A capacitance C charged to a potential difference V is discharged...

The initial energy stored in the capacitor is $U_i = \dfrac{1}{2}CV^2$.

During the discharging of a capacitor, the voltage across it at any time $t$ is given by $V(t) = V e^{-t/RC}$.

After one time constant $t = RC$, the voltage becomes $V(RC) = V e^{-1} = \dfrac{V}{e}$.

The energy stored in the capacitor at this time is $U_f = \dfrac{1}{2}C \left(\dfrac{V}{e}\right)^2 =$\dfrac{1}{2}$CV^2 \left(\dfrac{1}{e^2}\right).<br><br>By conservation of energy, the heat dissipated across the resistor is equal to the loss in the electrical potential energy of the capacitor.<br><br>$H = U_i - U_f = $\dfrac{1}{2}$CV^2 - $\dfrac{1}{2}$CV^2 \left(\dfrac{1}{e^2}\right) = \dfrac{1}{2}CV^2 \left(1 - \dfrac{1}{e^2}\right)$.$