A copper rod 2 m long has a circular cross-section of radius 1 cm. One end is kept at `100^@C` and the other at `0^@C`. The surface is insulated so that negligible heat is lost through the surface. In steady state, find
(a) the thermal resistance of the bar
(b) the thermal current H
(c) the temperature gradient `(dT)/(dx)` and
(d) the temperature at a distance 25 cm from the hot end.
Thermal conductivity of copper is 401 `W//m.K.`

Explanation

i) Thermal resistance, `R=l/(KA) = l/(K(pir^(2)))`
or `R=(2)/((401)(pi)(10^(-2)))^(2) =15.9kW^(-1)`
ii) Thermal current, `H=(DeltaT)/R = 100/15.9` or H=6.3 W
Temperature gradient
`(DeltaT)/(dx)=(0-100)/2 = -50 Km^(-1) = -50^(@)Cm^(-1)`
iv)
Let `T^(@)`C be the temperature at 25 cm from the hot end, then
T-100 = (Temperature gradient) x (distance)
or T-100 = `(-50)(0.25)`or `T=87.5^(@)`C