Answer
$\frac{dT}{dt} = 0.05(70-T)$, or, equivalently, $\frac{dT}{dt} = -0.05(T-70)$.
Work Step by Step
Newton's Law of Cooling states that the temperature of an object changes at a rate proportional to the difference between the temperature of the object and the temperature of its surroundings (the ambient air temperature in most cases). Suppose that the ambient air temperature is $70^{0}F$ and that the rate constant is $0.05(min)^{-1}$. Write a differential equation for the temperature of the object at any time. Note that the differential equation is the same whether the temperature of the object is above or below the ambient temperature.
$\textit{Solution}$
Let $T$ denote the temp. of the object.
Temp. of ambient, $T_{a}$: $70^{0}F.$
Rate constant: $0.05(min)^{-1}$.
According to Newton's Law of Cooling:
$\frac{dT}{dt}$ ~ $(T_{a} - T)$, where $t$ denotes time.
That is,
$\frac{dT}{dt} = k(T_{a} -T)$, where $k$ is the constant of proportionality, i.e., the rate constant.
So, in this problem,
$\frac{dT}{dt} = 0.05(70-T)$.
