Question 143: Thermodynamics & KTG

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 $$\sigma $$ = 5.7 $$ 	imes $$ 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

Accepted Answer

Effective area of incident light = $$\pi$$r
2
Total energy per second = 912 $$	imes$$ $$\pi$$r
2
From Stefan's law, we have
E = Ae $$\sigma$$ (T
4
$$-$$ T$$_0^4$$)
For black body emissivity, e = 1.
A = 4$$\pi$$r
2
912 $$	imes$$ $$\pi$$r
2
= 4$$\pi$$r
2
$$	imes$$ 5.7 $$	imes$$ 10
$$-$$8
[T
4
$$-$$ 300
4
]
$$({T^4} - {300^4}) = {{912} \over {4 	imes 5.7 	imes {{10}^{ - 8}}}} = 40 	imes {10^8}$$
$${T^4} = 40 	imes {10^8} + {3^4} 	imes {10^8}$$
$${T^4} = (90 + 81) 	imes {10^8} \Rightarrow 121 	imes {10^8}$$
$$T = {(121)^{1/4}} 	imes 100 = \sqrt {11}  	imes 100 \Rightarrow 3.3 	imes 100$$
T = 330 K

Answer

330 K

Answer

660 K

Answer

990 K

Answer

1550 K

Thermodynamics & KTG

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