Joule's effect is due to collision of free electrons with positive ions.

Let the specific resistance of the material be  $P$.

Joule heating effect is irreversible that means if the direction of current in a resistor is reversed, the cooling of resistor does not occur whereas heating of the resistor takes place.

Work done (heat produced) $= {I}^{2}Rt  \   Joule$

According to Joule's law of heating,  heat produced in a resistor     $H = I^2Rt$

According to Joule's law of heating, total heat energy supplied to the resistor, $H=i^{2}Rt$

The phenomenon of generating heat due to the resistance of the conductor when a current passes through it is known as heating effect of electricity.

$P=VI\quad .............(i)$

Joule heating (also referred to as resistive or ohmic heating) describes the process where the energy of an electric current is converted into heat as it flows through a resistance.

In particular, when the electric current flows through a solid or liquid with finite conductivity, electric energy is converted to heat through resistive losses in the material. The heat is generated on the microscale when the conduction electrons transfer energy to the conductor's atoms by way of collisions. It does not depend upon the direction of flow of current, but its amount.

As we know that power(P)=v$\times$I
so P=15$\times$220=3300 watt
cost per unit=2 rupees 
therefore,$\dfrac {3300\times2\times15\times30}{60\times1000}$=49.5 rupees.

Joules law can be stated as the quantity of heat generated (H) in a conductor of Resistance (R), when a current (I) flows through it for a time (t) is directly proportional to the square of the current, the resistance of the conductor, and the time for which the current flows.
So, $H=I^2Rt$
Now, for a conductor following Ohm's Law, $V=IR$, or, $I=\dfrac{V}{R}$
Therefore, $H=\dfrac{V^2}{R^2} \times R \times t=\dfrac{V^2}{R}t$

The loss in power in electric circuit due to heat produced in conductors due to their resistance is called $Joule's \hspace{2mm} heat$.

Laws of heating are given by Joule 

Amount of heat required to boil a given quantity of water in a constant

$\displaystyle P=\frac {W}{t}$ or $W = H = P \times t = \displaystyle \frac {V^2}{R} \times t = \displaystyle \frac {V^2t _1}{R _1} = \displaystyle \frac {V^2t _2}{R _2}$

$\displaystyle \Rightarrow \frac {t _2}{t _1} = \frac {R _2}{R _1} = \frac {l _2}{l _1}$

or $\displaystyle \frac {10}{12} = \frac {l _2}{l _1} \Rightarrow l _2 = \frac {10}{12} \times l _1$

$\displaystyle l _2 = \frac {5}{6}l$

Heat generated across the resistor is given by

At constant pressure,
$q=nCp  \Delta t$               $C _p =\frac{5R}{2}$
$=(0.25) \left ( \frac{5}{2} \times 8.314  \right ) (16)$
$=83.14 J$

From Joule's Law of heating-