Question

A cylindrical metalloic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with two reservoirs in time t?

Answer 1

Let thermal conductivity be K

Heat = Q

Time = t

Length = L

radius = r

Let ΔT be the difference of temperature of the two reservoirs.

Q = Kπr²(ΔT)t/L

Now the rod is melted radius is made ½ times the original radius

Let the new length of the rod = L’

Volume of the rod = V

V = πr²L = π(r/2)²L’

=> L’ = 4L

Now heat conducted by the new conductor is

Q’ = Kπ(r/2)²(ΔT)t/(4L)

= K πr²(ΔT)t/16L

= Q/16