As shown in the figure, an aluminium rod is joined to a copper rod....

Given:
Length of each rod, $L = 20 \text{ cm} = 0.2 \text{ m}$
Cross-sectional area, $A = 0.20 \text{ cm}^2 = 2 \times 10^{-5} \text{ m}^2$
Temperature of outer ends, T_{hot} = 80^{}^{{}^{\circ}} C
Temperature of junction, T_{cold} = 40^{}^{{}^{\circ}} C
Temperature difference across each rod, \Delta T = 80^{}^{{}^{\circ}} C - 40^{}^{{}^{\circ}} C = 40^{}^{{}^{\circ}} C

The rate of heat flow (heat current) through the aluminium rod is:
$H_{Al} = \dfrac{K_{Al} A \Delta T}{L} = \dfrac{200 \times 2 \times 10^{-5} \times 40}{0.2} = 0.8 \text{ W}$

The rate of heat flow through the copper rod is:
$H_{Cu} = \dfrac{K_{Cu} A \Delta T}{L} = \dfrac{400 \times 2 \times 10^{-5} \times 40}{0.2} = 1.6 \text{ W}$

The total rate at which heat reaches the cold junction is:
$H_{total} = H_{Al} + H_{Cu} = 0.8 + 1.6 = 2.4 \text{ W}$

To maintain the steady state, the heat extracted from the cold junction in one minute ($t = 60 \text{ s}$) is:
$Q = H_{total} \times t = 2.4 \times 60 = 144 \text{ J}$