Well, back when I was in grad school, and we had to make our water by hand, I could explain this. Let me see if I can now.

You can think of matter as a form of energy, and energy as a form of matter. The relationship is specified by Einstein’s famous E= m c^2 (that is "m c-squared). So when energy is radiated out from the black hole, the black hole has lost the amount of mass specified in that equation.

I have a question about the charge of a black hole. For some reason, I was pondering this very question a few days ago, and how convenient for someone else to ask it. First, to me the “existence” of virtual photons is somewhat nebulous. The arise in what are called “perturbation expansions” in quantum field theory. Basically, you start with an exact solution, and consider small deviations, or perturbations, from that. So they definitely exist as mathematical entities, but I really don’t know what it means to exist.

Consider, though, that every electron is emanating an electric (and gravitational) field as it crosses the event horizon. Any photon carrying the electrostatic field, that is leaving the electron, as it crosses the event horizon, is infinitely red shifted. But, a static field has no time dependence anyway, so what does the red shift mean? Is the electric field carried by photons that were emanated when the electron crossed the event horizon really the carriers of the electric field? Or are virtual photons crossing the event horizon and propagating to infinity (and beyond! - sorry Toy Story humor)???