Friday
May 24, 2013

# Homework Help: Physics

Posted by papito Urgent help needed on Wednesday, April 18, 2007 at 8:16pm.

Suppose the skin temperature of a naked person is 34°C when the person is standing inside a room whose temperature is 25°C. 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

No one has answered this question yet.

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