A body in a room cools from ${85}^{\circ }\text{c}$  to  ${80}^{\circ }\text{c}$ in 5 minutes. The time taken to cool from ${80}^{\circ }\text{c}$  to  ${75}^{\circ }\text{c}$ is (room temperature is 24⁰C

$\begin{array}{l}{\text{θ}}_{\text{1}}={85}^{\circ }{\text{c\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}θ}}_2={80}^{\circ }\text{c}\\ {\text{θ}}_2={80}^{\circ }{\text{c\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}θ}}_3={75}^{\circ }\text{c}\\ {\text{t}}_{\text{1}}\text{=5 minutes}\\ {\text{t}}_{\text{2}}=?\\ {\text{θ}}_0=\text{constant=temperature of surrounding}\end{array}$

$\begin{array}{rcl} & & from newton law of cooling\\ & & \frac{\text{dθ}}{{\text{dt}}_{\text{1}}}\text{=-k}\left(\frac{{\text{θ}}_{\text{1}}{\text{+θ}}_{\text{2}}}{\text{2}}{\text{-θ}}_{\text{0}}\right)\text{---(1)}\\ & & \text{substitute given values in above equation}\\ \frac{{\text{θ}}_{\text{1}}{\text{-θ}}_{\text{2}}}{5}& =& -k\left(\frac{\text{85+80}}{\text{2}}{\text{-θ}}_{\text{0}}\right)\\ \frac{\text{85-80}}{5}& =& -k\left(\frac{\text{82.5}}{\text{2}}{\text{-θ}}_{\text{0}}\right)\\ \frac{5}{5}& =& -k\left(\text{41.25-24}\right)---\left(2\right)\left(\because Room temperature is 24°C\right)\\ & & second part of question, given values substitute in eqn\left(1\right)\\ \frac{\text{80-75}}{t_2}& =& -k\left(\frac{\text{80+75}}{\text{2}}{\text{-θ}}_{\text{0}}\right)\\ \frac{\text{5}}{t_2}& =& -k\left(\text{38.75-24}\right)---\left(3\right)\\ & & eqn\left(2\right) ÷eqn\left(3\right)\\ & & t_2= 5.47min\end{array}$

${\text{t}}_{\text{2}}$is more than 5 minutes