Solving Radiation Heat Transfer with the Stefan-Boltzmann Law

Hello! Welcome to our next lesson in heat transfer.

So far, we've explored heat transfer through conduction (Fourier's Law) and convection (Newton's Law of Cooling). We've even combined them using the thermal resistance model. Today, we turn our attention to the third and final mode of heat transfer: radiation. This is the process that allows the Sun to heat the Earth across the vacuum of space.

Our goal for this lesson is to apply the Stefan-Boltzmann law to solve radiation heat transfer problems between bodies. We'll start with the fundamental equation, tackle the crucial distinction between total and net power radiated, and then apply these concepts to a classic JEE Advanced problem involving multiple interacting bodies.

1. The Stefan-Boltzmann Law: Fundamentals

Unlike conduction and convection, which require a medium, thermal radiation is the transfer of energy via electromagnetic waves. Any object with a temperature above absolute zero (0 K) emits thermal radiation. The Stefan-Boltzmann law quantifies this phenomenon.

To start, let's establish the key definitions and the main formula.

Stefan Boltzmann Law - Definition and Equation

This article from Vedantu provides an excellent overview of the Stefan-Boltzmann law, the concept of a blackbody, and the variables involved in the equation.

Please read the sections titled 'Blackbody Radiation,' 'What is Absolute Temperature?,' 'Application of Stefan-Boltzmann Law,' and 'What is Stephan-Boltzmann Constant?'. Focus on understanding: The definition of a blackbody (a perfect absorber and emitter). Why absolute temperature (Kelvin) is essential in this law. The formula for a blackbody versus a real object (grey body), and the role of emissivity. The definition, value, and units of the Stefan-Boltzmann constant (σ).

As you've just read, the law describes the total power radiated by an object. Let's formalize this.

Stefan-Boltzmann Law Equation Breakdown — This image clearly breaks down the Stefan-Boltzmann law, defining each variable: Power (P), emissivity (e), the Stefan-Boltzmann constant (σ), surface area (A), and absolute temperature (T).

The total power $P$ (energy per second, in Watts) radiated from the surface of an object is given by:
$P = ϵ σ A T^{4}$
Where:

$ϵ$ (epsilon) is the emissivity of the surface, a dimensionless number between 0 and 1. It represents how effectively the surface radiates energy compared to a perfect radiator. For a perfect blackbody, $ϵ = 1$ . For a real (or grey) body, $ϵ < 1$ .
$σ$ (sigma) is the Stefan-Boltzmann constant, approximately $5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ .
$A$ is the surface area of the object in $m^{2}$ .
$T$ is the absolute temperature of the surface in Kelvin.

A critical point that trips up many students is the $T^{4}$ dependence. A small change in temperature leads to a very large change in radiated power. Also, always remember to convert temperatures from Celsius to Kelvin ( $T_{K} = T_{C} + 273.15$ ).

2. Net Heat Transfer: The Key to Problem-Solving

An object doesn't just radiate energy; it also absorbs energy from its surroundings. The formula $P = ϵ σ A T^{4}$ only gives the power emitted by the object. In most problems, especially in JEE Advanced, you'll be asked for the net rate of heat transfer.

If an object at temperature $T$ is in an environment with a surrounding temperature of $T_{0}$ , it emits power based on $T^{4}$ and absorbs power based on $T_{0}^{4}$ .

According to Kirchhoff's law of thermal radiation, the absorptivity $α$ of a body is equal to its emissivity $ϵ$ at a given temperature. Therefore:

Power Emitted: $P_{e mi tt e d} = ϵ σ A T^{4}$
Power Absorbed: $P_{ab sor b e d} = α σ A T_{0}^{4} = ϵ σ A T_{0}^{4}$

The net rate of energy loss is the difference between these two:
$P_{n e t} = P_{e mi tt e d} - P_{ab sor b e d} = ϵ σ A (T^{4} - T_{0}^{4})$

This is one of the most important formulas in this topic. Forgetting the $- T_{0}^{4}$ term is a very common mistake.

Stefan Boltzmann Law - BYJU'S

Now, let's focus on this crucial concept of net heat transfer. This BYJU'S resource clearly presents the formula for net radiated power and provides a straightforward example.

Read the sections 'What Is Stefan Boltzmann Law?' and 'Problems with Stefan Boltzmann Law'. Pay close attention to equation (3) for net energy radiated, Δu = eσA [T⁴ – T₀⁴], and follow the first worked example that calculates this value.

Now, let's put this into practice.

Test your understanding!

A solid copper sphere (radius = 10 cm, emissivity = 0.8) is heated to 727°C and suspended in a large room where the walls are at 27°C. What is the initial net rate of heat loss from the sphere? ( $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ )

Show answer

First, convert all temperatures to Kelvin:

Temperature of the sphere, $T = 727 + 273 = 1000 K$
Temperature of the room (surroundings), $T_{0} = 27 + 273 = 300 K$

Next, calculate the surface area of the sphere:

$A = 4 π r^{2} = 4 π (0.1 m)^{2} = 0.04 π m^{2} \approx 0.1257 m^{2}$

Now, apply the net heat transfer formula:
$P_{n e t} = ϵ σ A (T^{4} - T_{0}^{4})$
$P_{n e t} = (0.8) \times (5.67 \times 1 0^{- 8}) \times (0.04 π) \times (100 0^{4} - 30 0^{4})$
$P_{n e t} \approx (4.536 \times 1 0^{- 8}) \times (0.1257) \times (1 \times 1 0^{12} - 8.1 \times 1 0^{9})$
$P_{n e t} \approx (5.70 \times 1 0^{- 9}) \times (1 \times 1 0^{12} - 0.0081 \times 1 0^{12})$
$P_{n e t} \approx (5.70 \times 1 0^{- 9}) \times (0.9919 \times 1 0^{12})$
$P_{n e t} \approx 5654 W or 5.65 kW$

The key was using the $(T^{4} - T_{0}^{4})$ term to find the net rate of heat loss.

3. Advanced Application: Heat Transfer Between Parallel Plates

Now let's apply this law to a classic multi-body problem you might encounter in the JEE Advanced exam: heat transfer between multiple parallel plates.

Imagine two large parallel plates at temperatures $T_{i}$ and $T_{f}$ . If we place several other parallel plates between them, what is the final steady-state heat flux, and what are the temperatures of the intermediate plates?

The key insight is that in steady state, the net heat flow $H$ (in Watts) must be the same through each gap between adjacent plates.

[JEE ADVANCE] BLACK BODY RADIATION AND PARALLEL PLATES [ADVANCE PROBLEMS IN SCHOOL PHYSICS]

This video from Indian School of Physics tackles the parallel plates problem head-on. It demonstrates a powerful and systematic method for solving it, which you'll find very useful.

Watch the video from 01:20 to 06:10. The core logic is that the net heat flow (H) is constant in every gap. Follow the presenter's derivation as he writes an equation for H for each gap and then sums them. This allows him to find the overall H and the temperature of any intermediate plate 'j'.

Let's break down the logic from the video, which is very similar to an algorithm. Consider two outer plates at $T_{i}$ and $T_{f}$ , with $n$ identical black plates ( $ϵ = 1$ ) placed in between. Let the temperatures of the intermediate plates be $T_{1}, T_{2}, ..., T_{n}$ .

The net heat flow $H$ in each of the $n + 1$ gaps is:
$H = σ A (T_{i}^{4} - T_{1}^{4}) (Gap 1)$
$H = σ A (T_{1}^{4} - T_{2}^{4}) (Gap 2)$
$⋮$
$H = σ A (T_{n}^{4} - T_{f}^{4}) (Gap n+1)$

Since $H$ is the same for all, we can write:
$\frac{H}{σ A} = T_{i}^{4} - T_{1}^{4}$
$\frac{H}{σ A} = T_{1}^{4} - T_{2}^{4}$
$⋮$
$\frac{H}{σ A} = T_{n}^{4} - T_{f}^{4}$

If we sum these $n + 1$ equations, all the intermediate temperature terms ( $T_{1}^{4}, T_{2}^{4}, ...$ ) cancel out telescopically:
$(n + 1) \frac{H}{σ A} = T_{i}^{4} - T_{f}^{4}$
Solving for the net heat flow $H$ :
$H = \frac{σ A}{n + 1} (T_{i}^{4} - T_{f}^{4})$

This is a powerful result. Inserting $n$ shields between the two plates reduces the heat transfer by a factor of $n + 1$ . This is directly analogous to adding $n + 1$ identical resistors in series in an electrical circuit, where the total resistance becomes $n + 1$ times the individual resistance.

To find the temperature of the $j$ -th plate, $T_{j}$ , you can sum just the first $j + 1$ equations (from the gap before plate 1 to the gap after plate j):
$(j + 1) \frac{H}{σ A} = T_{i}^{4} - T_{j + 1}^{4}$
You can then substitute the expression for $H$ and solve for the unknown temperature.

Conclusion

In this lesson, we've focused on quantifying heat transfer by radiation. You are now equipped to solve a wide range of problems using the Stefan-Boltzmann law.

Key Takeaways:

The total power radiated by a body is proportional to the fourth power of its absolute temperature: $P = ϵ σ A T^{4}$ .
For problems involving heat exchange with the surroundings, you must calculate the net power, which accounts for both emission and absorption: $P_{n e t} = ϵ σ A (T^{4} - T_{0}^{4})$ .
In steady-state problems with multiple bodies, the principle of constant heat flux provides a systematic way to set up and solve for unknown temperatures or heat flow rates.

Preview of the Next Lesson:
The Stefan-Boltzmann law tells us how much energy is radiated in total. However, it doesn't tell us anything about the distribution of this energy across different wavelengths (i.e., the "color" of the radiation). In our next lesson, we will explore Wien's displacement law, which answers this very question and explains why objects glow red-hot, then yellow-hot, then white-hot as their temperature increases. Many advanced problems require you to use both laws together.

Recommended resources

1. The Stefan-Boltzmann Law: Fundamentals

2. Net Heat Transfer: The Key to Problem-Solving

3. Advanced Application: Heat Transfer Between Parallel Plates

Conclusion