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Physics Question on thermal properties of matter

A wall is made of equally thick layers AA and BB of different materials. Thermal conductivity of AA is twice that of BB. In the steady state, the temperature difference across the wall is 36∘C36^{\circ}C. The temperature difference across the layer AA is

A

12 ∘C12\,^{\circ}C

B

18 ∘C18\,^{\circ}C

C

6 ∘C6\,^{\circ}C

D

24 ∘C24\,^{\circ}C

Answer

12 ∘C12\,^{\circ}C

Explanation

Solution

Here, KA=2KBK_A = 2K_B, TA−TB=36∘CT_A - T_B = 36^{\circ}C Let TT is the temperature of the junction. As (ΔTΔt)A=(ΔTΔt)B\left(\frac{\Delta T}{\Delta t}\right)_{A}=\left(\frac{\Delta T}{\Delta t}\right)_{B} ∴KAA(TA−T)x=KBA(T−TB)x\therefore \frac{K_{A}A\left(T_{A}-T\right)}{x}=\frac{K_{B}A\left(T-T_{B}\right)}{x} 2KB(TA−T)=KB(T−TB)2K_{B}\left(T_{A}-T\right)=K_{B}\left(T-T_{B}\right) 2(TA−T)=T−TB2\left(T_{A}-T\right)=T-T_{B} Add (TA−T)\left(T_{A}-T\right) on both sides, we get 3(TA−T)=TA−T+T−TB3\left(T_{A}-T\right)=T_{A}-T+T-T_{B} 3(TA−T)=TA−TB3\left(T_{A}-T\right)=T_{A}-T_{B} TA−T=TA−TB3T_{A}-T=\frac{T_{A}-T_{B}}{3} =363=12 ∘C=\frac{36}{3}=12\,^{\circ}C ∴\therefore Temperature difference across the layer A=TA−T=12 ∘CA=T_{A}-T=12\,^{\circ}C