Transfer of energy due to vibration and collision of medium particles without dislocation from their equilibrium position.

`text(Conduction in Variable State :)`
When we start heating one end of a rod, heat transfer begins from hot end to cold end. At th is time, every small part gains some energy from hotter part, it absorbs some energy and passes rest of the energy to next section. This process results in increased temperature of that small part. Heat transfer to th is small part can be defined using differential form of Ohm's Law

`(dQ)/(dt)=-KA(dT)/(dx)`

Here `dT//dx=` temperature gradient

In the figure small part 'dx' is taking `dQ_1` heat and transferring `dQ_2` heat to next section.

`dQ =dQ_1 - dQ_2` (In steady stated `dQ=0`, i.e. `dQ_1 =dQ_2`)

`:.` `mSdT=KA(dT_1)/(dx) dt- KA(dT_2)/(dx) dt`

`(dT)/(dt)=(KA)/(mS)((dT_1)/(dx)-(dT_2)/(dx))`

dT = Increase in temperature of the section in time dt

As the system approaches to steady state, temperature of all cross section become constant `=>` `(dT)/(dt)=0`

Therefore we can say following things in variable state.

(a) At a particular position 'x' (where x is the distance from heating end of the rod), as `tuparrow` `(dT)/(dt)downarrow`.

(b) If we move along the rod at a particular time 't', as `xuparrow` `Tdownarrow`.

`text(Thermal Conduction in Steady State :)`
In this state net `DeltaQ=0` and temperature gradient throughout the rod becomes constant i.e. `(dT)/(dt)=C`

Let the two ends of rod of length `l` is maintained at temp `T_1` and `T_2` then the rate of heat flow is-

Thermal current `(dQ)/(dt)=(KA(T_1-T_2))/L=(T_1-T_2)/R_(th)`

Where thermal resistance `R_(th)=l/(KA)`