# You place a cup of 205°F coffee on a table in a room that is 72°F, and 10 minutes later, it is 195°F. Approximately how long will it be before the coffee is 180°F?

May 20, 2017

Approximately $27$ minutes.

#### Explanation:

This is a Newton's Law of Cooling Problem.

Newton's Law of Cooling states that an object cools down by the formula $T \left(t\right) = {T}_{s} + \left({T}_{0} - {T}_{s}\right) {e}^{- k t}$, where ${t}_{0}$ is the initial temperature of the liquid, ${t}_{s}$ is the temperature of the surrounding environment, $t$ is the number of minutes elapsed and $T \left(t\right)$ is the temperature after $t$ minutes elapsed. $k$ is the constant, which will differ depending on the object.

We have to start by finding $k$.

$195 = 72 + \left(205 - 72\right) {e}^{- k \left(10\right)}$

$\frac{123}{133} = {e}^{- 10 k}$

$\ln \left(\frac{123}{133}\right) = - 10 k$

$k = - \frac{1}{10} \ln \left(\frac{123}{133}\right)$

So our formula is

$T \left(t\right) = 72 + 133 {e}^{\frac{1}{10} \ln \left(\frac{123}{133}\right) t}$

We're looking for the amount of time it takes for the coffee to cool to $180$ degrees fahrenheit, so we write the equation:

$180 = 72 + 133 {e}^{\frac{1}{10} \ln \left(\frac{123}{133}\right) t}$

$\frac{108}{133} = {e}^{\frac{1}{10} \ln \left(\frac{123}{133}\right) t}$

$\ln \left(\frac{108}{133}\right) = \frac{1}{10} \ln \left(\frac{123}{133}\right) t$

A good approximation gives

$t \approx 26.64$

Therefore, it will take approximately $27$ minutes for the coffee to cool to 180˚F.

Hopefully this helps!