Electronic – the relation between the power dissipation in resistor and the temperature rise

The J is probably a typo, unit of power is watt and not joule. To calculate signal power: \$P=\frac{1}{T}\int\limits_0^T{u(t)i(t)\,dt}=\frac{1}{T}\int\limits_0^T{\frac{u^2(t)}{R}dt}=\frac{4}{TR}\int\limits_0^{T/4}{(10t)^2dt}=\frac{4\cdot 100}{4\cdot 100}\int\limits_0^{1}{{t^2}\,dt}=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}-0=\frac{1}{3}\,\mathrm{W}\$

Basically, think of current as a flow of water in a pipe, current is the amount of water being moved, more current means more 'water' flowing.

Voltage is the force or pressure pushing the 'water' along, more voltage means the 'water' is pushed harder and consequently moves faster through the pipe (that's why cranking up the voltage usually results in more current as well).

Now power is the amount of effort required to do all this, think about how much effort it takes to suck water up a straw... now try with a 50mm drainage pipe, you'll be lucky to move anything at all, it takes more effort to move more water even if the pressure is the same (head height is one way of measuring water pressure).

And for completeness, resistance would be the size of the pipe, a tiny pipe means you need to push harder to move the same amount of water at the same rate (1m^3 a second with a 1m pipe is easy, the water'll flow at an easy going meter and a half a second, try achieving that with a garden hose and you'll find you need to break the sound barrier). Trying to achieve the same flow rate with a higher resistance means you need more pressure and as a result you need to expend more effort to do so. Volts (pressure) = Amps (flow) * Resistance and Power (effort) = Volts*Amps (or flow * change in pressure, which interestingly still gives you power in watts)