Question

Three rods A, B and C of same length and cross-sectional area are joined in series. The thermal conductivities are in the ratio 1 : 2 : 1.5. If the open ends of A and C are at 200°C and 18°C, respectively. At equilibrium, the temperature at the junction of A and B is

• A. 74°C
• B. 116°C
• C. 156°C
• D. 148°C

Answer 1

Answer: (B)

116°C

Answer 2

Let $L$ be the length and $A$ be the cross-sectional area of each rod. Let $k_A$, $k_B$, and $k_C$ be the thermal conductivities of rods A, B, and C, respectively.
Given that the thermal conductivities are in the ratio 1 : 2 : 1.5, we can write $k_A = k$, $k_B = 2k$, and $k_C = 1.5k = \frac{3}{2}k$ for some constant $k$.

The rods are joined in series. Let $T_A = 200^\circ$C be the temperature of the open end of rod A, and $T_C = 18^\circ$C be the temperature of the open end of rod C. Let $T_1$ be the temperature at the junction of rod A and rod B, and $T_2$ be the temperature at the junction of rod B and rod C.

At thermal equilibrium, the rate of heat flow ($H$) is the same through each rod. The rate of heat flow through a rod is given by Fourier's Law of Conduction: $H = \frac{kA \Delta T}{L}$.

For rod A: $H_A = \frac{k_A A (T_A - T_1)}{L} = \frac{k A (200 - T_1)}{L}$
For rod B: $H_B = \frac{k_B A (T_1 - T_2)}{L} = \frac{2k A (T_1 - T_2)}{L}$
For rod C: $H_C = \frac{k_C A (T_2 - T_C)}{L} = \frac{1.5k A (T_2 - 18)}{L}$

Since $H_A = H_B = H_C$, we have:
$\frac{k A (200 - T_1)}{L} = \frac{2k A (T_1 - T_2)}{L} = \frac{1.5k A (T_2 - 18)}{L}$

Cancelling $\frac{kA}{L}$ from all terms (assuming $k, A, L \ne 0$):
$200 - T_1 = 2(T_1 - T_2) = 1.5(T_2 - 18)$

This gives us two equations:

1. $200 - T_1 = 2(T_1 - T_2)$
   $200 - T_1 = 2T_1 - 2T_2$
   $200 = 3T_1 - 2T_2$ (Equation 1)

2. $2(T_1 - T_2) = 1.5(T_2 - 18)$
   $2T_1 - 2T_2 = 1.5T_2 - 27$
   $2T_1 = 3.5T_2 - 27$
   Multiply by 2:
   $4T_1 = 7T_2 - 54$ (Equation 2)

We need to find $T_1$. From Equation 1, $2T_2 = 3T_1 - 200$, so $T_2 = \frac{3T_1 - 200}{2}$.
Substitute this expression for $T_2$ into Equation 2:
$4T_1 = 7 \left( \frac{3T_1 - 200}{2} \right) - 54$
$4T_1 = \frac{21T_1 - 1400}{2} - 54$
Multiply the entire equation by 2:
$8T_1 = 21T_1 - 1400 - 108$
$8T_1 = 21T_1 - 1508$
$1508 = 21T_1 - 8T_1$
$1508 = 13T_1$
$T_1 = \frac{1508}{13}$

Performing the division:
$1508 \div 13 = 116$.

So, the temperature at the junction of A and B is $T_1 = 116^\circ$C.

Alternatively, using thermal resistance $R = \frac{L}{kA}$. Since $L$ and $A$ are constant, $R \propto \frac{1}{k}$.
The ratio of thermal conductivities is $k_A : k_B : k_C = 1 : 2 : 1.5 = 1 : 2 : \frac{3}{2}$.
The ratio of thermal resistances is $R_A : R_B : R_C = \frac{1}{1} : \frac{1}{2} : \frac{1}{3/2} = 1 : \frac{1}{2} : \frac{2}{3}$.
Multiplying by 6, the ratio is $6 : 3 : 4$.
Let $R_A = 6r$, $R_B = 3r$, $R_C = 4r$.

In series, the total resistance is $R_{total} = R_A + R_B + R_C = 6r + 3r + 4r = 13r$.
The total temperature difference is $\Delta T_{total} = T_A - T_C = 200 - 18 = 182^\circ$C.
The heat flow rate through the series is $H = \frac{\Delta T_{total}}{R_{total}} = \frac{182}{13r} = \frac{14}{r}$.

The temperature difference across rod A is $T_A - T_1 = H R_A$.
$200 - T_1 = \left(\frac{14}{r}\right) (6r) = 14 \times 6 = 84$.
$T_1 = 200 - 84 = 116^\circ$C.

The temperature at the junction of A and B is $116^\circ$C.