Heating and cooling based on ordinary differential equation

Question

The room temperature in your office is $70 \,°\rm F$. Experience has taught you that the temperature of a cup of coffee will drop from $120 \,°\rm F$ to $100 \,°\rm F$ in $10$ minutes. What should be the temperature of your cup of coffee if you want it to take $20$ minutes before it drops to $100 \,°\rm F$?

Answer 1

The Newton cooling law establishes

$$ T(t) = T_s + (T_i - T_s) e^{-K t} $$

where

$$ \begin{array}{rcl} T_s &=& \mbox{surrounding temperature}\\ T_i &=& \mbox{initial temperature}\\ K &=& \mbox{cooling constant}\\ t &=& \mbox{time} \end{array} $$

so we have

$$ 100^{\circ} = 70^{\circ}+\left(120^{\circ}-70^{\circ}\right)e^{-10K}\Rightarrow K $$

after computing $K$ we have

$$ 100^{\circ} = \left(T_i-70^{\circ}\right)e^{-20 K} $$

and then we can compute $T_i$