Question

A 19th century lab technician is testing possible metals for bulb filaments. A tantalum filament with a surface area of 0.370 mm2 and an emissivity of 0.965 radiates 0.80 W of light. Determine the filament's temperature (in K). The melting point for tantalum is 3269 K.

Answer 1

Explanation:

The given data is as follows.

       A = 0.370 $mm^{2}$ = $3.7 \times 10^{-7} m^{2}$

    emissivity ($\sigma$) = 0.965,     E = 0.80 W

As we know that,

            $\frac{E}{A} = \sigma \times s \times T^{4}$

where,   s = Stefan-Boltzmann constant = $5.6696 \times 10^{-8} w/m^{2} K$

Now, putting the given values into the above formula as follows.

        $\frac{E}{A} = \sigma \times s \times T^{4}$  

        $\frac{0.80}{0.370} = 0.965 \times 5.6696 \times 10^{-8} \times T^{4}$

            2.162 = $5.471164 \times 10^{-8} \times T^{4}$

          39516271.13 = $T^{4}$

             T = 79.28 K

Thus, we can conclude that the filament's temperature is 79.28 K.