This Resistance is soooo Hot.....

Firstly we will calculate charge on the capacitor when it is half charged i.e

$Q=\frac { CV }{ 2 }$ .

Then the work done by battery is :

${ W }_{ b }=V(\frac { CV }{ 2 } )=\frac { 1 }{ 2 } C{ V }^{ 2 }$

Energy in the capacitor : $U=\frac { { Q }^{ 2 } }{ 2C } =\frac { { (\frac { CV }{ 2 } ) }^{ 2 } }{ 2C } =\frac { 1 }{ 8 } C{ V }^{ 2 }$

Net heat released :

$H={W}_{b}-U=\frac { 3 }{ 8 } C{ V }^{ 2 }$

Now we know Current through the internal resistance and the external resistance is same hence power is directly proportional to their resistances. So we write :

${ P }_{ r }=kr,{ P }_{ R }=kR\quad and\quad { P }_{ Total }={ P }_{ r }+{ P }_{ R }=k(r+R)\\ \Rightarrow k=\frac { { P }_{ Total } }{ r+R } \quad \Rightarrow { P }_{ r }=\frac { { P }_{ Total }r }{ r+R }$

Using $R=2r$ we get Heat liberated in $r=\frac{1}{3}$ of total heat.

Heat liberated in $r = \frac { 1 }{ 8 } C{ V }^{ 2 }$

Put the values and get $\boxed{H=2 Joule}$