An object kept in a large room having air temperature of 25^ C...

Key Idea From Newton's law of cooling, the time taken $(t)$ by a body to cool from $T_1$ to $T_2$ when placed in a medium of temperature $T_0$ can be calculated from relation $$\frac{T_1-T_2}{t}=\frac{1}{K}\left(\frac{T_1+T_2}{2}-T_0\right)$$ When the object cool from $80^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 12 minutes, then from Newton's law of cooling, \begin{array}{r} \frac{80-70}{12}=\frac{1}{\mathrm{~K}}\left(\frac{80+70}{2}-25\right) \quad\left[\because \mathrm{T}_0=25^{\circ} \mathrm{C}\right] \\ \frac{5}{6}=\frac{1}{\mathrm{~K}} 50 \end{array} Similarly, when object cool from $70^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ we get $$\begin{gathered} \frac{70-60}{\mathrm{t}}=\frac{1}{\mathrm{~K}}\left(\frac{70+60}{2}-25\right) \\ \frac{10}{\mathrm{t}}=\frac{1}{\mathrm{~K}} 40 \end{gathered}$$ Divide Eq. (i) and (ii), we get $$\begin{gathered} \frac{5}{6} \times \frac{\mathrm{t}}{10}=\frac{50}{40} \\ \Rightarrow \quad \mathrm{t}=\frac{5}{4} \times 12=15 \text { minutes } \end{gathered}$$