Joule heating due to the (slow) electron drift velocity? I understand the concept of why the signal speed is higher than the electron drift velocity, but I can't understand the concept of joule heating. If electrons move slow then how do they produce a lot of  heat when they hit the nucleus.  Besides my friend once told me that the drift velocity is the net movement and electrons move fast in all directions, if that is the case why do they move like that?
 A: The absolute velocity of the electrons actually doesn't matter for joule heating.  Think about it this way, if there is no current flowing there wouldn't be any joule heating.  So, even if electrons are moving quickly and randomly when no current is flowing, we know no joule heating would occur and that joule heating is really about the net change in effect caused by the current.  That is, the base electron velocity doesn't have an effect.  All that matters is the $\Delta V$ over the base electron velocity which is given by the drift velocity.
Joule heating is really about electrical energy lost to heat due to resistance.  Even if the average drift velocity of an electron is tiny, there are so many electrons moving that the tiny energy loss to heat for each electron adds up.  As you know, current is the result of a huge number of moving electrons.  It's a numbers game.  The more electrons losing a tiny bit of energy there are, the more total heat is generated.
Via Ohm's Law you can see that $P_{ower} = I^2 R$ so it's no wonder that heat generation is proportional to $I^2 R$.
Also, in your question you mentioned an electron bumping into a nucleus.  That is not what's happening.  Electrons are colliding with the electron cloud of atoms and via electromagnetic repulsion are pushing the whole atom a bit, increasing its kinetic energy.  It's just free electrons interacting with bound electrons.
