Compute the heat generating while transferring 96000 coulombs of charge in one hour through a potential difference of 50V.

- Hint – In this question use the direct concept that the heat generated while transferring any q coulombs of charge at a v voltage is given as $H = Q \times V$joules. This will help approaching the problem statement.

Complete step-by-step solution -

As we know that heat (H) generated is = ${i^2} \times r \times t$ joule................. (1)
Where, i = current, r = resistance and t is the time interval (i.e. how long the device is run).
Now given data:
Amount of charge (Q) transferred = 96000 coulomb.
Now as we know that charge is the product of current and time.
$ \Rightarrow Q = i \times t$coulomb.................. (2)
Now it is also given that potential difference (V) = 50 V.
Now as we know that potential difference (V) is the product of current (i) and resistance (r).
$ \Rightarrow V = i \times r$ volt.................. (3)
Now we have to find out the heat generated in one hour.
$ \Rightarrow t = 1$ hour.
So from equation (1), (2) and (3) we have,
$ \Rightarrow H = Q \times V$ joules.
Now substitute the values we have,
$ \Rightarrow H = 96000 \times 50 = 4800000$ joules.
$ \Rightarrow H = 96000 \times 50 = 4.8$ MJ, $\left[ {\because 1M = {{10}^6}} \right]$
So this is the required amount of heat generated.

Note – The key concept is why heat is generated when a charge is displaced in a potential from one point to another. The potential difference acting between the points in which the charge is being displaced does some work over the charge and this work done constitutes heat formation of heat. The basic definition of potential difference itself suggests that it is the work done per unit charge.