A 200 W lamp loses all its energy by radiation from a filament of area 1.5 cm² and emissivity 0.5. Take σ = 5.67 × 1 0 − 8 W m⁻² K⁻⁴ and neglect absorption. Find the filament temperature, in K.

T 4 = e σ A P = 0.5 × 5.67 × 1 0 − 8 × 1.5 × 1 0 − 4 200 = 4.703 × 1 0 13 , so T = 2619 K.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoHeat TransferMedium2 min

PhysicsMediuminteger

A 200 W lamp loses all its energy by radiation from a filament of area 1.5 cm² and emissivity 0.5. Take $σ = 5.67 \times 1 0^{- 8}$ W m⁻² K⁻⁴ and neglect absorption. Find the filament temperature, in K.

Answer:

2619

Solution:

$T^{4} = \frac{P}{e σ A} = \frac{200}{0.5 \times 5.67 \times 1 0 ^{- 8} \times 1.5 \times 1 0 ^{- 4}} = 4.703 \times 1 0^{13}$ , so $T = 2619$ K.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Heat Transfer

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