How Does Human Skin Radiate Heat in a Cooler Room?

[owura143]

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

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

 

[mezarashi]

You should become a bit suspicious when you review your calculations and find that the given room temperature of 25 degrees is nowhere to be found!

The question was also framed asking specifically for the "net" loss, meaning, the total power emitted - total power absorbed.