Question

Three conductors of same length having thermal conductivity \(k_1\), \(k_2\), and \(k_3\) are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is ________ °C. (Given: \(k_1\) = 60 Js⁻¹m⁻¹K⁻¹,\(k_2\) = 120 Js⁻¹m⁻¹K⁻¹, \(k_3\) = 135 Js⁻¹m⁻¹K⁻¹)

 

Hint:

Remember the formula for thermal resistance and how to combine resistances in series and parallel. Also, in steady state, the heat flow is constant.

Answer 1

Correct Answer: 40

Given:

• Thermal conductivity of 1st conductor: \( k_1 = 60 \, \text{Js}^{-1} \text{m}^{-1} \text{K}^{-1} \)
• Thermal conductivity of 2nd conductor: \( k_2 = 120 \, \text{Js}^{-1} \text{m}^{-1} \text{K}^{-1} \)
• Thermal conductivity of 3rd conductor: \( k_3 = 135 \, \text{Js}^{-1} \text{m}^{-1} \text{K}^{-1} \)
• The area of cross-section of the 3rd conductor is twice that of the 1st conductor.

We are asked to find the value of \( \theta \) in the steady-state condition.

In steady state, the heat flow through all the conductors must be the same, since the system is in thermal equilibrium.

The formula for the heat transfer through a conductor is given by:

\[ Q = \frac{kA(T_1 - T_2)}{L} \]

Where:

• \( Q \) is the rate of heat transfer,
• \( k \) is the thermal conductivity,
• \( A \) is the cross-sectional area,
• \( T_1 \) and \( T_2 \) are the temperatures at the two ends, and
• \( L \) is the length of the conductor.

Since the lengths are the same and in steady state, the heat flow \( Q \) is constant, so we can write the equation for heat transfer for each conductor:

For the first conductor:

\[ Q_1 = \frac{k_1 A (T_1 - \theta)}{L} \]

For the second conductor:

\[ Q_2 = \frac{k_2 A (\theta - T_2)}{L} \]

For the third conductor (with double the cross-sectional area):

\[ Q_3 = \frac{k_3 (2A) (\theta - T_3)}{L} \]

Now, equating the heat transfer rates between the three conductors (since they are all in steady state and have the same heat flow), we get:

\[ Q_1 = Q_2 = Q_3 \]

Solving these equations, we find that the value of \( \theta \), which is the temperature at the junction of the conductors, is:

40°C

Answer:

The value of \( \theta \) is 40°C.

Answer 2

Step 2 — Write heat currents using thermal conductance: 
For a rod of thermal conductivity $k$, cross-sectional area $A$ and length $L$ between temperatures $T_a$ and $T_b$ the steady heat current is $$ \dot Q = G\,(T_a-T_b),\qquad G=\frac{kA}{L}\ \text{(thermal conductance)}. $$ Using this, \[ \dot Q_1 = G_1(100-\theta),\quad \dot Q_2 = G_2(100-\theta),\quad \dot Q_3 = G_3(\theta-0)=G_3\theta. \] Step 3 — Express conductances with given areas:
Since $A_1=A_2=A$ and $A_3=2A$ and all lengths equal, \[ G_1=\frac{k_1A}{L},\qquad G_2=\frac{k_2A}{L},\qquad G_3=\frac{k_3(2A)}{L}=\frac{2k_3A}{L}. \] Dividing the steady-state balance equation by the common factor $A/L$ eliminates geometry: $$ k_1(100-\theta)+k_2(100-\theta)=2k_3\,\theta. $$ Step 4 — Substitute numerical $k$ values:
Plug $k_1=60,\;k_2=120,\;k_3=135$ (units cancel in the algebraic ratio): \[ (60+120)(100-\theta)=2\times 135\;\theta. \] Simplify left and right: \[ 180(100-\theta)=270\theta. \] Step 5 — Solve for $\theta$:
\[ 18000-180\theta=270\theta \quad\Rightarrow\quad 18000=450\theta \quad\Rightarrow\quad \theta=\frac{18000}{450}=40. \] Final Answer: $\displaystyle \boxed{\theta=40^\circ\text{C}}$.