Solution:

Using Newton's law of cooling:

Given:
- Initial cooling: \( T_1 = 80^\circ \text{C} \) to \( T_2 = 79.9^\circ \text{C} \) in 5 min, surroundings \( T_s = 20^\circ \text{C} \).
- New cooling required: from \( 70^\circ \text{C} \) to \( 69.9^\circ \text{C} \).

For both cases, Newton's law states:
\[
\frac{\Delta T}{\Delta t} \propto (T - T_s)
\]

Since the temperature difference is small in both cases, we assume:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{T_1 - T_s}{T_3 - T_s}
\]

So:
\[
\frac{5}{\Delta t_2} = \frac{80 - 20}{70 - 20} \Rightarrow \Delta t_2 = 5 \times \frac{70 - 20}{80 - 20} = 5 \times \frac{50}{60} = 6 \text{ min}
\]