Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 411: 42

$5^{\circ}$

Here, $T-T_s=(T_0-T_s)e^{-kt}$ This implies that $35^{\circ}-65^{\circ}=(T_0-65^{\circ})e^{-10k}$ and $50^{\circ}-65^{\circ}=(T_0-65^{\circ})e^{-20k} $ This implies that $(T_0-65^{\circ})e^{-10k}=2(T_0-65^{\circ})e^{-20k}$ and $k=\dfrac{\ln 2}{10}$ Thus, $T_0=65^{\circ}-30^{\circ}(e^{\ln 2})=5^{\circ}$