Capacitance required to continue in a power lapse

[Jacob87411]

A capacitor is often used in electronics to keep energy flowing even if there is a momentary loss of power from the electric company. What capacitance would be required to a 150-watt televisiion (plugged into a standard 120 volt ac outlet) to protect it from a .1 second lapse in power?

I have done fine with every other question in this section but this one doesn't seem to be going as well. I am just curious how to begin. I know..

C=E0 x A/d (E0=permitivity of free space, 8.85x10^-12)
The time = .1 s

I just don't see how it all goes together

 

[Curious3141]

The energy released from a discharging capacitor that starts out with voltage V between its plates and continues to full discharge is given by : $$E = \frac{1}{2}CV^2$$. This is the total energy released by the discharging capacitor. The energy needed to tide the TV over the lapse time is given by $$E = Pt$$ where P is the power and t is the time. Equate the two and figure out C.

 

[Jacob87411]

Thanks that makes a lot of sense