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

Thermal Conduction Through Bars

Imagine holding a metal rod with one end in a fire. Within seconds, the other end gets hot — even though it never touched the flame. Something travelled through the rod. That something is heat, and the process is thermal conduction.

The Intuition: What's Actually Happening?

At the microscopic level, atoms in the hot end vibrate violently. These vibrations bump into neighbouring atoms, passing energy along like a line of dominoes. In metals, free electrons also carry energy quickly — that's why a metal spoon feels cold at first (it steals heat from your hand fast) and heats up fast at the other end.

The key idea: heat flows from the hotter region to the colder region, and the rate of flow depends on three things:

• How big the temperature difference is
• How thick the bar is (cross-sectional area)
• What the bar is made of (its thermal conductivity)

The Precise Statement: Fourier's Law of Heat Conduction

For a bar of uniform cross-section, the rate of heat transfer $Q/t$ (in joules per second, or watts) is given by:

Where:

• $Q/t$ = rate of heat flow (W)
• $k$ = thermal conductivity of the material (W/m·K) — a property like "how good is this at conducting heat"
• $A$ = cross-sectional area of the bar (m²)
• $T_1 - T_2$ = temperature difference between the hot end and cold end (K or °C)
• $L$ = length of the bar (m)

Why It Makes Sense

Think of the bar as a pipe for heat. A wider pipe (larger $A$) lets more heat through. A longer pipe (larger $L$) makes it harder for heat to travel — like walking a longer corridor. A bigger temperature difference ($T_1 - T_2$) is like a steeper hill — heat flows faster downhill.

The material constant $k$ is the "conductivity" of the bar. Copper has $k \approx 400$ W/m·K, wood has $k \approx 0.1$ W/m·K. That's why a copper rod feels cold to touch (it pulls heat from your hand) while wood at the same temperature feels neutral.

A Worked Example

A copper rod ($k = 400$ W/m·K) is 0.5 m long with cross-sectional area $2 \times 10^{-4}$ m². One end is at 100°C, the other at 20°C. Find the heat flow.

So 12.8 joules of heat flow through the rod every second. …