{e}^{-0.0123t}+35$.

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

$\begin{cases}70=130{e}^{-0.0123t}+35\hfill & \text{Substitute in 70 for }T\left(t\right).\hfill \\ 35=130{e}^{-0.0123t}\hfill & \text{Subtract 35}.\hfill \\ \frac{35}{130}={e}^{-0.0123t}\hfill & \text{Divide by 130}.\hfill \\ \mathrm{ln}\left(\frac{35}{130}\right)=-0.0123t\hfill & \text{Take the natural log of both sides}\hfill \\ t=\frac{\mathrm{ln}\left(\frac{35}{130}\right)}{-0.0123}\approx 106.68\hfill & \text{Divide by the coefficient of }t.\hfill \end{cases}$

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to $70^\circ\text{F}$.

### Try It 4

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?

Solution