Yes, if a black body absorbs all the radiation incident upon it and emits radiation according to its temperature, it will re-emit all the heat supplied to it through radiation and eventually reach a thermal equilibrium where the amount of absorbed and emitted energy is equal. In this equilibrium state, the black body will maintain a constant temperature.
A black body is an idealized concept in physics that absorbs all incident radiation, regardless of the wavelength or frequency. It is also a perfect emitter, meaning it radiates energy at all wavelengths according to its temperature. The relationship between the temperature of a black body and the amount of radiation it emits is described by Planck's law and the Stefan-Boltzmann law.
When a black body absorbs radiation, it converts the energy of the absorbed photons into internal energy, raising its temperature. Subsequently, it emits radiation based on its new temperature. The rate of energy absorption and emission will balance out when the black body reaches thermal equilibrium.
If the heat supplied to a black body is solely through radiation, and the black body absorbs all the incident radiation, then it will indeed re-emit all the heat supplied. The temperature of the black body will stabilize once the absorbed energy matches the emitted energy, resulting in a steady temperature.
It's worth noting that in practical situations, real objects may not behave as perfect black bodies. They may have varying degrees of reflectivity or emissivity at different wavelengths. Therefore, the energy absorption and emission characteristics might differ. However, the concept of thermal equilibrium and the exchange of energy through radiation still hold true.