Question

The surface area of a black body is \(5 \times 10^{-4}\) m\(^2\) and its temperature is 727\(^\circ\)C. The energy radiated by it per minute is (\(\sigma = 5.67 \times 10^{-8}\) J/m\(^2\)-s-K\(^4\))

• A. \(1.7 \times 10^3\) J
• B. \(2.5 \times 10^2\) J
• C. \(8 \times 10^3\) J
• D. \(3 \times 10^4\) J

Hint:

Always convert temperature to Kelvin before applying Stefan's law. \(T(\text{K}) = T(^\circ\text{C}) + 273\).

Answer 1

The Correct Option is A

Step 1: Basic Principle
By Stefan's law, power radiated: \(P = \sigma A T^4\). Energy in time \(t\) is \(E = Pt\).
Step 2: Solution Procedure:
\(T = 727 + 273 = 1000\) K, \(A = 5 \times 10^{-4}\) m\(^2\), \(t = 60\) s \[ P = 5.67 \times 10^{-8} \times 5 \times 10^{-4} \times (1000)^4 = 5.67 \times 10^{-8} \times 5 \times 10^{-4} \times 10^{12} = 28.35 \text{ W} \] \[ E = 28.35 \times 60 \approx 1701 \approx 1.7 \times 10^3 \text{ J} \]
Step 3: Required Answer:
Energy radiated per minute \(= \mathbf{1.7 \times 10^3}\) J.