A body cools from ${70}^{\circ }\text{c}$  to  ${50}^{\circ }\text{c}$ in 6 minutes. If temperature of surroundings is  ${30}^{\circ }\text{c}$.  What is the temperature after 12 minutes?

$\begin{array}{rcl} & & \begin{array}{l}{\text{initial temperature of the body is θ}}_{\text{1}}{\text{=70}}^{\text{o}}\text{c\hspace{0.17em} after 6 minutes}\\ {\text{temperature of body is θ}}_{\text{2}}{\text{=50}}^{\text{o}}{\text{c; surrounding temperature is \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}θ}}_{\text{0}}{\text{=30}}^{\text{o}}{\text{c;\hspace{0.17em}final temperature of body is\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}θ}}_{ }\text{=?}\\ \\ {\text{time t}}_{\text{1}}\text{=6 minutes}\\ {\text{time t}}_{\text{2}}\text{=12\hspace{0.17em}minutes}\\ from newton law of cooling\end{array}\\ & & \frac{\text{dθ}}{\text{dt}}\text{=-k}\left(\frac{{\text{θ}}_{\text{1}}{\text{+θ}}_{\text{2}}}{\text{2}}{\text{-θ}}_{\text{0}}\right)\text{---(1)}\\ & & \mathrm{substitute} \mathrm{given} \mathrm{values} \mathrm{in} \mathrm{eqn}\left(1\right)\\ & & \\ & & \frac{{\text{θ}}_{\text{1}}{\text{-θ}}_{\text{2}}}{time}\text{=}-k\left(\frac{{\text{θ}}_{\text{1}}{\text{+θ}}_{\text{2}}}{\text{2}}{\text{-θ}}_{\text{0}}\right)\\ \frac{70-50}{6}& =& -k\left(\frac{70+50}{\text{2}}\text{-30}\right)\\ \frac{20}{6}& =& -k\left(\text{60-30}\right)\\ k& =& -\frac{1}{9}---\left(2\right)\\ & & \mathrm{substitute} \mathrm{second} \mathrm{part} \mathrm{of} \mathrm{question}, \mathrm{given} \mathrm{values} \mathrm{and} \mathrm{eqn}\left(2\right) \mathrm{in} \mathrm{eqn}\left(1\right)\\ \frac{50-\theta }{12}& =& -\left(-\frac{1}{9}\right)\left(\frac{50+\theta }{\text{2}}\text{-30}\right)\\ & \Rightarrow & \theta ={34}^{0}C\end{array}$