A black hole doesn't need mass per se. Its a hole in the fabric of space and time. But it does release energy in the form of Hawking Radiation. Which (According to people much smarter than I) is a form of thermal radiation, it loses mass until it does BURST in a gamma ray explosion. or maybe it just goes POOF and gone.

Check it:

http://en.wikipedia.org/wiki/Hawking_radiation#Black_ho...
(WARNING you may need a doctorates in physics)

ds^2 = -\left(1-{2M\over r}\right)dt^2 + {1\over 1-{2M\over r}} dr^2 + r^2 d\Omega^2 \;

The black hole is the background spacetime for a quantum field theory.

The field theory is defined by a local path integral, so if the boundary conditions at the horizon are determined, the state of the field outside will be specified. To find the appropriate boundary conditions, consider a stationary observer just outside the horizon at position r = 2M + u2 / 2M. The local metric to lowest order is:

ds^2 = - {u^2\over 4M^2} dt^2 + 4 du^2 + dX_\perp^2 = - \rho^2 d\tau^2 + d\rho^2 + dX_\perp^2 \;

which is Rindler in terms of τ = t / 4M and ρ = 2u. The metric describes a frame that is accelerating to keep from falling into the black hole. The local acceleration diverges as u\rightarrow 0.

The horizon is not a special boundary, and objects can fall in. So the local observer should feel accelerated in ordinary Minkowski space by the principle of equivalence. The near-horizon observer must see the field excited at a local inverse temperature

\beta(u)=2\pi \rho = (4\pi) u = 4\pi \sqrt{2M(r-2M)} \;,

the Unruh effect.

The gravitational redshift is by the square root of the time component of the metric. So for the field theory state to consistently extend, there must be a thermal background everywhere with the local temperature redshift-matched to the near horizon temperature:

\beta(r') = 4\pi \sqrt{2M(r-2M)} \sqrt{1-{2M\over r'} \over 1-{2M\over r}} \;

The inverse temperature redshifted to r' at infinity is

\beta(\infty) = (4\pi)\sqrt{2Mr} \;

and r is the near-horizon position, near 2M, so this is really:

\beta = 8 \pi M \;

So a field theory defined on a black hole background is in a thermal state whose temperature at infinity is:

T_H = {1 \over 8 \pi M} \;

which can be expressed more cleanly in terms of the surface gravity of the black hole, the parameter that determines the acceleration of a near-horizon observer.

T_H = \frac{\kappa}{2 \pi} \;

in natural units with G, c, \hbar and k equal to 1, and where κ is the surface gravity of the horizon. So a black hole can only be in equilibrium with a gas of radiation at a finite temperature. Since radiation incident on the black hole is absorbed, the black hole must emit an equal amount to maintain detailed balance. The black hole acts as a perfect blackbody radiating at this temperature.

In engineering units, the radiation from a Schwarzschild black hole is black-body radiation with temperature:

T = {\hbar \, c^3 \over 8 \pi G M k_b} \;\quad(\approx {1.227 \times 10^{23}\; kg \over M}\; K)\;

where \hbar is the reduced Planck constant, c is the speed of light, kb is the Boltzmann constant, G is the gravitational constant, and M is the mass of the black hole.

From the black hole temperature, it is straightforward to calculate the black hole entropy. The change in entropy when a quantity of heat dQ is added is:

dS = {dQ\over T} = 8\pi M dQ \;

the heat energy that enters serves increases the total mass:

dS = 8 \pi M dM = d(4 \pi M^2) \;.

The radius of a black hole is twice its mass in natural units, so the entropy of a black hole is proportional to its surface area:

S = \pi R^2 = {A \over 4} \;.