ordinary differential equations
A $200°F$ cup of tea is left in a $65°F$ room. At time $t=0$ the tea is cooling at $5°F$ per minute. Write an initial-value problem (differential equation with an initial condition) that models the temperature $T$ of the tea. Assume Newton's Law of Cooling. (Not required to solve the differential equation)
I'm struggling with this problem but I know newton's law of cooling is: $$\frac{dT}{dt}=-k(T-Ta)$$
I shall use $u$ for the temperature as indicated in the problem. Formulating the statement, we have $$\frac{{du}}{{dt}} = - h(u - {u_{{\rm{external}}}})$$ which could easily be solved (first order ODE) to yield $$u(t) = {u_{{\rm{external}}}} + A{e^{ - ht}}$$ where $A$ is a constant. To obtain this constant ($A$) and also the conduction constant ($h$), note that (we have chosen the unit of time to be hour) $${u_{{\rm{external}}}} = 10,u(0) = 70,u(2) = 40$$ so that $$\begin{array}{l}70 = 10 + A\\40 = 10 + A{e^{ - 2h}}\end{array}$$ which would yield $$A = 60,h = \frac{{\ln 2}}{2}$$ Summarizing, the answer could be stated as $$u(t) = 10 + 60{e^{ - \frac{{\ln 2}}{2}t}}$$ Below I have plotted the variation of the inside temperature with time from the moment of heating failure to 20 hours later 
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
$$t=\frac{1}{k}\left(\frac{190-30}{135-30}\right)=\frac{1}{\frac{1}{5}\ln\left(\frac{59}{54}\right)}\ln\left(\frac{32}{21}\right)$$ $$t=\frac{5\ln\left(\frac{32}{21}\right)}{\ln\left(\frac{59}{54}\right)}\approx \color{red}{\color{}{23.78\ minutes}}$$
Best Answer
$T_a$ in Newton's law is a temperature of room; $T_a = 65$. So, equation for modeling is $$ \frac{dT}{dt} = -k(T-65). $$ Now we should to determine $k$. "At time $t=0$ the tea is cooling at $5^\circ$F per minute". Ok, we have $$ \left.\frac{dT}{dt}\right|_{t=0} = -k(200-65)=135k = 5\Longrightarrow k =\frac{1}{27} $$ (time in minutes, of course).
UPDATE (Answer to OP question in comment)
No, $5^\circ$F per minute is a speed; it is derivative of $T$ (LHS of Newton's law). And by Newton's law, it's not a constant. You can solve equation above: $$ T = 65 + 135e^{-t/27}. $$ It's exponent, not a linear function (as you assummed).
And how you can see from it (or from law directly), if $T=65$, $dT/dt$ is zero; if the tea has cooled, it is no longer cool.