Question

Two plates $A$ and $B$ of equal surface area are placed one on top of the other to form a composite plate of the same surface area. The thickness of $A$ and $B$ are $4.0$cm and $6.0$cm respectively. The temperature of the exposed surface of plate $A$ is $-10$${}^o$C and that of the exposed surface of plate B is $10$${}^0$C. Neglect heat loss from the edges of the composite plate, the temperature of the contact surface is $T_1$ if the plates A and B are made of the same material and $T_2$ if their thermal conductivities are in the ratio $2:3$ then

• A. $T_1=-4^{0}C$
• B. $T_1=-2^{0}C$
• C. $T_2=-3^{0}C$
• D. $T_2=0^{0}C$

Answer 1

Answer: (B), (D)

The correct options are
B $T_1=-2^{0}C$
D $T_2=0^{0}C$

We use the resistance approach for this problem.

The heat resistance is given by,

$R=\frac{L}{KA}$

If the two plates are made of the same plate,

$\frac{K_{a}A(T_1+10)}{4}=-\frac{K_{b}A(T_1-10)}{6}$

Solving this for $T_1$ gives, it equal to -2

Similarly, if the two plates are made of materials with thermal conductivities in the ratio, 2:3,

$\frac{K_{a}A(T_1+10)}{4}=-\frac{K_{b}A(T_1-10)}{6}$

With $\frac{K_a}{K_b}=\frac{2}{3}$

Solving this, gives $T_2=0$