Electrical – Energy stored in capacitors connected in series

Solve the problem by first assuming that the capacitors are connected through a resistor with some resistance, say \$R\$ ohms. As long as there is any difference between the voltages across the capacitors, current will flow through that resistor and thus transfer charge from the first capacitor into the second capacitor. So the first capacitor discharges slowly (exponential decay!) into the second capacitor and the current ceases at \$t = \infty\$ when the two capacitors each have charge \$Q/2\$, and thus voltage \$V/2\$. Half the energy stored in the first capacitor is lost since when the discharge of one into the other is complete, they each have one-fourth of the initial energy. Where did this energy go? Will the answer change if we use a different resistor in this experiment?

After this, think about what happens if \$R\$ is infinitesimally small.

The 5-second overload rating tells you a couple of things. One, as you note, is an implied "total Joule" rating, which says something about the thermal characteristics of the resistor assembly overall.

The other is that the manufacturer expects that the resistance wire can handle about 700 mA for at least that amount of time. You're asking it to briefly carry twice that current, which may or may not push the resistance wire into a nonlinear region, which could lead to localized failures. With a time constant of 1.65 seconds, it takes about 1.2 seconds for the current to drop from 1.46 A to 700 mA.

I would expect that you'll be fine, but it would be worthwhile to do some tests of your own, running short pulses of current in the range of 1A - 2A through the resistor and monitoring for anomalous resistance changes (i.e., excessive voltage drop).

Or you could just ask the manufacturer directly ...