3.19 Solution of the cooling problem

The differential equation is

$${{dS}\over{dx}}=-kS,$$

where the negative sign (-) indicates that the coffee is cooling. At $x=0$, we have $S(0)=100C-20C=80C$, since water boils at $100C.$ Now the solution of the differential equation is

$$S(x)=S(0)e^{-kx},$$

so $S(x)=80e^{-0.01x}$. We look for $x$ such that $S(x)=60-20=40$, so

$$40=80e^{-0.01x},$$

$$x=100\log 2=69.31seconds;$$

so after a minute or so, the coffee is drinkable.