A thin wire is heated to 627°C. Its radiating surface area is 2 × 1 0 − 4 m² and its emissivity is 0.3. Ignore the surroundings and find the electric power needed, in W.

P = e σ A T 4 = 0.3 × 5.67 × 1 0 − 8 × 2 × 1 0 − 4 × 90 0 4 = 2.23 W.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoHeat TransferEasy2 min

PhysicsEasynumerical

A thin wire is heated to 627°C. Its radiating surface area is $2 \times 1 0^{- 4}$ m² and its emissivity is 0.3. Ignore the surroundings and find the electric power needed, in W.

Answer:

2.23

Solution:

$P = e σ A T^{4} = 0.3 \times 5.67 \times 1 0^{- 8} \times 2 \times 1 0^{- 4} \times 90 0^{4} = 2.23$ W.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Heat Transfer

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