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 $\sigma=5.7 \times 10^{-8} Wm^{-2}K^{-4}$ 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

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Heat & Thermodynamics

Parallel rays of light of intensity

I = 912 W m^{- 2}

are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant

σ = 5.7 \times 1 0^{- 8} W m^{- 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

Answer: (A)

At steady state, energy absorbed equals energy radiated. The sphere absorbs radiation over its cross-sectional area

π r^{2}

and radiates over its surface area

4 π r^{2}

.

Energy absorbed per unit time:

I \times π r^{2} = 912 π r^{2}

.

Energy radiated per unit time:

σ T^{4} \times 4 π r^{2} - σ T_{0}^{4} \times 4 π r^{2}

, where

T_{0} = 300

K.

At equilibrium:

\frac{912}{4} = σ (T^{4} - 30 0^{4})

228 = 5.7 \times 1 0^{- 8} (T^{4} - 81 \times 1 0^{8})

T^{4} = \frac{228}{5.7 \times 1 0 ^{- 8}} + 81 \times 1 0^{8} = 40 \times 1 0^{8} + 81 \times 1 0^{8} = 121 \times 1 0^{8}

T = 4 121 \times 1 0^{8} = 330

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option