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pentagon [3]
3 years ago
10

A small surface is placed inside an isothermal enclosure. The enclosure provides an irradiation of 6300 W/m2 . The surface absor

bs 5600 W/m2 . After a long time has elapsed (assume steady state): (a) What is the temperature of the surface? (b) What is the emissivity of the surface?
Physics
1 answer:
irina1246 [14]3 years ago
4 0

Answer with Explanation:

We are given that

Irradiation,G=6300W/m^2

Surface A absorb.G_A=5600 W/m^2

a.G=\sigma T^4

T^4=\frac{G}{\sigma}

T=(\frac{G}{\sigma})^{\frac{1}{4}}

Where \sigma =5.67\times 10^{-8}=Boltzman's constant

T=(\frac{6300}{5.67\times 10^{-8}})^{\frac{1}{4}}=577.4 K

Temperature of surface =577.4 K

b.E_{in}=E_{out}

E_{emitted}=\sigma E_A T^4

5600=5.67\times 10^{-8}E_A(577.4)^4

E_A=\frac{5600}{5.67\times 10^{-8}\times (577.4)^4}=0.89

Hence, the emissivity of  surface =0.89

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