Using 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(t)=ae^{kt}+T_s

This formula is derived as follows:

\begin{array}{ll}T(t)=A b^{t t}+T_{s} & \\T(t)=A e^{\ln \left(b^{d}\right)}+T_{s} & \text { Laws of logarithms. } \\T(t)=A e^{c t \ln b}+T_{s} & \text { Laws of logarithms. } \\T(t)=A e^{k t}+T_{s} & \text { Rename the constant } c\text { ln } b, \quad \text { calling it } k .\end{array}

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(t)=A e^{k t}+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(t)=Ae^{kt}+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 5

Using Newton's Law of Cooling

A cheesecake is taken out of the oven with an ideal internal temperature of 165°F, and is placed into a 35°F refrigerator. After 10 minutes, the cheesecake has cooled to 150°F. If we must wait until the cheesecake has cooled to 70°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(t)=Ae^{kt}+35

We know the initial temperature was 165, so T(0)=165.

\begin{array}{ll}165=A e^{k0}+35 & \text { Substitute (0, 165).} \\ A = 130 & \text { Solve for A. } \end{array}

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

\begin{array}{ll}150=130 e^{k 10}+35 & \text { Substitute }(10,150) \\115=130 e^{k 10} & \text { Subtract } 35 . \\\frac{115}{130}=e^{10 k} & \text { Divide by } 130 . \\\ln \left(\frac{115}{130}\right)=10 k & \text { Take the natural log of both sides. } \\k=\frac{\ln \left(\frac{115}{100}\right)}{10} \approx-0.0123 & \text { Divide by the coefficient of } k .\end{array}

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

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

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

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70°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?
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