Thermal Radiation Formula in 60 Sec ⚡

 

❓ Concept Question

Why does a hot filament emit large amount of radiation, and how do we calculate the power radiated from a hot body?

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🖼 Concept Image

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✍️ Short Concept

Any hot body continuously emits thermal radiation.

The amount of power emitted depends strongly on temperature and emissivity.

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🔷 Step 1 — Blackbody Radiation Idea 💯

Every hot body emits energy in the form of radiation.

Power emitted is proportional to:

$$T^4$$

So,

Small temperature increase → huge increase in radiation.

This explains why filaments glow intensely.

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🔷 Step 2 — Stefan–Boltzmann Law

Radiated power from a surface:

$$P = e \sigma A T^4$$

Where:

$$e = \text{emissivity}$$
$$\sigma = 5.67 \times 10^{-8}$$
$$A = \text{surface area}$$
$$T = \text{absolute temperature (Kelvin)}$$

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🔷 Step 3 — Role of Emissivity

Emissivity measures how efficiently a surface radiates energy.

$$e = 1$$

Perfect blackbody.

For real objects:

$$e < 1$$

Example: bulb filament.

👉 Real materials never behave like ideal blackbodies.

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🔷 Step 4 — Radiating Area of Wire

For a thin cylindrical wire:

Radiating surface area:

$$A = \pi d l$$

Where:

d = diameter
l = length

⚠️ Ends of wire are usually ignored.

Only curved surface area contributes significantly.

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🔷 Step 5 — JEE Golden Traps

❌ Temperature in Celsius

❌ Forgetting unit conversions

❌ Assuming emissivity = 1

Correct approach:

👉 Always convert temperature to Kelvin.

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✅ Final Takeaway

Thermal radiation follows:

$$\boxed{P = e \sigma A T^4}$$

Higher temperature → dramatically larger radiation.

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⭐ Golden JEE Insight

Temperature dependence is extremely strong:

$$T^4$$

If temperature doubles:

Radiated power increases by 16 times.