A body at 40 0 ∘ C radiates 6 J c m − 2 s − 1 . A star radiates 6 × 1 0 4 J c m − 2 s − 1 . Find the temperature of the star in ∘ C (nearest whole number).

The power rises by 1 0 4 , so T rises 10 times. T = 673 × 10 = 6730 K = 645 7 ∘ C.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoHeat TransferMedium2 min

PhysicsMediuminteger

A body at $40 0^{\circ}$ C radiates 6 J $c m^{- 2}$ $s^{- 1}$ . A star radiates $6 \times 1 0^{4}$ J $c m^{- 2}$ $s^{- 1}$ . Find the temperature of the star in $^{\circ}$ C (nearest whole number).

Answer:

6457

Solution:

The power rises by $1 0^{4}$ , so $T$ rises 10 times. $T = 673 \times 10 = 6730$ K = $645 7^{\circ}$ C.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Heat Transfer

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