Why does the thermal conductivity of metals decrease with an increase in temperature?
Thermal conductivity of any material is dependent on two things:i. Motion of free electronsii. Molecular vibrationsFor metals, the thermal conductivity is mainly a function of the motion of free electrons. As the temperature increases, the molecular vibrations increase (in turn decreasing the mean free path of molecules). So, they obstruct the flow of free electrons, thus reducing the conductivity. In case of non metals, there are no free electrons. So, only the molecular vibrations are responsible for conduction of heat and hence for non metals the conductivity increases with increase in temperature.
A copper rod of length 0.30 m and cross-sectional area 9.00 10-2 cm2 is connected to an iron rod with the same?
Good problem. I would approach it as two baths of water attached with a super insulated copper rod from the boiling bath and a iron rod from the ice bath. The rods are insulated from the air so no heat transfer occurs around the rods at all, only heat transfer has to be longitudinally. Heat moves from hot toward cold. But the iron rod is in contact directly with the ice bath and the ice should bring a perfectly insulated iron rod down to 0C. But from the opposing direction, the same occurs except the heat is driven toward the ice bath and copper is about 5 times more conductive to heat than iron. And since only the end of the iron is in the ice bath, then that will be the cold sink. Any iron out of the ice bath will have heat going to it from the boiling bath. All the iron does is slow the heat transfer from the hot bath to the ice bath. So treat the whole rod as one conductor going toward the ice bath so that the junction temperature will be the same as boiling bath since the rod is insulated from air. All the junction does is add an equation to lengthen the heat transfer from the boiling bath to the ice bath. Heat conduction Q/ Time = (Thermal conductivity) x (Area) x (Thot - Tcold)/Thickness thermal conductivity of copper = 385 W/m K thermal conductivity of iron = 79.5 W/m K For copper heat conduction: Q/t = 385 * .000009 m^2 * (100C - 0C) / 30 cm thickness Q/t = 1.155 watts conduction heat loss up to iron junction For iron conduction: Q/t = 79.5 * .000009 m^2 * ( 100C - 0C) / 30 cm = 0.2385 watts heat conduction from junction to ice bath. Now I think, and might be wrong here, is that you just add the two transfer rates and divide by two and get an average thermal conduction power rate: (1.155 watts + 0.2385 watts) / 2 = 0.70 watts heat transfer rate