First slide
Conduction
Question

Two conducting slabs having same shape and size  of heat conductivities k1 and k2 joined as shown in figure. The temperature at ends of the slabs are θ1 and θ2 (θ1>θ2). The final temperature of junction (θm ) is:
 

Moderate
Solution

Assuming both slabs have identical thicknesses, Qt=k1A(θ1-θm)l=k2A(θm-θ2)lθm=k1θ1+k2θ2k1+k2

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