Q.An iron bar ($L_1 = 0.1\ \text{m}$, $A_1 = 0.02\ \text{m}^{2}$, $K_1 = 79\ \text{W m}^{-1}\ \text{K}^{-1}$) and a brass bar ($L_2 = 0.1\ \text{m}$, $A_2 = 0.02\ \text{m}^{2}$, $K_2 = 109\ \text{W m}^{-1}\ \text{K}^{-1}$) are soldered end to end as shown in Fig. 10.16. The free ends of the iron bar and brass bar are maintained at $373\ \text{K}$ and $273\ \text{K}$ respectively. Obtain expressions for and hence compute

The problem gives two rods with lengths in the ratio $1:2$, thermal conductivities $K_1$ and $K_2$, and the same temperature difference $\Delta T = T_2 - T_1$ across each. The heat flow rates are equal. We need the ratio of their cross-sectional areas.

Concept & Intuition

Fourier’s law of heat conduction says the rate of heat flow through a rod is proportional to its cross-sectional area and thermal conductivity, and inversely proportional to its length. So if two rods have the same temperature difference and the same heat flow, the rod with larger length or smaller conductivity must compensate with a larger area. Here, rod B is twice as long as rod A, so to keep heat flow equal, its area must be adjusted accordingly.

1. Write Fourier’s law for each rod For steady-state conduction, the heat flow rate $Q$ is

where $K$ is thermal conductivity, $A$ is cross-sectional area, $L$ is length, and $(T_2 - T_1)$ is the temperature difference.

2. Set the heat flows equal Let rod A have length $L$, area $A_A$, conductivity $K_1$. Let rod B have length $2L$, area $A_B$, conductivity $K_2$. Both have the same $\Delta T = T_2 - T_1$. Equality of heat flow gives:

3. Cancel common factors …

The key concept here is Fourier’s law of heat conduction. For a rod of length $L$, cross-sectional area $A$, and thermal conductivity $k$, the rate of heat transfer (heat current) when a temperature difference $\Delta T$ is maintained across its ends is:

Since both rods have the same length $L$ and the same temperature difference $\Delta T = 30^\circ\text{C}$, the only factors that can differ are $k$ and $A$. The problem states the heat conduction rates are equal, so we set the expressions equal and solve for the unknown area.

1. Write the equality of heat currents For rod 1: $\displaystyle \frac{Q}{t} = \frac{k_1 A_1 \Delta T}{L}$ For rod 2: $\displaystyle \frac{Q}{t} = \frac{k_2 A_2 \Delta T}{L}$ Since the rates are equal:

2. Cancel common factors $L$ and $\Delta T$ are the same and non-zero, so they cancel:

3. Rearrange for $A_2$

4. Insert the given ratio …

The key concept here is steady-state heat conduction in series. When two materials are placed end-to-end and the system reaches a steady temperature profile, the rate of heat flow (heat current) through each material must be identical — otherwise heat would accumulate at the interface. For a slab of thickness $L$, cross-sectional area $A$, and thermal conductivity $k$, the heat current is given by Fourier’s law:

Since the problem states both blocks have the same cross-sectional area and the same thickness, the only variables are the conductivities and the temperature differences across each block.

1. Identify the temperature drops. For block A: hot end at $80^\circ\text{C}$, interface at $60^\circ\text{C}$, so

For block B: interface at $60^\circ\text{C}$, cold end at $20^\circ\text{C}$, so

2. Write the heat current for each block. Let the common cross-sectional area be $A$ and the common thickness be $L$. Then

where $k_B$ is the unknown conductivity of metal B.

3. Set the currents equal (steady state). Because the blocks are in series and no heat is lost at the interface, $$$$

In steady state, the heat current through slab 1 equals that through slab 2 (same area $A$):

$K_1 d_2(\theta_1-\theta)=K_2 d_1(\theta-\theta_2).$ …

Heat flows along the cylinder axis, so the inner cylinder and the outer shell are thermal conductors in parallel (same length $L$, same $\Delta T$).

• Core area: $A_1=\pi r^2$, conductivity $K_1$.
• Shell area: $A_2=\pi(2r)^2-\pi r^2=3\pi r^2$, conductivity $K_2$.

Parallel combination weights conductivity by area: …

Concept: two slabs of equal thickness in SERIES (heat flows through one and then the other) - thermal resistances add.

R_total = L/(K1 A) + L/(K2 A) = 2L/(K_eq A)

2/K_eq = 1/K1 + 1/K2

K_eq = 2 K1 K2 / (K1 + K2) …