
#1
Apr1807, 07:12 PM

P: 12

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 stefanboltzmann 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 



#2
Apr1807, 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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