Let's consider the power $$P$$ dissipated by a single heater filament with resistance $$R$$ when connected to a voltage $$V$$.

The power dissipated by the heater filament is given by:

$$P = \frac{V^2}{R}$$

When the two identical heater filaments are connected in parallel, the equivalent resistance $$R_P$$ is given by:

$$\frac{1}{R_P} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \Rightarrow R_P = \frac{R}{2}$$

The total power dissipated $$P_P$$ by the two filaments connected in parallel is:

$$P_P = \frac{V^2}{R_P} = \frac{V^2}{\frac{R}{2}} = 2V^2 \cdot \frac{1}{R} = 2P$$

When the two identical heater filaments are connected in series, the equivalent resistance $$R_S$$ is given by:

$$R_S = R + R = 2R$$

The total power dissipated $$P_S$$ by the two filaments connected in series is:

$$P_S = \frac{V^2}{R_S} = \frac{V^2}{2R} = \frac{1}{2}V^2 \cdot \frac{1}{R} = \frac{1}{2}P$$

Now, let's find the ratio of the heat produced in the same time for parallel to series connection:

$$\frac{P_P}{P_S} = \frac{2P}{\frac{1}{2}P} = \frac{2P \times 2}{P} = \frac{4P}{P} = 4 : 1$$