Radiation - solving for Q - Stefan-Boltzmann law

[sportzmaniac]

Homework Statement

A person is standing outdoors in the shade where the temperature is 27°C.
(a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is 160 cm2 and its emissivity is 0.85.

(b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were 0.66?

e = .85 for a, .66 for b
sigma = 5.67*10^-8
A = 1.6m squared
T = 300.15K (27 + 273.15)

Homework Equations

q = e*sigma*T^4*A*t

The Attempt at a Solution

By plugging in the known variables, I got

Q/(1 sec) = .85*(5.67x10^-8)*((300.15) ^4)*1.6m squared

Q =625.86 J/s

Q/(1 sec) = .66*(5.67x10^-8)*((300.15) ^4)*1.6m squared

Q = 485.96 J/s


I was told by webassign these answers were wrong. Am I not solving just for Q here? I can't think of any other way besides solving for Q. Also, I'm not sure on the temperature. Do I just take the 27 and convert to kelvin or do I have to do something else? thanks in advance for the help

 

[Delphi51]

A meter squared is 100 cm by 100 cm = 10 000 cm².
So 160 cm² is not 1.6 m².

 

[sportzmaniac]

A meter squared is 100 cm by 100 cm = 10 000 cm².
So 160 cm² is not 1.6 m².

thank you. By converting it right I got the right answer