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: Use Stefan's law.
A black body radiates power $P = \sigma A T^{4}$, so the energy loss per second grows fast with temperature.

Step 2: A blackened wire is a black body.
Its emissivity is about 1, so the full Stefan formula applies. The rate of losing energy equals this radiated power.

Step 3: List the values.
$A = 10^{-5}$ m$^{2}$, $T = 3000$ K, $\sigma = 5.67\times 10^{-8}$ W m$^{-2}$ K$^{-4}$.

Step 4: Work out $T^{4}$.
\[ (3000)^{4} = 81\times 10^{12} \]
Step 5: Multiply everything.
\[ P = 5.67\times 10^{-8}\times 10^{-5}\times 81\times 10^{12} \]\[ = 5.67\times 81\times 10^{-1} = 459.27\times 10^{-1} \]
Step 6: Final value.
\[ P \approx 45.9 \text{ W} \approx 46 \text{ W} \]So the wire loses energy at about $46$ W, which is option 2.
\[ \boxed{P \approx 46 \text{ W}} \]