Question

A hot bath at Th 82.2 °C is connected to a cold bath at T 13.9°C by an insulated metal rod as shown in the diagram (Figure 1). The metal rod is made from two metals, aluminium is next to the hot bath and has a length 11-65.2 cm, steel makes up the remainder of the rod. The junction between the aluminium and steel has a temperature T 55.4 °C. The rate of energy flow through the rods is P= 191 W Cross-sectional view Aluminium Steel Thermal T insulation Diagram not to scale Figure 1: Schematic of the described hot and cold baths at temperatures Th and Te, respectively, connected via an insulated metal rod consisting of aluminium of length l1 next to the hot bath, and steel of length 2 next to the cold bath. The temperature of the junction between aluminium and steel is T Part 1) What is the cross sectional area of the rod? Part 2) The steel rod has the same cross sectional area as the aluminium one. What is the length, l2, of the steel rod? CIII Part 3) The steel rod is now replaced with a copper rod with identical length and cross sectional area. What is the temperature of the junction between the aluminium and copper?

Answer 1

Given T_h = 82.2 degree C, T_c = 13.9 degree C l_1 = 65.2 cm = 0.652 m T_j = 55.4 degree C P = 191 w Part - (a) for aluminium rod thermal conductivity K_Al = 205 W/m - k Now, for Al rod P = K_Al A (T_h - T_T_j)/l_1 191 = 205 times A times (82.2 - 55.4)/0.652 = 0.0226 m^2 A = 226 cm^2Part - (b) For steel rod, thermal conductivity K_s = 46 w/m - k (from web) P = K_s A(T_j - T_c)/l_2 l_2 = 46 times 0.0226 times (55.4 - 13.9)/191 = 0.225 m l_2 = 22.5 cm Part - (c) In that case P_Al = P_c_4 K_Al A(T_h - T_j)/l_1 = K_C_4 A(T_j - T_c)/l_2 205 times A times (82.2 - T_j)/0.652 = 365 times A times (T_j - 13.9)/0.225 T_j = 24.5 degree C