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Exercise:
A typical incandescent light bulb with a power of PO has a tungsten filament with length LO and diameter dO. abcliste abc When the light bulb is on the electric power is converted to radiation. Calculate the temperature of the filament that allows for an equlibrium. The emissivity of a tungsten filament is about epsO. abc How long does it take to heat the filament to this temperature? Ase that the radiation can be neglected during the heat-up time. abcliste

Solution:
abcliste abc The power emitted as radiation is given by P I A where I is the ensity of the radiation and A the surface area of the filament. The ensity is given by I epsilonsigma T^ where epsilon is the emissivity of tungsten sigma the Stefan-Boltzmann constant and T the surface temperature of the filament. The surface area surface of an open cylinder can be expressed as A pi d L It follows for the power P epsilon sigma T^ pi d L Solving for the temperature yields T TF leftfracPeps times ncS times pi times d times Lright^/ T approx resultTP abc The energy absorbed by the filament can be expressed as P Delta t c m Delta T c sscrhoW pi d/^ L Delta T where c is the specific heat of tungsten sscrhoW the density of tungsten and Delta T the temperature increase. Solving for the time yields Delta t tF fraccWo times RWo times pi times d^ times L times T - Tr times P t approx resulttP- abcliste
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Exercise:
A typical incandescent light bulb with a power of PO has a tungsten filament with length LO and diameter dO. abcliste abc When the light bulb is on the electric power is converted to radiation. Calculate the temperature of the filament that allows for an equlibrium. The emissivity of a tungsten filament is about epsO. abc How long does it take to heat the filament to this temperature? Ase that the radiation can be neglected during the heat-up time. abcliste

Solution:
abcliste abc The power emitted as radiation is given by P I A where I is the ensity of the radiation and A the surface area of the filament. The ensity is given by I epsilonsigma T^ where epsilon is the emissivity of tungsten sigma the Stefan-Boltzmann constant and T the surface temperature of the filament. The surface area surface of an open cylinder can be expressed as A pi d L It follows for the power P epsilon sigma T^ pi d L Solving for the temperature yields T TF leftfracPeps times ncS times pi times d times Lright^/ T approx resultTP abc The energy absorbed by the filament can be expressed as P Delta t c m Delta T c sscrhoW pi d/^ L Delta T where c is the specific heat of tungsten sscrhoW the density of tungsten and Delta T the temperature increase. Solving for the time yields Delta t tF fraccWo times RWo times pi times d^ times L times T - Tr times P t approx resulttP- abcliste
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Branches
Electrodynamics
Tags
heat, power, radiation, temperature
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Difficulty
(4, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
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Decoration
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