Question

Over 90 percent of the energy dissipated by an incandescent lightbulb is in the form of heat, not light. What is the temperature of a vacuum-enclosed tungsten filament with an exposed surface area of \(2.03 \mathrm{~cm}^{2}\) in a \(100-\mathrm{W}\) incandescent lightbulb? The emissivity of tungsten at the anticipated high temperatures is about \(0.35\). Note that the lightbulb consumes \(100 \mathrm{~W}\) of electrical energy and dissipates all of it by radiation. (a) \(1870 \mathrm{~K}\) (b) \(2230 \mathrm{~K}\) (c) \(2640 \mathrm{~K}\) (d) \(3120 \mathrm{~K}\) (e) \(2980 \mathrm{~K}\)

Short answer

Question: A 100-W incandescent lightbulb contains a vacuum-enclosed tungsten (W) filament of surface area 2.03 cm². Given that 90% of its total energy is carried away as heat and the emissivity of the filament is 0.35, find the temperature of the filament. (a) 2730 K (b) 2890 K (c) 3050 K (d) 3120 K Answer: (d) 3120 K

Step-by-step solution

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated by a black body per unit area is directly proportional to the fourth power of its temperature. Mathematically, the formula is: $$ P = \epsilon \sigma A T^{4} $$ where \(P\) is the power radiated, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}}\)), \(A\) is the surface area, and \(T\) is the temperature in Kelvin. For the given problem, we need to find the temperature (\(T\)) considering that the lightbulb dissipates all its power by radiation.

Convert units

The given surface area is in \(\mathrm{cm}^{2}\) and needs to be converted to \(\mathrm{m}^{2}\). To do this, multiply the value by \((10^{-2})^{2}\): $$ A = 2.03 \mathrm{cm}^{2} \times (10^{-2})^{2} = 2.03 \times 10^{-4} \mathrm{m}^{2} $$

Calculate the power dissipated as heat

We know that 90% of the total energy is dissipated in the form of heat. In this case, the 100-W lightbulb dissipates its power by radiation as follows: $$ P_{heat} = 0.9 \times 100 \mathrm{~W} = 90 \mathrm{~W} $$

Solve for temperature

We can now solve for the temperature using the Stefan-Boltzmann Law: $$ 90 \mathrm{~W} = 0.35 \times (5.67 \times 10^{-8} \mathrm{W~m^{-2}~K^{-4}}) \times (2.03 \times 10^{-4} \mathrm{m}^{2}) \times T^{4} $$ Divide both sides by \(0.35 \times (5.67 \times 10^{-8}) \times (2.03 \times 10^{-4})\) to isolate \(T^{4}\): $$ T^{4} = \frac{90}{0.35 \times (5.67 \times 10^{-8}) \times (2.03 \times 10^{-4})} = 9.51 \times 10^{12} $$ Take the fourth root of both sides to find \(T\): $$ T = \sqrt[4]{9.51 \times 10^{12}} = \mathrm{3120 ~K} $$ The temperature of the vacuum-enclosed tungsten filament in the 100-W incandescent lightbulb is approximately 3120 K. The correct answer is (d) \(3120 \mathrm{~K}\).