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 0.20 cm2. 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 KAl = 200 W m–1°C–1andKCu = 400 W m–1 °C–1.
Given –
Length of each rod of 20 cm = 0.2m
Area of cross-section 0.20 cm2
= 2 × 10-5 m2
Junction temperature = 40°C
End temperature = 80°C
The conductivities of Aluminum and copper,KAl=200 W m–1 °C–1
And KCu = 400 W m–1 °C–1.
Now ,total heat drawn per second –
= Heat drawn due to copper rod + heat drawn due to Aluminium rod
= QAl + QCu
We know, rate of heat absorption by the rod of length l, area A is given by
Q = 
Where 
is the change in temperature
= 
+
⇒ Q = 2.4 J
Heat drawn in 1 minute = 2.4 × 60 = 144J
Hence, amount of heat taken out from the cold junction in one minute after the steady state is reached is 144J