Newton's Law of Cooling

Task

A cup of hot coffee will, over time, cool down to room temperature. The principle of physics governing the process is Newton's Law of Cooling. Experiments with a covered cup of coffee show that the temperature (in degrees Fahrenheit) of the coffee can be modelled by the following equation $$f(t) = 110e^{-0.08t}+75.$$ Here the time $t$ is measured in minutes after the coffee was poured into the cup.

1. Explain, using the structure of the expression $110e^{-0.08t} + 75$, why the coffee temperature decreases as time elapses.
2. What is the temperature of the coffee at the beginning of the experiment?
3. After how many minutes is the coffee $140$ degrees? After how many minutes is the coffee $100$ degrees?

IM Commentary

The coffee cooling experiment is a popular example of an exponential model with immediate appeal. The model is realistic and provides a good context for students to practice work with exponential equations.

Solution

1. The only part of the expression representing $f(t)$ involving the variable $t$ is the exponential $e^{-0.08t}$. Since the coefficient of $t$ in the exponent, $-0.08$, is negative, the values of $f(t)$ decrease as time passes. Thus the temperature of the coffee decreases as time elapses. This is appropriate as we know from experience that over time the coffee will cool down to room temperature: In this case, the $75$ in the expression representing $f(t)$ is the ambient room temperature in degrees Fahrenheit.
2. The beginning of the experiment is when the time variable $t$ takes the value zero: Since $$f(0) = 110e^{-0.08 \cdot 0} + 75 = 110 + 75 = 185,$$ the initial temperature of the coffee is $185$ degrees Fahrenheit, a little less than the boiling temperature of water.
3. To find when the coffee is $140$ degrees we want to solve $$f(t) = 110e^{-0.08t} + 75 = 140.$$ Subtracting $75$ from both sides and then dividing both sides by $110$ gives $$e^{-0.08t} = \frac{65}{110}.$$ By the definition of the natural logarithm, this gives $$-0.08t = \ln{\left(\frac{65}{110}\right)}.$$ Dividing both sides by $-0.08$ gives a value of about $6.6$ minutes for the coffee to cool to $140$ degrees.

   The same reasoning gives, as an expression for the time when the coffee has cooled to $100$ degrees, $$\frac{\ln{\left(\frac{25}{110}\right)}}{-0.08}.$$ This is about $18.5$ minutes.