In the given circuit diagram (figure), switch $S_W$  is shifted from position 1 to position 2. Then  

Initially, when the switch is closed on position 1, the capacitor C is connected in series with batteries $E_{1}and{E}_2$. From KVL, we have

$\frac{Q_i}{C}-E_2+E_1=0$

or $Q_i=(E_2-E_1)C\mathrm{..........}\left(i\right)$

Depending upon the sign of $(E_2-E_1),$charge $Q_i$on the left plate may be positive if  $(E_2>E_1) ,ornegative (if{E}_2<E_1);$charge on right plate would be equal and positive. When the switch is moved to position 2, the left plate (earlier having charge $+Q_i$ will now have charge 

$Q_f=-E_{1}C\mathrm{.....................}\left(ii\right)$

The net charge flow through the circuit is

$\Delta Q=Q_f-Q_1=[-E_1(E_2-E_1)]C=-E_{2}C$

We can say that a net positive charge equal to $E_{2}C$ is pulled by the battery of emf $E_1$ from the left plate of the capacitor, which flows through battery $E_1$ and is transferred to the right plate of the capacitor. Work done by battery $E_1$ in the process of charge transfer is

$\Delta W=E_1{E}_{2}C\mathrm{....................}\left(iii\right)$

A part of this work changes the energy of the capacitor

$\Delta W_c=\frac{Q_f^2}{2C}-\frac{Q_i^2}{2c}=\frac{1}{2}{E}_1^{2}C-\frac{1}{2}{(E_2-E_1)}^{2}C$

$=\frac{1}{2}(2E_1{E}_2-E_2^2)C$

And the remaining part is lost as Joule heat:

$H=\Delta W-\Delta W_C=\frac{1}{2}{E}_2^{2}C$