Related Formula
Stefan-Boltzmann Law for radiated thermal power:
P = ε σ A T⁴$$P = \epsilon \sigma A T^4$$
Surface area A$A$ of a cylindrical wire of diameter d$d$ and length L$L$ is:
A = π d L$A = \pi d L$
Core Logic
Convert given parameters to standard SI units:
- L = 10 ~cm = 0.1 ~m$L = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
- d = 0.5 ~mm = 0.5 × 10⁻³ ~m$d = 0.5 \mathrm{~mm} = 0.5 \times 10^{-3} \mathrm{~m}$
- T = 1727 + 273.15 = 2000 ~K$T = 1727 + 273.15 = 2000 \mathrm{~K}$
- P = 94.2 ~W$P = 94.2 \mathrm{~W}$
- σ = 6.0 × 10⁻⁸ ~W · m⁻² · K⁻⁴$\sigma = 6.0 \times 10^{-8} \mathrm{~W \cdot m^{-2} \cdot K^{-4}}$
Step 1: Express Radiated Power and Emissivity
First, calculate the surface area A$A$:
A = 3.14 × (0.5 × 10⁻³) × 0.1 = 1.57 × 10⁻⁴ ~m²$$A = 3.14 \times (0.5 \times 10^{-3}) \times 0.1 = 1.57 \times 10^{-4} \mathrm{~m}^2$$
Substitute A$A$, σ$\sigma$, and T$T$ into Stefan's formula:
94.2 = ε × (6.0 × 10⁻⁸) × [3.14 × (0.5 × 10⁻³) × (10 × 10⁻²)] × (2000)⁴$$94.2 = \epsilon \times (6.0 \times 10^{-8}) \times [3.14 \times (0.5 \times 10^{-3}) \times (10 \times 10^{-2})] \times (2000)^4$$
Calculate temperature term:
(2000)⁴ = 1.6 × 10¹²$$(2000)^4 = 1.6 \times 10^{12}$$
94.2 = ε × (6 × 10⁻⁸) × (1.57 × 10⁻⁴) × (16 × 10¹²)$$94.2 = \epsilon \times (6 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (16 \times 10^{12})$$
94.2 = ε × 6 × 1.57 × 16 × 1 = ε × 150.72$$94.2 = \epsilon \times 6 \times 1.57 \times 16 \times 1 = \epsilon \times 150.72$$
ε = (94.2)/(150.72) = (5)/(8)$$\epsilon = \frac{94.2}{150.72} = \frac{5}{8}$$
Thus, (x)/(8) = (5)/(8) x = 5$\frac{x}{8} = \frac{5}{8} \implies x = 5$.
Pattern Recognition
Sees: Bulb filament heat radiation equation.
Shortcut: Simplify the multiplication with factors of 10 first. T=2000 ~K$T=2000 \mathrm{~K}$ has four zeros, so T⁴$T^4$ adds 10¹²$10^{12}$ which cancels the 10⁻⁸$10^{-8}$ and 10⁻⁴$10^{-4}$ from area and constant. The coefficient equation directly yields the ratio ε = 5/8$\epsilon = 5/8$.
Chapter Mix
Class 11 Physics: Thermal Properties of Matter
