A body cools from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 min . In the next time of ' $t$ ' min, the body continues to cool from $50^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The total time taken by the body to cool from $80^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$ is [The temperature of the surroundings is $20^{\circ} \mathrm{C}$.]

The temperature of the surroundings is $T_s = 20^{\circ}\mathrm{C}$. Newton's Law of Cooling in approximate finite difference form states:

$\frac{T_1 - T_2}{\Delta t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$

First, determine the cooling constant $K$ using the cooling from $80^{\circ}\mathrm{C}$ to $50^{\circ}\mathrm{C}$ in $5\,\mathrm{min}$:

$\frac{80 - 50}{5} = K \left( \frac{80 + 50}{2} - 20 \right) \Rightarrow 6 = K (45) \Rightarrow K = \frac{2}{15}\,\mathrm{min}^{-1}$

Now, apply this $K$ to find the time $t$ to cool from $50^{\circ}\mathrm{C}$ to $30^{\circ}\mathrm{C}$:

$\frac{50 - 30}{t} = \frac{2}{15} \left( \frac{50 + 30}{2} - 20 \right) \Rightarrow \frac{20}{t} = \frac{2}{15} (20) \Rightarrow t = 7.5\,\mathrm{min}$

The total time to cool from $80^{\circ}\mathrm{C}$ to $30^{\circ}\mathrm{C}$ is $5 + 7.5 = 12.5\,\mathrm{min}$.