By newton's law of cooling (with approximation), 
$\frac{\mathrm{\Delta }T}{\mathrm{\Delta }t}=-C\left(T_{\mathrm{ag}}-T_{\mathrm{s}}\right)$

$$\begin{gathered} 1^{\mathrm{st} } \\ \frac{-10 °\mathrm{C}}{5 \min }=-C\left(70 °\mathrm{C}-25 °\mathrm{C}\right) \end{gathered}$$

$\Rightarrow C=\frac{2}{45}{ \min }^{-1}$

$$\begin{gathered} 2^{\mathrm{nd} } \\ \frac{T-65}{5 \min }=-C\left(\frac{T+65}{2}-25\right)=-\left(\frac{2}{45}\right)\left(\frac{T+15}{2}\right) \end{gathered}$$

$\Rightarrow 9( T-65)=-(T+15)$
$\Rightarrow 10 \mathrm{T}=570$
$\Rightarrow T=57 °\mathrm{C}$