You visited us 1 times! Enjoying our articles? Unlock Full Access!

Conduction Law

Two walls of ...

Question

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 temperature at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature at the contact of the walls will be

\frac{K_{1} T_{1} + K_{2} T_{2}}{d_{1} + d_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{(K_{1} d_{1} + K_{2} d_{2}) T_{1} T_{2}}{T_{1} + T_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{K_{1} d_{1} T_{1} + K_{2} d_{2} T_{2}}{K_{1} d_{1} + K_{2} d_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$

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

$\Rightarrow K_{1} T d_{2} - K_{1} T_{1} d_{2} = K_{2} T_{2} d_{1} - K_{2} T d_{1}$

$\Rightarrow T = \frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$

Suggest Corrections

Similar questions

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 temperature at the outer surfaces are

T_{1}

and

T_{2}

, the temperature at the contact of the walls will be

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

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 temperature at the outer surfaces are

T_{1}

and

T_{2}

, the temperature at the contact of the walls will be

Q. Two walls of thickness

l_{1}

and

l_{2}

and thermal conductivities

k_{1}

and

k_{2}

are in contact. In the steady state, if the temperature at the outer faces are

T_{1}

and

T_{2}

, the temperature at the common wall is :

Two blocks of the same thickness and thermal conductivities $k_{1} a n d k_{2}$ are placed in contact. Their outer surfaces are kept at constant temperatures $T_{1} a n d T_{2}$ . If $T_{2}$ > $T_{1}$ , and $k_{2}$ > $k_{1}$ the temperature of the interfaces in the steady state will be

Join BYJU'S Learning Program

Explore more

Conduction Law

Standard XI Physics

Join BYJU'S Learning Program

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

The correct option is B K1T1d2+K2T2d1K1d2+K2d1 Heat flow across both the walls will be same. −K1Ad1(T−T1)=−K2Ad2(T2−T) ⇒K1Td2−K1T1d2=K2T2d1−K2Td1 ⇒T=K1T1d2+K2T2d1K1d2+K2d1

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 temperature at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature at the contact of the walls will be