electric-currentpowerthermodynamicsvoltage
120 volts x 20 amps = 2,400 Watts
However, if I increased the voltage and lowered the current, you can also use a smaller wire size (more inexpensive), also have less heat and achieve the same watt Power.
1,000 volts x 2.4 amps = 2,400 Watts
More the power, more the heat generated.
$$P(\textrm{Power})=V\cdot I$$ $$H(\textrm{in joules})\propto V\cdot I\cdot t\,(\textrm{in s})$$ $$ \therefore H(\textrm{in joules})\propto P\cdot t (\textrm{in s})$$
The 11.1 volt gloves does not give greater heat just because it has more Potential difference, but because $\textrm{Power}$ is greater in it.
Your initial circuit is like this:
So you get 2A flowing and a power of 20W dissipated in the resistor.
Then you double the voltage and the resistance:
The two batteries add up to single source of 20V. The two resistors add up to a total resistance of 10$\Omega$. So, as you correctly state, the current is the same as before (2A). Therefore the power is now 40W
But this power is shared between the two resistors: 20W each, exactly as before.
You might also note that the potential at the point between the resistors is 10V, so each resistor has 10V across it, exactly as before.
Really all you've done is doubled up the circuit so that you have twice of what you had before.
Best Answer
The term you're looking for is Joule Heating.
The heat dissipated in a conductor is proportional to $I^2 R$ where $I$ is the current and $R$ is the resistance. Heating happens when moving charge (electrons) collide with the molecules in the conductor inelastically (that is, they transfer some kinetic energy to the molecule).
Remember, current is defined as the amount of charge passing a given spot per unit of time. It makes sense then that the more charge passing a spot, the more collisions occur and therefor the more heat is dissipated.
This is why power lines use high voltages for transmission so that they can provide more power with less current.