An aluminium rod and a copper rod of equal length 1 ·0 m and cross-sectional area 1 cm^2 are w

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An aluminium rod and a copper rod of equal length 1 ·0 m...

An aluminium rod and a copper rod of equal length $1 \cdot 0 \mathrm{~m}$ and cross-sectional area $1 \mathrm{~cm}^{2}$ are welded together as shown in figure (28-E3). One end is kept at a temperature of $20^{\circ} \mathrm{C}$ and the other at $60^{\circ} \mathrm{C} .$ Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium $=200 \mathrm{Wm}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and of copper $=390 \mathrm{~W} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}$

Key Concepts

Composite Conduction (Parallel Conduction Paths)

When two or more conductive elements are arranged between the same temperature reservoirs, they act in parallel, and the total rate of heat transfer is the sum of the rates through each individual path. This concept is crucial in analyzing systems where different materials contribute simultaneously to the overall heat conduction.

Fourier’s Law of Heat Conduction

This law states that the heat transfer rate through a material is proportional to the temperature gradient and the cross-sectional area, and inversely proportional to the length of the material. It is commonly expressed as Q/t = (kA ?T) / L, where Q/t is the rate of heat transfer, k is the thermal conductivity, A is the cross-sectional area, ?T is the temperature difference, and L is the length over which the temperature difference exists.

Thermal Conductivity

Thermal conductivity is a material property that indicates the ability of a material to conduct heat. Materials with higher thermal conductivity transfer heat more efficiently, meaning that for a given temperature difference and geometry, they will allow a greater rate of heat flow compared to materials with lower thermal conductivity.

Rate of Heat Transfer

The rate of heat transfer represents the amount of heat energy that is conducted per unit time, typically measured in watts. In conduction problems, it is calculated using Fourier’s law and depends on factors such as the thermal conductivities of the materials, the cross-sectional areas, the temperature difference across the materials, and the distance between the hot and cold regions.

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