Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 411: 41

a) 17.5 minutes b) $13.26$ minutes

Here, $T-T_s=(T_0-T_s)e^{-kt}$ This implies that $60-20=70e^{-10k} \implies k \approx 0.05596$ a) We have $T-T_s=(T_0-T_s)e^{-kt}$ $(35-20)=70e^{-0.05596t} \approx 27.5 min$. This is the total time and it will take (27.5-10)=17.5 minutes longer to reach the temperature $35^{\circ}$ b) $T-T_s=(T_0-T_s)e^{-kt}$ This implies that $35+15=105e^{-0.05596 t} \implies t \approx 13.26$ minutes