The amount of heat flows in time t through a cylindrical metallic rod of length L and uniform area of cross-section $A\left( { = \pi {R^2}} \right)$ with its ends maintained at temperatures T₁ and T₂ (T₁ > T₂) is given by

$Q = {{KA\left( {{T_1} - {T_2}} \right)t} \over L}$   ...(i)

where K is the thermal conductivity of the material of the rod.
Area of cross-section of new rod

$A' = \pi {\left( {{R \over 2}} \right)^2} = {{\pi {R^2}} \over 4} = {A \over 4}$   ...(ii)

As the volume of the rod remains unchanged

$\therefore$ AL = A'L'

where L' is the length the new rod

$\Rightarrow L' = L{A \over {A'}}$   ...(iii)

= 4L     (Using (ii))

Now, the amount of heat flows in same time t in the new rod with its ends maintained at the same temperatures T₁ and T₂ is given by

$Q' = {{KA'\left( {{T_1} - {T_2}} \right)t} \over {L'}}$   ...(iv)

Substituting the values of A' and L' from equations (ii) and (iii) in the above equation, we get

$Q' = {{K(A/4)\left( {{T_1} - {T_2}} \right)t} \over {4L}} = {1 \over {16}}{{KA\left( {{T_1} - {T_2}} \right)t} \over L} = {1 \over {16}}Q$
   (Using (i))