It's not about heat dissipation, it's the fact that if you assume the base resistors are unloaded (like they seem to), but then you load them significantly, the transistor won't be biased the way you planned for and might not work right.
Here's a usual way to bias a common-emitter amp which takes base current properly into consideration:
0. The power supply voltage is VCC.
1. Choose the collector voltage VC (typically VCC/2) and current IC (your choice).
2. The collector resistor is therefore (VCC - VC)/IC.
3. If you will place a capacitor on the emitter, choose the emitter resistor to drop 5% to 10% of VCC, at the same current as the collector resistor. Otherwise, select a gain A for your amplifier, and A = RC/RE (so RE = RC/A).
4. Calculate the base current IB, which is IC/hFE. hFE is the DC current gain of the transistor, found in the datasheet.
5. Calculate the base voltage VB, which is VBE (about 0.65V) + VE.
6. The lower base resistor has VB across it and should have about 5 IB through it, so RBL = VB/(5IB).
7. The upper base resistor has (VCC-VB) across it and should have about 6 IB through it (one more because it supplies the actual base current), so RBH = (VCC-VB)/(6IB)
There is no reason for a signal amplifier like this to have any substantial heat dissipation. Keep all the currents low and you don't have to care about heat.