The radiant flux received at a distance of 2.0 × 1 0 11 m from a star of radius 5.0 × 1 0 8 m is 900 W m⁻². Treat the star as a black body ( σ = 5.67 × 1 0 − 8 W m⁻² K⁻⁴) and find its surface temperature, in K.

σ T 4 = S ( R d ) 2 = 900 × 40 0 2 = 1.44 × 1 0 8 W/m². T = ( 5.67 × 1 0 − 8 1.44 × 1 0 8 ) to the power one quarter = 7099 K.

Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoHeat TransferMedium2 min

PhysicsMediuminteger

The radiant flux received at a distance of $2.0 \times 1 0^{11}$ m from a star of radius $5.0 \times 1 0^{8}$ m is 900 W m⁻². Treat the star as a black body ( $σ = 5.67 \times 1 0^{- 8}$ W m⁻² K⁻⁴) and find its surface temperature, in K.

Answer:

7099

Solution:

$σ T^{4} = S (\frac{d}{R})^{2} = 900 \times 40 0^{2} = 1.44 \times 1 0^{8}$ W/m². $T = (\frac{1.44 \times 1 0 ^{8}}{5.67 \times 1 0 ^{- 8}})$ to the power one quarter = $7099$ K.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Heat Transfer

⏱ 2m

Doubts & Discussion

Loading discussions...