Ok.
Let's imagine a 6V relay, with a 75Ω coil, that starts to operate at 4.8V. So you need at least 64mA of current running through the coil.
Taking β=100, you need at least 640 uA at the base. => (Vc - 0.7)/5000 >= 640uA
Vc >= 3.9V
How long does it take to the cap discharge from 6V to 3.9V?
3.9 = 6*e^-t/5
-t/5 = ln(0.65)
t =~ 2.15 s, not 15 s.
You would need a RC constant of 35 to get about 15 seconds. For a 5k resistor, you would need a 7 mF (70,000 µF) capacitor.
The best thing is to use a MOSFET, with despresible gate current, you can choose the RC constant that would fit for you.
And yes, you can use the power from the waste disposer, but that would be a transformless power supply (not good).