Question

A blackened platinum wire of surface area $10^{-5}\text{m}^{2}$ is maintained at temperature of 3000K. At what rate the wire is loosing energy [Stefan-Boltzmann constant $\sigma=5.67\times10^{-8}\text{W m}^{-2}\text{K}^{-4}$]

• A. 50 W
• B. 46 W
• C. 76 W
• D. 38 W

Hint:

Pay careful attention to powers of 10: $(3000)^4 = 3^4 \times 10^{12} = 81 \times 10^{12}$. Keeping track of exponents saves computation time!

Answer 1

The Correct Option is B

Step 1: Concept
According to Stefan's Law, the total energy radiated per second (power $P$) by a perfectly black body is directly proportional to the fourth power of its absolute temperature: $P = \sigma A T^4$, where $A$ is the surface area, $T$ is the absolute temperature, and $\sigma$ is the Stefan-Boltzmann constant.

Step 2: Meaning
A "blackened" wire behaves as an ideal blackbody radiator with an emissivity value $e \approx 1$. The rate of losing energy corresponds exactly to the radiant thermal power.

Step 3: Analysis
Substitute the given numeric values into Stefan's formula: $A = 10^{-5} \text{ m}^2$, $T = 3000 \text{ K}$, and $\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2}\text{K}^{-4}$. Therefore, $P = (5.67 \times 10^{-8}) \times 10^{-5} \times (3000)^4 = 5.67 \times 10^{-13} \times (81 \times 10^{12}) = 5.67 \times 81 \times 10^{-1} = 459.27 \times 10^{-1} \approx 45.92 \text{ W}$, which rounds closely to 46 W.

Step 4: Conclusion
Hence, the precise rate of thermal energy emission from the wire is 46 W.

Final Answer: (B)