Question

The ends of an element of zinc wire are kept at a small temperature difference \[\Delta T\]and a small current \[I\] is passed through the wire then the heat developed per unit time:
\[(A)\]Is proportional to and \[I\]
\[(B)\]Is proportional to\[{I^3}\]and \[\Delta T\]
\[(C)\]Is proportional to Thomson Coefficient of the metal
\[(D)\]Is proportional to \[\Delta T\] only

Answer 1

Hint: In this question our mind goes to the idea of heating effect of current and loss of heat by cooling effect. These both processes occur simultaneously and affect each other so we will see how the rate of production of heat and rate of loss through cooling.

Formula Used: -
1. When electric current \[I\] is flowing in the wire under a potential difference \[V\]for time \[\Delta t\] then
Heat developed per second \[\dfrac{{\Delta H}}{{\Delta t}} = V_{I}\]
Where \[\Delta H\] is heat produced

2. According to Newton’s Law of cooling
Rate of heat loss \[ - \dfrac{{\Delta H}}{{\Delta t}} = k\Delta T\] where \[k\]is constant and \[\Delta T\]is a temperature difference.
Step by step solution: -
There are two process cooling and production of heat occurs simultaneously so
We have the following expression below
for production of heat
\[\dfrac{{\Delta H}}{{\Delta t}} = V_{I}\] ………………………………. (1)
For cooling
\[ - \dfrac{{\Delta H}}{{\Delta t}} = k\Delta T\]………………………….. (2)
From two above equation (1) and (2) we can conclude that
Rate of heat produced \[ \propto I\]……………………………. From (1) equation
Rate of heat loss \[ \propto \Delta T\]……………………………. From (2) equation
Hence option \[(A)\]is correct.

Note: In this question we must have the knowledge about the heating effect of current and the Newton’s Law of cooling with proper understanding. Here the heating of wire due to current in wire and loss of heat by cooling come into effect simultaneously both the processes are opposite to each other.