Question

An spherical object of radius 0.8 m and emissivity 0.5 is at a
temperature of 90

Answer 1

rate of energy transer H = dQ/dt = Ae (sigma) T^4

where A = area = 4pir^2, e= emissitivity = (1 for black bdoy)

sigma = constant = 5.67*10^-8

T = tempoeratrue in kelvin

so H = dq/dt = 4*3.14* 0.8*0.8* 5.67*10^-8 * (363^4 -292^4)

H =2332.889 W W = option2

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Answer 2

power emitted = sigma*A*e*T^4

=5.67*10^-8 * 4*pi*0.8^2*0.5*(90-19)^4

=5.79 W