Two walls of thickness d1 and d2, thermal conductivities K1 and K2 are in contact. In the steady state if the temperatures at the outer surfaces are T1 and T2, the temperature at the common wall will be

Question

Two walls of thickness d1 and d2, thermal conductivities K1 and K2 are in contact. In the steady state if the temperatures at the outer surfaces are T1 and T2, the temperature at the common wall will be (A) (K1 T1+K2 T2/d1+d2) (B) (K1 T1 d2+K2 T2 d1/K1 d2+K2 d1) (C) ((K1 d1+K2 d2) T1 T2/T1+T2) (D) (K1 d1 T1+K2 d2 T2/K1 d1+K2 d2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b24339b60ff9c6d9b6b1b4b53a0d82b7-.png" /> Heat flow across both the walls will be same. (K1 A/d1)(T-T1)=(K2 A/d2)(T2-T) K1 T d2-K1 T1 d2 =K2 T2 d1-K2 T d1 T=(K1 T1 d2+K2 T2 d1/K1 d2+K2 d1)

Q. Two walls of thickness $d_{1}$ and $d_{2}$ , thermal conductivities $K_{1}$ and $K_{2}$ are in contact. In the steady state if the temperatures at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature at the common wall will be

$\frac{K _{1} T _{1} + K _{2} T _{2}}{d _{1} + d _{2}}$

$\frac{K _{1} T _{1} d _{2} + K _{2} T _{2} d _{1}}{K _{1} d _{2} + K _{2} d _{1}}$

$\frac{( K _{1} d _{1} + K _{2} d _{2} ) T _{1} T _{2}}{T _{1} + T _{2}}$

$\frac{K _{1} d _{1} T _{1} + K _{2} d _{2} T _{2}}{K _{1} d _{1} + K _{2} d _{2}}$

Heat flow across both the walls will be same.
$\frac{K _{1} A}{d _{1}} (T - T_{1}) = \frac{K _{2} A}{d _{2}} (T_{2} - T)$
$K_{1} T d_{2} - K_{1} T_{1} d_{2}$
$= K_{2} T_{2} d_{1} - K_{2} T d_{1}$
$T = \frac{K _{1} T _{1} d _{2} + K _{2} T _{2} d _{1}}{K _{1} d _{2} + K _{2} d _{1}}$

Q. Two walls of thickness d1​ and d2​, thermal conductivities K1​ and K2​ are in contact. In the steady state if the temperatures at the outer surfaces are T1​ and T2​, the temperature at the common wall will be

d1​+d2​K1​T1​+K2​T2​​

K1​d2​+K2​d1​K1​T1​d2​+K2​T2​d1​​

T1​+T2​(K1​d1​+K2​d2​)T1​T2​​

K1​d1​+K2​d2​K1​d1​T1​+K2​d2​T2​​