Exponential growth and decay

Question

There's a cup of coffee made with boiling water standing at room where room temperature is $20ºC$. If $H(t)$ is the temperature of this cup of coffee at the time $t$, in minutes, explain what the differential equation says in everyday terms. What is the sign of $k$?

$$\frac{dh}{dt} = -k(H-20)$$

Then solve the differential equation for 90ºC in 2 minutes and how long it will take to cool to 60ºC

Observing $dh/dt = 0$ we find that $H=20$ this means that the function stops changing at the room temperature $H=20$. As $t$ is implied to be $H = 20 + Ae^{-kt}$ as $t$ approaches infinity $H=20$.

Answer 1

Note: You boiled the coffee, let's say $H(0)\approx 100^o C$

Now you know that the coffee would continuously lose heat until it comes to the room temperature, never going below it. So $\frac{dh}{dt}$ is negative. $(H-20)$ is always positive. So, k is positive

You are right when you say the solution is $H(t)=20+Ae^{-kt}$

Now your problem says in two minutes, the temperature becomes $90^o$, i.e. $H(2)=90^o$

$90=20+Ae^{-2k}$

So $A=70e^{2k}$

Your equation now changes (when you substitute the value of A)

Now all you need to do is find $t$ given $H(t)=60$

Answer 2

I suppose that $h$ is $H$ in the equation. So, what you know is that $$\frac{dH}{dt} = -k(H-20)$$ If you remember what is a derivative, in plain words, it seems to mean that, during a short period of time, the temperature of water changes proportionally to the difference between water and room temperature.

Suppose that $H_0$ be the temperature at time $t_0$. For time $t$, you can write that $$\frac{dH}{dt}\approx \frac{H-H_0}{t-t_0}=-k(H_0-20)$$ that is to say that, for a small interval of time, $$H\approx H_0-k(H_0-20)(t-t_0)$$

Answer 3

I think you can use separation of variables, so we have $$ \frac{dH}{dt} = -k(H-20) \\ \frac{dH}{H-20} = -kdt \\ \int{\frac{dH}{H-20}} = \int{-kdt} \\ \ln(H-20) = -kt+C \\ e^{\ln{(H-20)}} = e^{-kt+C} \\ H-20 = e^{-kt+C} \\ H-20 = e^{-kt}e^{C} $$

Now, solve for $e^C$, setting $t = 0$ gives

$$ H_0 - 20 = e^C $$ which finally gives $$ H(t) = (H_0 - 20)e^{-kt} + 20 \quad\quad (1) $$

Now we need to calculate $k$ using equation $(1)$ and the information given in the problem: Assume $H_0 = 100$, then

$$ 90 = (100 - 20)e^{-2k} + 20 \\ 70 = 80e^{-2k} \\ \frac{70}{80} = e^{-2k} \\ \ln\left(\frac{70}{80}\right) = -2k \\ \ln(7/8) = -2k \\ k = -\frac{\ln(8/7)}{2} \\ k \approx 0.0667656963123 $$

so $$ t = \frac{ln(2)}{k} \\ t \approx \frac{0.6931471805599}{0.0667656963123} \\ t \approx 10.3817861393628 $$

So it takes about 10 minutes for the cup of coffee to cool to $60^{\circ}$ C.