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Question

Physics Question on Heat Transfer

A body at a temperature of 728C728^{\circ} C and has surface area 5cm25\, cm ^{2}, radiates 300J300\, J of energy each minute. The emissivity is (Given Boltzmann constant =5.67×108Wm2K4=5.67 \times 10^{-8} Wm ^{2} K ^{4} )

A

e =0.18

B

e=0.02

C

e=0.2

D

e=0.15

Answer

e =0.18

Explanation

Solution

a=EAt=eσ(T4T04)Ata=E A t=e \sigma\left(T^{4}-T_{0}^{4}\right) A t where, t=t= time T0=T_{0}= temperature of surrounding when T>T0,Q=eσT4AtT>T_{0}, Q=e \sigma T^{4} A t 300=e×(5.67×108)(1000)4(5.00×104)300 =e \times\left(5.67 \times 10^{-8}\right)(1000)^{4}\left(5.00 \times 10^{-4}\right) e=0.18e=0.18