Thermal Energy Transfer in Physics
Physics Formula Heat Transfer: Conduction, Convection, and Radiation
Heat transfer is a fundamental concept in physics that describes how thermal energy moves from one object or substance to another. There are three primary mechanisms of heat transfer: conduction, convection, and radiation. Each process operates through different physical principles and has specific formulas associated with it.
Understanding the formulas of heat transfer is crucial in engineering, meteorology, environmental science, and everyday applications like cooking and insulation. This article will explain the formulas, provide real-life examples, and demonstrate how to solve problems using each heat transfer method.
1. Heat Transfer by Conduction
Definition
Conduction is the transfer of heat through a material without any movement of the material itself. It usually occurs in solids, where particles vibrate and pass kinetic energy to neighboring particles.
Formula
The rate of heat transfer through conduction is given by Fourier’s law:
$$ Q = \frac{k \cdot A \cdot \Delta T \cdot t}{d} $$
Where:
- \( Q \): Heat transferred (Joules)
- \( k \): Thermal conductivity of the material (W/m·K)
- \( A \): Cross-sectional area (m²)
- \( \Delta T \): Temperature difference across the material (K or °C)
- \( d \): Thickness or length of heat path (m)
- \( t \): Time in seconds (s)
Example
Suppose a copper rod (thermal conductivity \( k = 400 \, \text{W/m·K} \)) has a cross-sectional area of \( 0.005 \, \text{m}^2 \), thickness \( 0.02 \, \text{m} \), and is subjected to a temperature difference of \( 50^\circ \text{C} \) for 10 minutes.
Convert time to seconds: \( t = 600 \, \text{s} \)
Now calculate: $$ Q = \frac{400 \cdot 0.005 \cdot 50 \cdot 600}{0.02} = \frac{60,000}{0.02} = 3,000,000 \, \text{J} $$
So, 3 MJ of heat is transferred through the rod.
2. Heat Transfer by Convection
Definition
Convection is the transfer of heat by the movement of fluids (liquids or gases). It can be natural (due to density differences) or forced (using a fan or pump).
Formula
The convective heat transfer formula is:
$$ Q = h \cdot A \cdot \Delta T \cdot t $$
Where:
- \( Q \): Heat transferred (Joules)
- \( h \): Convective heat transfer coefficient (W/m²·K)
- \( A \): Surface area in contact with fluid (m²)
- \( \Delta T \): Temperature difference between surface and fluid (K or °C)
- \( t \): Time (s)
Example
Air at 25°C flows over a hot plate maintained at 75°C. The convective heat transfer coefficient is \( h = 30 \, \text{W/m}^2·\text{K} \), and the surface area is \( A = 1 \, \text{m}^2 \). Calculate the heat transferred in 5 minutes.
Time: \( t = 5 \times 60 = 300 \, \text{s} \)
Temperature difference: \( \Delta T = 75 - 25 = 50^\circ \text{C} \)
Apply the formula: $$ Q = 30 \cdot 1 \cdot 50 \cdot 300 = 450,000 \, \text{J} $$
So, 450 kJ of heat is transferred via convection.
3. Heat Transfer by Radiation
Definition
Radiation is the transfer of heat through electromagnetic waves, typically infrared. Unlike conduction and convection, radiation does not require a medium—it can occur in a vacuum.
Formula
Heat transfer by radiation is described by the Stefan-Boltzmann law:
$$ Q = \epsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \cdot t $$
Where:
- \( Q \): Heat transferred (Joules)
- \( \epsilon \): Emissivity of the surface (0 to 1)
- \( \sigma \): Stefan-Boltzmann constant \( = 5.67 \times 10^{-8} \, \text{W/m}^2·\text{K}^4 \)
- \( A \): Surface area (m²)
- \( T \): Temperature of the object (K)
- \( T_0 \): Temperature of surroundings (K)
- \( t \): Time in seconds (s)
Example
A black body (\( \epsilon = 1 \)) with a surface area of \( 0.1 \, \text{m}^2 \) is at \( 500 \, \text{K} \), and the surrounding is at \( 300 \, \text{K} \). Calculate the heat lost by radiation in 2 minutes.
Convert time: \( t = 120 \, \text{s} \)
Apply the formula: $$ Q = 1 \cdot 5.67 \times 10^{-8} \cdot 0.1 \cdot (500^4 - 300^4) \cdot 120 $$ Calculate: $$ 500^4 = 6.25 \times 10^{10}, \quad 300^4 = 8.1 \times 10^9 \\ \Delta T^4 = 5.44 \times 10^{10} $$ Now: $$ Q = 5.67 \times 10^{-8} \cdot 0.1 \cdot 5.44 \times 10^{10} \cdot 120 \\ Q \approx 3693.4 \, \text{J} $$ So approximately 3.7 kJ is radiated.
Comparison of the Three Methods
| Method | Medium | Formula | Example |
|---|---|---|---|
| Conduction | Solids | \( Q = \frac{kA\Delta T t}{d} \) | Metal rod heating |
| Convection | Liquids, Gases | \( Q = hA\Delta T t \) | Boiling water |
| Radiation | Vacuum or air | \( Q = \epsilon \sigma A (T^4 - T_0^4)t \) | Sunlight warming a surface |
Applications in Real Life
- Conduction: Cooking utensils, thermal insulation, engine parts.
- Convection: Air conditioning, ocean currents, heating systems.
- Radiation: Solar panels, thermal cameras, greenhouse effect.
Practice Problem
A stainless-steel pan with bottom area \( A = 0.03 \, \text{m}^2 \), thickness \( d = 0.005 \, \text{m} \), and thermal conductivity \( k = 16 \, \text{W/m·K} \) is placed on a stove. The bottom of the pan is at \( 180^\circ \text{C} \), and the inside is at \( 100^\circ \text{C} \). How much heat is transferred in 5 minutes?
Convert time: \( t = 300 \, \text{s} \)
Temperature difference: \( \Delta T = 80^\circ \text{C} \)
Using: $$ Q = \frac{kA\Delta T t}{d} = \frac{16 \cdot 0.03 \cdot 80 \cdot 300}{0.005} = 2,304,000 \, \text{J} $$ So, about 2.3 MJ of heat is conducted.
Conclusion
The study of heat transfer is vital for understanding how thermal energy moves in different environments and materials. Conduction, convection, and radiation each have unique characteristics, formulas, and applications. By mastering these concepts and practicing the formulas, one can analyze and solve real-world thermal problems effectively.
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