As shown in the figure, an aluminium rod and a copper rod, each having a length of 1.0 m and a cross-sectional area of 1 cm ^2 , are welded together. One end of the combination is maintained at 20^ C while the other end is kept at 60^ C . Determine the amount of heat taken out per second from the hot end. The thermal conductivity of aluminium is 200 W m ⁻¹ ^ C ⁻¹ and that of copper is 390 W m ⁻¹ ^ C ⁻¹ .

Step-by-step solution

From the given figure, the aluminium and copper rods are connected in parallel. The total heat current (heat taken out per second) is the sum of the heat currents through each rod: H = H_ Al + H_ Cu Using the formula for heat conduction, H = KA T L : H = K_ Al A T L + K_ Cu A T L H = (K_ Al + K_ Cu ) A T L Given values are: K_ Al = 200 W m ⁻¹ ^ C ⁻¹ K_ Cu = 390 W m ⁻¹ ^ C ⁻¹ A = 1 cm ^2 = 10⁻⁴ m ^2 T = 60^ C - 20^ C = 40^ C L = 1.0 m Substituting the values into the equation: H = (200 + 390) 10⁻⁴ 40 1.0 H = 590 40