Solveeit Logo
Login

Question

Question: A wall is made of equally thick layers A and B of different materials. Thermal conductivity of A is ...

A wall is made of equally thick layers A and B of different materials. Thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36°C. The temperature difference across the layer A is

A

12 °C

B

18 °C

C

6 °C

D

24 °C

Answer

12 °C

Explanation

Solution

Here, KA=2Kn,TA−TB=36∘CK_{A} = 2K_{n},T_{A} - T_{B} = 36{^\circ}C

Let T is the temperature of the junction.

As (ΔTΔt)A=(ΔTΔt)n\left( \frac{\Delta T}{\Delta t} \right)_{A} = \left( \frac{\Delta T}{\Delta t} \right)_{n}

∴KAA(TA−T)x=KBA(T−TB)x\therefore\frac{K_{A}A(T_{A} - T)}{x} = \frac{K_{B}A(T - T_{B})}{x}

2KB(TA−T)=KB(T−TB)2K_{B}(T_{A} - T) = K_{B}(T - T_{B})

2(TA−T)=T−TB2(T_{A} - T) = T - T_{B}

Add (TA−TT_{A} - T) on both sides, we get

3(TA−T)=TA−T+T−TB3(T_{A} - T) = T_{A} - T + T - T_{B}

3(TA−T)=TA−TB3(T_{A} - T) = T_{A} - T_{B}

TA−T=TA−TB3=363=12∘CT_{A} - T = \frac{T_{A} - T_{B}}{3} = \frac{36}{3} = 12{^\circ}C

∴\thereforeTemperature difference across the layer A

=TA−T=12∘C= T_{A} - T = 12{^\circ}C