Parallel rays of light of intensity are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant and assume that the energy exchange with the surroundings is only through radiation. The final steady st

Question

Parallel rays of light of intensity $I = 912 {Wm}^{- 2}$ are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant $σ = 5.7 \times 10^{- 8} {Wm}^{- 2} K^{- 4}$ and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to [2014]

Smart Achievers · Accepted Answer

(1)

Let T be the final steady state temperature of the black body.

In steady state,

Energy lost = Energy gained

$σ (T^{4} - T_{0}^{4}) \times 4 π R^{2} = I (π R^{2})$

$∴$ $5.7 \times 10^{- 8} [T^{4} - {(300)}^{4}] \times 4 = 912$

$∴$ $T = 330 K$

Solution

1 330 K

2 660 K

3 990 K

4 1550 K

Ans.
(1)

Let T be the final steady state temperature of the black body.

In steady state,

Energy lost = Energy gained

$σ (T^{4} - T_{0}^{4}) \times 4 π R^{2} = I (π R^{2})$

$∴$ $5.7 \times 10^{- 8} [T^{4} - {(300)}^{4}] \times 4 = 912$

$∴$ $T = 330 K$

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