Question

An incandescent light bulb has a tungsten filament that is heated to a temperature $3 \times {10^3}K$ when an electric current passes through it. If the surface area of the filament is approximately ${10^{ - 4}}{m^2}$ and it has an emissivity of $0.32$ , the power radiated by the bulb is:
A. $147W$
B. $175W$
C. $200W$
D. $225W$

Answer 1

To solve this question, first we will rewrite the given facts of the question, and then write the power formulae with respect to the given information. We use Stefan–Boltzmann law in this question to find the power radiated by the bulb.

Complete step by step solution:
Given that-
Tungsten filament heated to a temperature, $T = 3 \times {10^3}K$
Surface area of the filament, $A = {10^{ - 4}}{m^2}$
Filament has an emissivity, $e = 0.32$
Now, according to the Stefan–Boltzmann constant;
The constant of proportionality, $\sigma = 5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$ .
Now, the power radiated by the bulb:
$P = eA\sigma {T^4} \\\ \Rightarrow P = 0.32 \times {10^{ - 4}} \times 5.67 \times {10^{ - 8}}{(3000)^4} \\\ \therefore P = 147W \\\$
Therefore, the power radiated by the bulb is $147W$ .
Hence, the correct option is A. $147W$

Note:
Stefan-Boltzmann Law The thermal energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by. For hot objects other than ideal radiators, the law is expressed in the form: where $e$ is the emissivity of the object ( $e = 1$ for ideal radiator).