Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 411: 43

$-3^{\circ}$

Here, $T-T_s=(T_0-T_s)e^{-kt}$ This implies that $39^{\circ}-T_s=(46^{\circ}-T_s)e^{-10k}$ and $33^{\circ}-T_s=(46^{\circ}-T_s)e^{-20k}$ This implies that $\dfrac{39^{\circ}-T_s}{46^{\circ}-T_s}=(\dfrac{39^{\circ}-T_s}{46^{\circ}-T_s})^2$ Thus, $T_s=-3^{\circ}$