A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s is _________ J.

We start with the basic electrical power-energy relations. The instantaneous electrical power converted to heat in a resistor is given by the formula $$P = I^{2} R$$, where $$P$$ is power in watts, $$I$$ is current in amperes and $$R$$ is resistance in ohms. The thermal energy (heat) produced in a time interval $$t$$ is the product of power and time, i.e. $$E = P\,t$$.

From the data provided for the first situation: energy dissipated $$E_{1} = 192\ \text{J}$$ in time $$t_{1} = 1\ \text{s}$$ with current $$I_{1} = 4\ \text{A}$$.

Using $$E = P\,t$$, we obtain the initial power:

$$P_{1} = \frac{E_{1}}{t_{1}} = \frac{192\ \text{J}}{1\ \text{s}} = 192\ \text{W}.$$

Now we employ $$P = I^{2} R$$ to determine the resistance of the resistor. Substituting $$P_{1} = 192\ \text{W}$$ and $$I_{1} = 4\ \text{A}$$:

$$192 = (4)^{2} R.$$

Since $$(4)^{2} = 16$$, we get

$$192 = 16\,R.$$

Solving for $$R$$:

$$R = \frac{192}{16} = 12\ \Omega.$$

Next, the current is doubled, so the new current is

$$I_{2} = 2\,I_{1} = 2 \times 4\ \text{A} = 8\ \text{A}.$$

With the same resistor, the new power $$P_{2}$$ is again obtained from $$P = I^{2} R$$:

$$P_{2} = I_{2}^{2} R = (8)^{2} \times 12.$$

Calculating $$(8)^{2} = 64$$, we have

$$P_{2} = 64 \times 12 = 768\ \text{W}.$$

This power acts for a new time interval of $$t_{2} = 5\ \text{s}$$. The thermal energy produced in this interval is therefore

$$E_{2} = P_{2}\,t_{2} = 768\ \text{W} \times 5\ \text{s}.$$

Multiplying, we obtain

$$E_{2} = 768 \times 5 = 3840\ \text{J}.$$

So, the answer is $$3840\ \text{J}$$.