(a) Write the mathematical expression for Joule's law of heating.

(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V.

The amount of work done when an electric charge, move against a potential difference which is denoted by V is given by;

W=V x Q …….. (1)

From the above expression,

V=W/Q

We know that,

V=IR …….. (2)

Q=It……… (3)

Therefore,

I=Q/t

Let us substitute (2) and (3) in (1) we get,

W=(IR)(It)

W=H

Where H is the amount of heat produced, it is expressed in Joules J and I is expressed in terms of Ampere.

Therefore, we can write

H = I²Rt

(b) Given

Charge given, Q = 96000 C

Time given (t) = 2 hour = 2 × 60 × 60 s =7200 s

Voltage (V) = 40 V

We know that, current can be found out by using the formula,

I = Q/t

I = 96000/7200 = 40/3 A

Amount of heat generated can be found out using the formula,

H = VIt = 40 × 40/3 × 7200 = 3840 kJ

Thus, the “amount of heat” generated is 3840 kJ