Question

Two slabs $A$ and $B$ of equal surface area are placed one over the other such that their surface are completely in contact. The thickness of slab $A$ is twice that of $B$. The coefficient of thermal conductivity of slab $A$ is twice that of $B$. The first surface of slab $A$ is maintained at $100$ $^\circ$C, while the second surface of slab $B$ is maintained at $25$ $^\circ$C. The temperature at the contact of their surface is.

• A. 62.5 $^\circ$C
• B. 45 $^\circ$C
• C. 55 $^\circ$C
• D. 85 $^\circ$C

Answer 1

Answer: (A)

62.5 $^\circ$C

Answer 2

Let $L_A$ and $L_B$ be the thicknesses of slabs A and B, respectively, and $K_A$ and $K_B$ be their thermal conductivities. Let $A$ be the surface area of each slab. We are given that $L_A = 2 L_B$ and $K_A = 2 K_B$. The first surface of slab A is maintained at $T_1 = 100$ $^\circ$C, and the second surface of slab B is maintained at $T_2 = 25$ $^\circ$C. Let $T_i$ be the temperature at the contact surface between slabs A and B.

In the steady state, the rate of heat transfer through slab A is equal to the rate of heat transfer through slab B. The rate of heat transfer through a slab is given by Fourier's law of conduction: $Q = \frac{K A \Delta T}{L}$.

For slab A, the temperature difference is $T_1 - T_i$. The heat flow rate through slab A is $Q_A = \frac{K_A A (T_1 - T_i)}{L_A}$.

For slab B, the temperature difference is $T_i - T_2$. The heat flow rate through slab B is $Q_B = \frac{K_B A (T_i - T_2)}{L_B}$.

Since the slabs are in series and in steady state, $Q_A = Q_B$.

$\frac{K_A A (T_1 - T_i)}{L_A} = \frac{K_B A (T_i - T_2)}{L_B}$

Substitute the given relationships $L_A = 2 L_B$ and $K_A = 2 K_B$:

$\frac{(2 K_B) A (100 - T_i)}{2 L_B} = \frac{K_B A (T_i - 25)}{L_B}$

The term $\frac{2 K_B A}{2 L_B}$ simplifies to $\frac{K_B A}{L_B}$. So the equation becomes:

$\frac{K_B A (100 - T_i)}{L_B} = \frac{K_B A (T_i - 25)}{L_B}$

Assuming $K_B \neq 0$, $A \neq 0$, and $L_B \neq 0$, we can cancel the term $\frac{K_B A}{L_B}$ from both sides:

$100 - T_i = T_i - 25$

Now, we solve for $T_i$:

$100 + 25 = T_i + T_i$

$125 = 2 T_i$

$T_i = \frac{125}{2}$

$T_i = 62.5$ $^\circ$C

Alternatively, we can use the concept of thermal resistance. The thermal resistance of a slab is $R = \frac{L}{KA}$.

For slab A, $R_A = \frac{L_A}{K_A A} = \frac{2 L_B}{(2 K_B) A} = \frac{L_B}{K_B A}$.

For slab B, $R_B = \frac{L_B}{K_B A}$.

We observe that $R_A = R_B$.

When two thermal resistances are in series, the temperature drop across each resistance is proportional to its value. The heat flow rate is $Q = \frac{\Delta T}{R}$.

$Q = \frac{T_1 - T_i}{R_A} = \frac{T_i - T_2}{R_B}$.

Since $R_A = R_B$, we have $\frac{T_1 - T_i}{R_A} = \frac{T_i - T_2}{R_A}$, which implies $T_1 - T_i = T_i - T_2$.

$T_1 + T_2 = 2 T_i$

$T_i = \frac{T_1 + T_2}{2}$

Substitute the values $T_1 = 100$ $^\circ$C and $T_2 = 25$ $^\circ$C:

$T_i = \frac{100 + 25}{2} = \frac{125}{2} = 62.5$ $^\circ$C.

The temperature at the contact of their surface is 62.5 $^\circ$C.