Energy dissipation in RC circuit given AC sinewave input
First you decide what the output load will be - if it's open circuit i.e. a simple RC low pass filter you need to calculate the current flow thru the series R and C using an AC version of ohms law: -

I = V/R becomes I = V/Z where Z, in your example is 

\$R +\dfrac{1}{j\omega C}\$.

Once you have that current then the power dissipation in the resistor is current squared multiplied by resistance.

You could also calculate Z (without using complex numbers) using: -

\$Z_{IN} = \sqrt{R^2 + X_C^2}\$ where \$X_C\$ is capacitive reactance.

If there is an output load resistor (\$R_L\$) then input impedance changes to: -

\$Z_{IN} = R + \dfrac{\frac{R_L}{j\omega C}}{R_L + \dfrac{1}{j\omega C}}\$

or

\$Z_{IN} = R + \dfrac{R_L}{R_L + j\omega CR_L}\$

You can take this further to but it's become maths rather than EE now.