Question

A spherical perfect black body of radius 10 cm is maintained at \( 727^{\circ}C \). The total power radiated from it is (approximately) Stefan-Boltzmann constant, \( \sigma=5.67\times10^{-8} \, Wm^{-2}K^{-4} \)

• A. \( 7120 \, W \)
• B. \( 7270 \, W \)
• C. \( 1000 \, W \)
• D. \( 7000 \, W \)

Hint:

In radiation problems, always combine powers of 10 first (like \( (10^3)^4 = 10^{12} \)). It reduces calculation errors and saves time in exams.

Answer 1

The Correct Option is A

Step 1: Recall the radiation law.
A perfect black body radiates power given by the Stefan Boltzmann law: \[ P = \sigma A T^{4} \] Here $A$ is the surface area and $T$ is the absolute temperature in kelvin.
Step 2: Change temperature to kelvin.
\[ T = 727 + 273 = 1000\ \text{K} \]
Step 3: Find the surface area of the sphere.
With radius $R = 10$ cm $= 0.1$ m: \[ A = 4\pi R^{2} = 4\pi(0.1)^{2} = 0.04\pi\ \text{m}^{2} \]
Step 4: Note the fourth power of temperature.
\[ T^{4} = (10^{3})^{4} = 10^{12} \]
Step 5: Put everything into the formula.
\[ P = (5.67\times10^{-8})(0.04\pi)(10^{12}) = 5.67 \times 0.04\pi \times 10^{4} \]
Step 6: Work out the number.
Using $0.04\pi \approx 0.1256$: \[ P = 5.67 \times 0.1256 \times 10^{4} \approx 7120\ \text{W} \] \[ \boxed{P \approx 7120\ \text{W}} \]