JEE Advance - Physics (2014 - Paper 2 Offline - No. 2)
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–2K–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
330 K
660 K
990 K
1550 K
Explanation
Effective area of incident light = $$\pi$$r2
Total energy per second = 912 $$\times$$ $$\pi$$r2
From Stefan's law, we have
E = Ae $$\sigma$$ (T4 $$-$$ T$$_0^4$$)
For black body emissivity, e = 1.
A = 4$$\pi$$r2
912 $$\times$$ $$\pi$$r2 = 4$$\pi$$r2 $$\times$$ 5.7 $$\times$$ 10$$-$$8 [T4 $$-$$ 3004]
$$({T^4} - {300^4}) = {{912} \over {4 \times 5.7 \times {{10}^{ - 8}}}} = 40 \times {10^8}$$
$${T^4} = 40 \times {10^8} + {3^4} \times {10^8}$$
$${T^4} = (90 + 81) \times {10^8} \Rightarrow 121 \times {10^8}$$
$$T = {(121)^{1/4}} \times 100 = \sqrt {11} \times 100 \Rightarrow 3.3 \times 100$$
T = 330 K
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