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define-conduction-and-thermal-conductivity

🚀 Conduction is the process of heat transfer through a material without any movement of the material itself. It occurs when there is a temperature difference within a solid or between solids in contact. Thermal conductivity is a property of materials that quantifies their ability to conduct heat. It is defined as the amount of heat that passes through a unit area of a material in a unit time when there is a temperature gradient across that area.

Theory Explanation

Understanding Conduction

Conduction occurs at the microscopic level when fast-moving particles collide with slower-moving particles, transferring energy. This process continues until thermal equilibrium is reached, meaning the temperature is uniform throughout the material.

Defining Thermal Conductivity

Thermal conductivity (k) is defined as the quantity of heat (Q) that passes through a material of thickness (d) in a time (t) when there is a temperature difference (ΔT) across it. The formula is given by Fourier's law of heat conduction: Q = k * A * (ΔT/d), where A is the area through which heat is conducted.

\[ Q = k \cdot A \cdot \frac{\Delta T}{d} \]

Key Points

  • 🎯 Conduction is a mode of heat transfer that occurs in solids.
  • 🎯 Thermal conductivity is a material property that indicates how well a material conducts heat.
  • 🎯 The higher the thermal conductivity, the better the material is at conducting heat.

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Examples:💡

Example 1: Calculate the heat conducted through a metal rod of length 2 m, cross-sectional area 0.01 m², with a thermal conductivity of 200 W/m·K, and a temperature difference of 50 K between its ends.

Solution:

Step 1: Identify the given values: Length (d) = 2 m, Area (A) = 0.01 m², Thermal conductivity (k) = 200 W/m·K, Temperature difference (ΔT) = 50 K.

Step 2: Use Fourier's law to calculate the heat conducted: Q = k * A * (ΔT/d).

\[ Q = 200 \cdot 0.01 \cdot \frac{50}{2} = 500 W \]

Example 2: A copper wire of length 1.5 m and cross-sectional area 0.0005 m² has a thermal conductivity of 385 W/m·K. If one end is heated to 100°C and the other end is at 20°C, calculate the heat transfer per second.

Solution:

Step 1: Given: Length (d) = 1.5 m, Area (A) = 0.0005 m², Thermal conductivity (k) = 385 W/m·K, Temperature difference (ΔT) = 100 - 20 = 80 K.

Step 2: Apply Fourier's law: Q = k * A * (ΔT/d).

\[ Q = 385 \cdot 0.0005 \cdot \frac{80}{1.5} = 10.27 W \]

Common Mistakes

  • Mistake: Confusing conduction with convection and radiation, which are other modes of heat transfer.

    Correction: Remember that conduction occurs in solids and involves direct contact, while convection involves fluid movement and radiation involves energy transfer through electromagnetic waves.

  • Mistake: Incorrectly applying the formula for thermal conductivity, especially mixing up the units.

    Correction: Always check that the units are consistent, especially for area (m²), length (m), and temperature difference (K).

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