A parallel beam of radius is incident on a highly conducting sphere of unknown emissivity. The intensity of incident radiation on sphere is 912 W/m2 . Take Stefan–Boltzman constant σ=5.7×10−8Wm−2k−4 and surrounding temperature is 300 K. Assume that the energy exchange with surrounding is only through radiation. Find the final steady state temperature of the body (in 0C ). [Take 11= 3.32 and 7= 2.65]

In steady state912×a πr2=σ×e×4 πr2(T4−Ts4) 9125.7×10−8×4=T4−Ts4 T4=40×108+(300)4 =40×108+81+108 =121×108 T2=11×104 T=11×102 = 332 K = 590C

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A parallel beam of radius is incident on a highly conducting sphere of unknown emissivity. The intensity of incident radiation on sphere is 912 $W / m^{2}$ . Take Stefan–Boltzman constant $σ = 5.7 \times 10^{- 8} W m^{- 2} k^{- 4}$ and surrounding temperature is 300 K. Assume that the energy exchange with surrounding is only through radiation. Find the final steady state temperature of the body (in $^{0} C$ ). [Take $\sqrt{11}$ = 3.32 and $\sqrt{7}$ = 2.65]

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answer is 59.

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Detailed Solution

In steady state

$912 \times a π r^{2} = σ \times e \times 4 π r^{2} (T^{4} - T_{s}^{4}) \frac{912}{5.7 \times 10^{- 8} \times 4} = T^{4} - T_{s}^{4} T^{4} = 40 \times 10^{8} + {(300)}^{4} = 40 \times 10^{8} + 81 + 10^{8} = 121 \times 10^{8} T^{2} = 11 \times 10^{4} T = \sqrt{11} \times 10^{2} = 332 K = 59^{0} C$