Example 5 from Algebra and Trigonometry, 6.7 Exponential and Logarithmic Models

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?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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

$T (t) = A e^{k t} + 35$
We know the initial temperature was 165, so $T (0) = 165 .$

$\begin{array}{l} 165 = A e^{k 0} + 35 & Substitute (0, 165) . \\ A = 130 & Solve for A . \end{array}$
We were given another data point, $T (10) = 150,$ which we can use to solve for $k .$
$\begin{array}{l} 150 = 130 e^{k 10} + 35 & Substitute (10, 150) . \\ 115 = 130 e^{k 10} & Subtract 35 . \\ \frac{115}{130} = e^{10 k} & Divide by 130 . \\ \ln (\frac{115}{130}) = 10 k & Take the natural log of both sides . \\ k = \frac{\ln (\frac{115}{130})}{10} \approx - 0.0123 & Divide by the coefficient of k . \end{array}$
This gives us the equation for the cooling of the cheesecake: $T (t) = 130 e^{- 0.0123 t} + 35 .$

Now we can solve for the time it will take for the temperature to cool to 70 degrees.
$\begin{array}{l} 70 = 130 e^{- 0.0123 t} + 35 & Substitute in 70 for T (t) . \\ 35 = 130 e^{- 0.0123 t} & Subtract 35 . \\ \frac{35}{130} = e^{- 0.0123 t} & Divide by 130 . \\ \ln (\frac{35}{130}) = - 0.0123 t & Take the natural log of both sides \\ t = \frac{\ln (\frac{35}{130})}{- 0.0123} \approx 106.68 & Divide by the coefficient of t . \end{array}$
It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to $70°F .$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

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?

How did it go?

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

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

Your turn

Solution

Your turn