Emissivity in 60 Sec 🔥 | Stefan Law Shortcut

 

❓ Question

A wire of:

Length = 10 cm
Diameter = 0.5 mm
Temperature = 1727°C
Power radiated = 94.2 W

Find the emissivity (e) of the wire.

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🖼 Question Image

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✍️ Short Solution

This is a Stefan–Boltzmann Law question.

Golden formula:

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

👉 Convert temperature to Kelvin
👉 Find curved surface area
👉 Substitute and solve for e

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🔷 Step 1 — Convert All Units (MOST IMPORTANT 💯)

Length:

$$L = 10\,\text{cm} = 0.1\,\text{m}$$

Diameter:

$$d = 0.5\,\text{mm} = 0.0005\,\text{m}$$

Radius:

$$r = 0.00025\,\text{m}$$

Temperature:

$$T = 1727 + 273 = 2000\,K$$

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🔷 Step 2 — Surface Area of Wire

Wire is cylindrical → Use curved surface area:

$$A = 2\pi r L$$
$$A = 2\pi (0.00025)(0.1)$$
$$A = 1.57 \times 10^{-4} \, m^2$$

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🔷 Step 3 — Apply Stefan’s Law

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

Given:

$$P = 94.2$$
$$\sigma = 5.67 \times 10^{-8}$$
$$T^4 = (2000)^4 = 16 \times 10^{12}$$

Substitute:

$$94.2 = e \times (5.67 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (16 \times 10^{12})$$

Simplify RHS:

$$\approx e \times 142.3$$

So,

$$e = \frac{94.2}{142.3}$$
$$e \approx 0.66$$

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✅ Final Answer

$$\boxed{e \approx 0.66}$$

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⭐ Golden JEE Insight

Whenever filament or wire radiates:

1️⃣ Always convert °C → K
2️⃣ Use curved surface area only
3️⃣ Remember $T^4$ grows VERY fast

⚡ Small temperature change → huge power change.