Question

The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the water temperature, A is the room temperature, and k is a positive constant.
If the coffee cools from 180°F to 100°F in 10 minutes at a room temperature of 75°F, how long (to the nearest minute) will it take the water to cool to 80°F?

Answer 1

This is a doozy.  I actually have another form of the equation for cooling, and it's in Celcius (so I convert) and seconds (we can switch back to minutes in the end).  The Celcius and Fahrenheit don't matter because it's the time we are interested in and as long as we use the C and F equivalents, we're fine.  Promise.  The formula is as follows: $T(t)= T_{s}+( T_{0}- T_{s})e ^{-kt}$.  It's really easy to use.  T(t) is the ending temperature; $T_{s}$ is the room temp; $T_{0}$ is our initial temperature of the liquid; e is Euler's number, k is the constant of variation, and t is the time in seconds.  In Celcius, our temps are as follows:  180 F = 82.2 C; 100 F = 37.8 C; 75 F = 23.9 C; 80 F = 26.7 C.  10 minutes is 600 seconds.  Filling in our formula with all of that, we will solve for k.  Then we will use that k value to solve for the time in the question we are told to answer.  Let's do this.  $37.8=23.9+(82.2-23.9)e ^{-600k}$.  Simplifying a bit we have $37.8=23.9+(58.3)e ^{-600k}$.  Subtracting 23.9 from both sides we have $13.9=58.3e ^{-600k}$.  divide both sides by 58.3 to get $.2384219=e ^{-600k}$.  At this point in your mathematical career, I'm assuming that since you're doing this, you're studying natural and common logs.  The base of a natural log is e, therefore, if we take the natural log of e, they both "undo" each other and cancel each other out completely.  So let's do that. $ln(.2384219)=-600k$.  Simplifying on the left gives us $-1.433713=-600k$.  Dividing both sides by -600 gives us a k value of .002389.  Now we will do the equation again, using that k value, to solve for how long it takes for the coffee to cool to 80 F (26.7 C).$26.7=23.9+(82.2-23.9)e ^{-.002389t}$.  Simplifying a bit again gives us $26.7=23.9+(58.32)e ^{-.002389t}$.  Subtracting 23.9 from both sides gives us $2.8=(58.32)e ^{-.002389t}$.  We will now divide both sides by 58.32 to get $.0480109=e ^{-.002389t}$.  Again we will take the natural log of both sides to "undo" the e: $ln(.0480109)=-.002389t$.  The natural log of .0480109 is -3.036325671.  So, dividing both sides by the negative decimal on the right gives us $\frac{-3.036325671}{-.002389}=t$ and t = 1270.9 seconds.  Divide that by 60 to get the time in minutes.  t = 21.1826...so 21 minutes.  Phew!!!

Answer 2

ANSWER
21 minutes

EXPLANATION
This method involves solving the given first-order differential equation.
We can note that we are given a separable differential equation with given values.

   $\dfrac{dT}{dt} = -k(T - A)$

If the room temperature is 75, then A = 75 and we have

   $\dfrac{dT}{dt} = -k(T - 75)$

We can solve by moving the differential dT with the factor containing T and the other terms with the differential dt.

   $\begin{aligned} \dfrac{dT}{dt} &= -k(T - 75) \\ dT &= -k(T - 75)dt \\ \dfrac{dT}{T-75} &= -k \,dt \end{aligned}$

Integrate both sides of the equation

   $\begin{aligned} \int \dfrac{dT}{T-75} &= \int -k \,dt \\ \ln|T - 75| &= -k t + C \end{aligned}$

To solve for the constant of integration, C, we make note of detail that says at time t = 0 minutes, we have T = 180

   $\begin{aligned} \ln|180 - 75| &= -k (0) + C \\ C &= \ln(105) \end{aligned}$

To solve for the constant from Newton's law of cooling, k, we make note of the detail that says at time t = 10 minutes, we have T = 100. Since we know that C = ln(105):

   $\begin{aligned} \ln|100 - 75| &= -k (10) + \ln(105) \\ \ln(25) &= -10k + \ln (105) \\ \ln(25) - \ln(105) &= -10k \\ k &= -\tfrac{1}{10}\ln(25/105) \end{aligned}$

Therefore, the equation now becomes

   $\begin{aligned} \ln|T - 75| &= \tfrac{1}{10}\ln(25/105) \cdot t + \ln (105) \end{aligned}$

Use definition of logarithm
   $\ln x = y \iff \log_e x = y \iff e^y = x$
to convert this into an exponential equation:

   $\begin{aligned} \ln|T - 75| &= \tfrac{t}{10}\ln(25/105) + \ln (105) \\ |T-75| &= e^{(t/10)\ln(25/105) + \ln (105)} \\ T - 75 &= \pm e^{(t/10)\ln(25/105) + \ln (105)} \\ T &= 75 \pm e^{(t/10)\ln(25/105) + \ln (105)} \end{aligned}$

Note that since $e^{(t/10)\ln(25/105) + \ln (105)}$ is always positive, we need to choose the (+) case in order to get to temperatures of 180°F and 100°F. So T as a function of time, t, is

   $T = 75 + e^{(t/10)\ln(25/105) + \ln (105)}$

We are looking for the value the time t for which T = 80°F. Set T = 80 and solve for t

   $\begin{aligned} 80 &= 75 + e^{(t/10)\ln(25/105) + \ln (105)} \\ 5 &= e^{(t/10)\ln(25/105) + \ln (105)} \\ \ln(5) &= \tfrac{t}{10}\ln(25/105) + \ln(105) && (\text{\footnotesize logarithmic form}) \\ \ln(5) - \ln(105) &= \tfrac{t}{10}\ln(25/105) \\ \ln(5/105) &= \tfrac{t}{10}\ln(25/105) \\ \frac{\ln(5/105)}{\ln(25/105)} &= \tfrac{t}{10} \\ t&= \frac{10\ln(5/105)}{\ln(25/105)} \\ t &\approx 21.215 \end{aligned}$

Rounded to the nearest minute, it will take 21 minutes for the "water" to cool to 80°F