The amount of heat taken out from the cold junction in one minute after reaching steady state is approximately 1.33 Joules.
In a bimetallic rod composed of aluminum and copper with lengths of 20 cm and cross-sectional areas of 0.20 cm² each, the junction is maintained at a constant temperature of 40 degrees Celsius, while the two ends are kept at 80 degrees Celsius.
Once the system reaches a steady state, the heat transfer can be calculated using the formula Q= k⋅A⋅ΔT⋅t/L, where Q is the heat transfer, k is the material's thermal conductivity, A is the cross-sectional area, ΔT is the temperature difference, t is time, and L is the length of the rod. For aluminum (k Al =200Wm^−1 C ^−1) and copper (k Cu =400Wm^−1 C^−1), the calculation yields an approximate heat transfer of 1.33 Joules in one minute after reaching the steady state.
Complete question:
Figure shows an Aluminium rod joined to a copper rod each of the rods has a length of 20 cm and area of cross section 0.20cm^2. the junction is maintained at a constant temperature 40 degree Celsius and the two ends are maintained at 80 degree Celsius. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached . the conductivities are K al=200Wm^-1C^-1 and K cu=400Wm^-1C^-1.