Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are 'θ2' and ' θ1' respectively, (θ2 θ1). The temperature at the interface is :-

Question

Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are 'θ2' and ' θ1' respectively, (θ2 > θ1). The temperature at the interface is :- (A) (θ2 + θ1/2)  (B)  (θ1/10) + (9θ2/10) (C)  (θ1/3) + (2θ2/3) (D)  (θ1/6) + (5θ2/6)

Tardigrade Admin · Accepted Answer

Let the temperature of interface be " θ " i1 = i2 Steady state conduction (3KA(θ2-θ)/d) = (KA(θ - θ1)/3d) θ = (9 θ2/10) + (θ1/10)

$\frac{θ _{2} + θ _{1}}{2}$

$\frac{θ _{1}}{10} + \frac{9 θ _{2}}{10}$

$\frac{θ _{1}}{3} + \frac{2 θ _{2}}{3}$

$\frac{θ _{1}}{6} + \frac{5 θ _{2}}{6}$

Let the temperature of interface be " $θ$ "
$i_{1} = i_{2}$ {Steady state conduction}
$\frac{3 K A ( θ _{2} - θ )}{d} = \frac{K A ( θ - θ _{1} )}{3 d}$
$θ = \frac{9 θ _{2}}{10} + \frac{θ _{1}}{10}$

2θ2​+θ1​​

10θ1​​+109θ2​​

3θ1​​+32θ2​​

6θ1​​+65θ2​​

Let the temperature of interface be " θ " i1​=i2​ {Steady state conduction} d3KA(θ2​−θ)​=3dKA(θ−θ1​)​ θ=109θ2​​+10θ1​​

$\frac{θ _{2} + θ _{1}}{2}$

$\frac{θ _{1}}{10} + \frac{9 θ _{2}}{10}$

$\frac{θ _{1}}{3} + \frac{2 θ _{2}}{3}$

$\frac{θ _{1}}{6} + \frac{5 θ _{2}}{6}$

Let the temperature of interface be " $θ$ "
$i_{1} = i_{2}$ {Steady state conduction}
$\frac{3 K A ( θ _{2} - θ )}{d} = \frac{K A ( θ - θ _{1} )}{3 d}$
$θ = \frac{9 θ _{2}}{10} + \frac{θ _{1}}{10}$