Problem with differential equation 3 I'm new with differential equation and I can't figure out how to solve this problem:
The rate of change of temperature of an object is proportional
the difference between the object temperature and the temperature of the environment
(Newton's law). In addition, heat flows from hot to cold.
Boiled water in a pan and then it is removed from the heating element,
so that the initial temperature of water is 100 degrees Celsius, while
the surrounding room temperature is 20 degrees Celsius and will be assumed
constant.
Knowing that 5 minutes later the water temperature is 60 degrees Celsius,
how much longer will he wait for the temperature of the water
reaches 25 degrees C?
I'm stuck, so any tip will be helpful
Thanks in advance
 A: Wealll . . .
letting $T$ be the temperature of the heated object, and $T_0$ the ambient temperature, the difference is obviously $T - T_0$.  So the rate of change of $T$,  being $\frac{dT}{dt}$, being proportional to $T- T_0$, must satisfy an equation of the form
$\frac{dT}{dt} = c(T - T_0), \tag{1}$
for some constant $c$.  Now the key thing to note is  that, since as you so eloquently put it, "heat flows from hot to cold", we must have $c < 0$; otherwise a hot object would continue to get hotter, even in a cooler environment.  So let's just say $c = -k$ for some $ k > 0$.  So (1) becomes 
$\frac{dT}{dt} = -k(T - T_0). \tag{2}$
How to solve (2)?  Suppose we define a new variable 
$\Delta = T - T_0, \tag{3}$
then (2) becomes
$\frac{d \Delta}{dt} = -k \Delta, \tag{4}$
since
$\frac{d \Delta}{dt} = \frac{dT}{dt}, \tag{5}$
by virtue of the fact that $T_0$ is constant.   Thus we must have
$\Delta(t) = \Delta(t_0) e^{-k(t - t_0)}, \tag{6}$
since that's the only solution to (4).  In these equations, $t$ represents time, so that $t_0$ is the time the cooling starts, e.g. when the pot is taken off the stove.  (I'm guessing/hoping you already know that (6) is the only solution to (4); saves me a lot of one-finger 'droid typing!)
So now we plug in the initial conditions etc., taking $t_0 = 0$.  Thus
$\Delta(t_0) = 80° \text{C}, \tag{7}$
and if $t = 5 \, \text{min.}$, we get
$60° \text{C} =  \Delta(5 \, \text{min.}) = (80° \text{C}) e^{-k (5 \,\text {min.})}, \tag{8}$
or
$\frac{3}{4} = e^{-k (5 \text{min.})}, \tag{9}$
or
$k = \frac{-\ln(\frac{3}{4})}{5} \text {min.}^{-1} \approx 0.0575364 \, \text {min.}^{-1}. \tag{10}$
So having $k$, we use our solution again, but this time to figure out just when $\Delta = 25° \text{C}$.  This time we plug everything we know back into (6), using the new data, but now the unkown is $t - t_0$; so now we have $\Delta(t_0) = 40^{\circ} \, \text C$ and $\Delta(t) = 5^{\circ} \, \text C$, and the new value of $t_0$ is $5 \text{min.}$, so
$5° \text{C} =  \Delta(t \, \text{min.}) = (60° \text{C}) e^{-0.0575364 ((t - 5) \,\text {min.})}, \tag{11}$
or
$\frac{1}{12} =  e^{-0.0575364 ((t - 5) \,\text {min.})}, \tag{12}$
or
$(t - t_0)  \, \text{min.}= \frac{- \ln \frac{1}{12}}{0.0575364}  \,\text {min.}, \tag{13}$
or
$(t - t_0)  \, \text{min.} \approx 43.2 \, \text{min.}. \tag{14}$
Apparently, we need to wait an additional $43.2$ minutes for the water to reach $25^{\circ} \, \text C$.
Cheerio!
Fiat Lux
A: Water's initial temperature is $T(0)=T_0=100^{\circ}C$
Room temperature is $T_e=20^{\circ}C$
The general solution is:$\hspace{10 mm}T(t)=T_e+(T_0-T_e)e^{-kt}$
After five minutes the temperature is $T(5)=20+80e^{-5k}=60^{\circ}C$;$\hspace{20 mm}e^{-5k}=0.5=\lambda$
Now, we want to know how many minutes more we'll have to wait for the water to reach $25^{\circ}C$
$T(t)=20+80e^{-kt}=25^{\circ}C$;$\hspace{20 mm}e^{-kt}=\frac{1}{16}=\lambda^{\frac{t}{5}}=0.5^{\frac{t}{5}}$
$ln(\frac{1}{16})=\frac{t}{5}ln(\frac{1}{2})$;$\hspace{20 mm}t=20 $ min
So, he will have to wait $15$ min more for the water to be $25^{\circ}C$
