° C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = . J s – m – K – ; and of copper = J s – m – K – ).
Fig. . Steady state heat flow by conduction in a bar with its two ends maintained at temperatures T C and T D ; (T C > T D ). Fig.
. Answer The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross section of the rod.
In the steady state, heat flowing into the element must equal the heat flowing out of it; otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross section of the rod is the same at every point along the length of the combined steel-copper rod. Let T be the temperature of the steel-copper junction in the steady state. Then, K A K A T – − where and refer to the steel and copper rod respectively.
For A = A , L = . cm, L = . cm, K = . J s – m – K – , K = J s – m – K – , we have .
× − which gives T = . ° C Example . An iron bar ( L = . m, A = .
m , K = W m – K – ) and a brass bar ( L = . m, A = . m , K = W m – K – ) are soldered end to end as shown in Fig. .
. The free ends of the iron bar and brass bar are maintained