Heat flow

Consider a path $\Delta T$ $\Delta L$

Using

$Q=\kappa A \frac{\Delta T}{\Delta L}t$

$\frac {dQ}{dt}=\kappa A \frac {dT}{dL}$

Also by definition, heat flow at any place L remains same, so we let

$\frac{dQ}{dt}=C_1$

for some constant $C_1$ . The value of $C_1$ will be our answer.

Let's define something,

When $R=R_1$ , $L=0$ , when $R=R_2$ , $L=L_0$ .

So $R(L)=\frac{R_2-R_1}{L_0}L+R_1$

$A(L)=\pi R^2=\pi (\frac{R_2-R_1}{L_0}L+R_1)^2$

So

$T(L)=\int \frac{C_1}{\kappa A}dL$

$T(L)=- \frac{C_1 L_0}{\pi \kappa (R_2-R_1)}\frac{1}{\frac{R_2-R_1}{L_0}L-R_1}+C_2$

for some constant $C_2$

Compute all of the value of them and $T(0)=322$ and $T(5)=422$

leads to $\frac{dQ}{dt}=C_1=\boxed{20}$

let the net resistance be R.

now we can see that the radius at a distance x from one end is $r = R1 + \frac{R2-R1}{L}x$ .......(1)

then Resistance dR of a small element at a distance dx from one end will be -

$dR = \frac{dx}{k(\pi)((r)^2)}$

putting the value of r from (1) and integrating

we get $R = \frac{L}{k(\pi)(R1)(R2)}$

therefore rate of heat flow = heat current(i) = $\frac{T2-T1}{R}$ = 20 SI Units