A body initially at 60°C cools to 50°C in 10 min. What will be its temperature at the end of next 10 min. if the room temperature is 25°C. Assume Newton's law of cooling

According to Newton s law of cooling $\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}\left(\mathrm{T}-{\mathrm{T}}_0\right)$
where T = temperature of body and T_o = temperature of surroundings
Now   ${\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2} \frac{\mathrm{dT}}{\left(\mathrm{T}-{\mathrm{T}}_0\right)}=-\mathrm{k}{\int }_0^{\mathrm{t}} \mathrm{dt}$
      ${\left[{\log }_{\mathrm{e}}\left(\mathrm{T}-{\mathrm{T}}_0\right)\right]}_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}=-\mathrm{k}[\mathrm{t}{]}_0^{\mathrm{t}}$
or   ${\log }_{\mathrm{e}} \left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_0}{{\mathrm{T}}_1-{\mathrm{T}}_0}\right)=-\mathrm{kt}$
or    ${\log }_{\mathrm{e}}\left(\frac{60-25}{50-25}\right)=-\mathrm{k}\times (10\times 60) ...\left(1\right)$
Let T' be the temperature at the end of 10 min. In this case,
     ${\log }_{\mathrm{e}}\left(\frac{50-25}{{\mathrm{T}}^{\mathrm{\prime }}-25}\right)=-\mathrm{k}\times (10\times 60) ...\left(2\right)$
Solving eqs. (1) and (2), we get
    ${\mathrm{T}}^{\mathrm{\prime }}=42\cdot {85}^{\circ }\mathrm{C}$