A composite slab consists of two slabs A and B of different materials but of the same thickness placed in contact as shown in figure. The thermal conductivity of A and B are $$k1$$ and $$k2$$ respectively. A steady temperature difference of $$12$$∘C is maintained across the composite slab. If $$k1=k22$$, the temperature difference across slab A will be:

Question

A composite slab consists of two slabs  A and  B of different materials but of the same thickness placed in contact as shown in figure. The thermal conductivity of  A and B are  $$k1$$ and $$k2$$ respectively. A steady temperature difference of $$12$$∘C is maintained across the composite slab. If  $$k1=k22$$, the temperature difference across slab  A will be:

Infinity Learn · Accepted Answer

Let temperature of junction be  To∘CTemperature difference across slab  $$A=T1$$−To∘CTemperature difference across slab  $$B=To$$−$$T2$$∘CSince both slabs are connected in series, heat current through both will be equal $$k1AT1$$−$$T0l2=k2ATo$$−$$T2l2$$ As $$k1=k22$$⇒$$k22T1$$−$$To=To$$−$$T2k2$$⇒$$T1$$−$$To=2To$$−$$2T2$$⇒$$T1+2T2=3T0$$….(1) As $$T1$$−$$T2=12$$∘C⇒$$T2=T1$$−$$12$$…..(2) From (1) and (2), we get ⇒$$T1+2T1$$−$$12=3To$$⇒$$3T1$$−$$24=3T0$$⇒$$T1$$−$$T0=8$$∘C

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!