Newton's law of cooling, soup Newton's law of cooling states that the temperature $T(t)$ of an object at time $t > 0$ changes at a rate proportional to the difference between the temperature of the object and the temperature $T_S$ of its surroundings, provided that this difference is not too large. That is, $T(t)$ satisfies
$\quad T'(t) = k (T(t) - T_s)$
where k is a constant.
Suppose that the temperature of a cup of soup obeys Newton's law of cooling. If the soup has a temperature of $\; 190^\circ\, F$ when served to a customer, and 5 minutes later has cooled to $\; 180^\circ\, F$ in a room at $\; 72^\circ\, F$, how much longer must it take the soup to reach a temperature of$ \; 135^\circ\, F$? 
Answer (1) in additional minutes = ?
If the same cup of $190^\circ\, F$ soup is instead placed into a freezer set at $30^\circ\, F,$ what is the time required for the soup to cool from $190^\circ\, F to 135^\circ\, F$ in this situation?
Answer (2) in minutes = ?
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After watching the tutorial on the Cooling Law, I tried to use formula:
$ T(t) = T_{surrounding} - Ce^{-kt}$ (general formula for cooling?)
$C = (T - T_s)$
$T(t) = 72 - 118e^{-kt}$ (for my case)
To find k:
$180 = 72 - (190-72) \cdot e^{-k \frac{5}{60}}$
$180 = 72 - 118e^{-k \frac{1}{12}}$
$ k = -12 \cdot ln(\frac{-54}{59})$
Then to get answer to (1):
$135 = 72 - 118e^{-kt}$
$ 135 = 72 - 118e^{12 \cdot ln(\frac{-54}{59}) \cdot t}$
$ t = \frac{ln(\frac{-63}{118})}{12 \cdot ln(\frac{-54}{59}) \cdot t} $
But that's not the correct answer!
I then tried a different formula to get answer to (1):
$T(t) = (T - T_s) e^{kt} + T_s$
$T(t) = 118e^{kt} + 72$
$180 = 118e^{k \cdot 5/60} + 72$
$k = 12 ln(\frac{54}{59})$
But even with this k value I get the wrong answer, why?
And yes, I absolutely have to assume that t should be in hours format.
 A: Notice, in newton's law of cooling, the temperature $T(t)$ is assumed to be decreasing with respect to the time hence, rate of change of temperature $T'(t)=\frac{dT}{dt}$ is taken as negative hence we have the following equation of cooling $$T'(t)=-k(T(t)-T_s)$$
$$\frac{dT}{dt}=-k(T-T_s) \implies \frac{dT}{(T-T_s)}=-kdt$$
$$\int \frac{dT}{(T-T_s)}=-k\int dt$$ $$\ln(T-T_s)=-kt+C$$$$\implies T=e^{-kt+C}+T_s\tag 1$$
Now, setting $T(t)=\text{initial temperature}=T_0$ at $t=0$, we get $$T_0=e^{0+C}+T_s\implies C=\ln(T_0-T_s)$$
Hence, setting the value of $C$ in (1), the equation of cooling is given as 
$$kt=\ln\left(\frac{T_0-T_s}{T-T_s}\right)\tag 2$$
$\color{blue}{\text{Given condition}}$: The temperature falls from $\color{red}{(T_0=\color{blue}{190^\circ\ F})\to (T=\color{blue}{180^\circ\ F})}$ in time $\color{red}{t=\color{blue}{5\ minutes}}$ at room/surrounding temperature $\color{red}{T_s=\color{blue}{72^\circ \ F}}$ hence setting all the values in (2), we get 
$$k(5)=\ln\left(\frac{190-72}{180-72}\right)\implies k=\color{blue}{\frac{1}{5}\ln\left(\frac{59}{54}\right)}$$
Now, the time $t$ required to fall temperature from $\color{red}{(T_0=\color{blue}{190^\circ\ F})\to (T=\color{blue}{135^\circ\ F})}$ hence setting all the values in (2), we get 
$$t=\frac{1}{k}\left(\frac{190-72}{135-72}\right)=\frac{1}{\frac{1}{5}\ln\left(\frac{59}{54}\right)}\ln\left(\frac{118}{63}\right)$$ $$t=\frac{5\ln\left(\frac{118}{63}\right)}{\ln\left(\frac{59}{54}\right)}\approx \color{red}{\color{}{35.43\ minutes}}$$
Now, the time $t$ required to fall temperature from $\color{red}{(T_0=\color{blue}{190^\circ\ F})\to (T=\color{blue}{135^\circ\ F})}$ at surrounding temperature $\color{red}{T_s=\color{blue}{30\circ \ F}}$ (placed into a freezer) hence setting all the values in (2), we get 
$$t=\frac{1}{k}\left(\frac{190-30}{135-30}\right)=\frac{1}{\frac{1}{5}\ln\left(\frac{59}{54}\right)}\ln\left(\frac{32}{21}\right)$$ $$t=\frac{5\ln\left(\frac{32}{21}\right)}{\ln\left(\frac{59}{54}\right)}\approx \color{red}{\color{}{23.78\ minutes}}$$
A: Firstly, it is better to set exponent  negative ( factor k positive) so  no error would occur in sign as happened here.
$ \quad T'(t) = -k (T(t) - T_s)$ 
After all, it is not heating..
Also
$$ (T(t)- T_s)  =  (T_{hot}-  T_s)  e^{- k t } $$  
$$ {- k\,t }= log_e \dfrac  { T(t)- T_s} { T_{hot}-  T_s}    $$  
which helps to calculate constant $k$
A: $\dot T=k(T_s-T)$
$\dot T=k(72-T)$
$\dot T+kT=72k$
Homogeneous solution
$T_h=Ce^{-kt}$
Particular solution
$T=72$
General solution, by superposition
$T=72+Ce^{-kt}$
$T_0=T(0)=190$
$190=72+Ce^{-k(0)}$, $C=118$
$180=72+118e^{-k(5)}$, $k=(1/5)ln(59/54)$
Now we can write the solution
$T=72+118e^{-(1/5)ln(59/54)t}$
And for $T=135$
$135=72+118e^{-(1/5)ln(59/54)t}$
We find $t=35.43$ min
Now, answer (1), the additional minutes are
$35.43 - 5.00 = 30.43$ min    
For answer (2):
$\dot T=k(30-T)$
$\dot T+kT=30k$
Homogeneous solution
$T_h=Ce^{-kt}$
Particular solution
$T=30$
General solution, by superposition
$T=30+Ce^{-kt}$
$T_0=T(0)=190$
$190=30+Ce^{-k(0)}$, $C=160$
$k$ is a constant already found: $k=(1/5)ln(59/54)$
Now we can write the solution
$T=30+160e^{-(1/5)ln(59/54)t}$
And for $T=135$
$135=30+160e^{-(1/5)ln(59/54)t}$
We find $t=23.78$ min  
