Urgent help needed please. Radiation question


by owura143
Tags: radiation, urgent
owura143
owura143 is offline
#1
Apr18-07, 07:12 PM
P: 12
Suppose the skin temperature of a naked person is 34C when the person is standing inside a room whose temperature is 25C. The skin area of the individual is 2.0 m2

a) Assuming the emissivity is 0.80, find the net loss of radiant power from the body

b) Determine the number of food Calories of energy (1 food Calorie = 4186 J) that is lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.


I used
a)
Q/t = emissivity x stefan-boltzmann constant x T^4 x A

= 0.8 X 5.67^-8 X 298.15^4 X 2

=719.277


b) 1 watt per hour = 3600J
total joules = 3600 X 719.277 = 2589399.087


2589399.087 /4186 = total calories


Answers are wrong
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marcusl
marcusl is offline
#2
Apr18-07, 08:02 PM
Sci Advisor
PF Gold
P: 2,020
To get the net power loss from the body, you need to compute the power radiated by the warm skin, then subtract the power radiated by the room that is absorbed by the skin. Can you write the equation for that?

BTW, not sure why you used 25.15C degrees instead of 25C in your calc...


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