Figure shows an aluminium rod joined to a copper rod. Each of the rods has a length of `20cm` and area of cross section `0.20cm^(2)` . The junction is maintained at a constant temperature `40^(@)C` and the two ends are maintained at `80^(@)C` . 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^(-1).^(@)C^(-1)` . and `K_(Cu)=400Wm^(-1).^(@)C^(-1)` .

Explanation

`R_(A1) = (l)/(K_(A1)A) = (20 xx 10^(-2))/(200 xx 0.2 xx 10^(-4)) = 50^(@)C//W`
`R_(Cu) = (l)/(K_(Cu) A) = 25^(@)C//W`
`i_(1) = (80 - 40)/(R_(A1)) = (40)/(50) = 0.8 W`
`i_(2) = (80 - 40)/(R_(A1)) = (40)/(25) = 1.6 W`
Heat current reaching at cold junction
`i= i_(1) + i_(2) = 0.8 + 1.6 = 2.4 W`
Heat taken out at cold juction in
one minute `= it = 2.4 xx 60 = 144 J`