Physical explanation of Joule heating The heat $Q$ generated in a wire, for a current $I$ flowing through a wire of a given resistance $R$, for a time $t$ is given by $Q=\mathscr{k}I^2Rt$ where $\mathscr{k}$ is the proportionality constant. For a given wire the resistance R is fixed. Is it possible to explain physically why $Q$ is proportional to the square of $I$?
 A: The pd, V, across the wire tells you how much energy is transferred to thermal in the wire per coulomb flowing. But the current, I tells you the rate of flow of coulombs. So VI tells you the energy transfer per second. Now here's the key thing: in a metal wire at constant temperature, I is proportional to V, that is $V=IR$ in which R is a constant. So we have:$$\text{energy transfer per second}=I\ \times IR =I^2R.$$ But you might object: the wire gets hot! In that case R isn't constant (it increases)! However if the wire is made of an alloy (as most heating 'elements' are), then R doesn't change much with temperature.
A: Joule heating occurs when the electrons carrying current in a wire lose energy to the metal atoms of the lattice. If there is a current $I$ being driven by a potential difference $V$,  that power is $P=VI$.
For a linear resistor, by Ohms law,  $I=V/R$ captures how much of the electrical energy is "lost" to heating, through a macroscopic constant $R$. Plug them together and $P=I^2R$. Maybe the water analogy helps, more current means more "turbulence" , i.e. resistance to flow. But the square just falls out of the math.
