Simple Capacitor Wattage Ratings
The basic formula for a capacitor is: -

Q = \$C\cdot V\$ and this can be differentiated to \$\dfrac{dQ}{dt} = C\cdot\dfrac{dV}{dt}\$

rate of change of charge, \$\dfrac{dQ}{dt}\$, is current therefore: -

I = \$C\cdot\dfrac{dV}{dt}\$

Power\$^1\$ is I (current) x V (volts) therefore power taken by a capacitor is \$C\cdot V\dfrac{dV}{dt}\$.

But V multiplied by its differential will always average to zero therefore a capacitor doesn't get warm theoretically. However, due to losses in the capacitor (seen as either an equivalent series resistor or parallel resistor), it will get warm if pushed hard enough. 

This is usually due to dielectric losses or the effective series resistance (ESR) of the capacitor.

If the current is 25mA and the power source is 50 or 60Hz then the capacitor is unlikely to get measurably warm.

\$^1\$ If you integrate this formula with respect to time, power becomes energy and you get the hopefully recognizable formula seen below: -

\$E = \dfrac{C\cdot V^2}{2}\$