Q. 25

# 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

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