708. Problem 49.14 (RHK) The filament of a particular 100W light bulb is cylindrical wire of tungsten 0.280 mm in diameter and 1.80 cm long. Assuming an emissivity of unity and ignoring absorption of energy by the filament from the surroundings, we have to calculate (a) the operating temperature of the filament; (b) the time it will take filament to cool by after the bulb is switched off. 
Solution: Click For PDF Version (a) We will determine first the operating temperature of the filament. The surface area of the filament will be , where is the diameter and is the length of the filament. As , and , the surface area of the filament is . Let the temperature of the filament of the light bulb when it is radiating at 100 W be . Using the StefanBoltzmann law we write the equation from which temperature of the filament can be determined. (b) For answering the second part of the problem, we will use the following physical properties of tungsten: Density, , Specific heat, . When the bulb has been switched off and the temperature of the filament is , let the temperature of the filament drop by in . The change in the heat energy of the filament will be Using the StefanBoltzmann law, we write . We thus obtain the following differential equation giving change in temperature of the filament with time. We integrate this equation, From the above equation we can calculate the time taken by the filament of the bulb to cool down by from when the temperature of the filament was 3250 K. We have the data
Therefore,
