Saturday

December 20, 2014

December 20, 2014

Posted by **papito Urgent help needed** on Wednesday, April 18, 2007 at 8:22pm.

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

A food calorie is 1000energy calories.

ok. what about the (a) part?

**Answer this Question**

**Related Questions**

Physics - Suppose the skin temperature of a naked person is 34 degree celsius ...

Physics - Suppose the skin temperature of a naked person is 34°C when the person...

physics - A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...

physics - A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...

physics - A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...

PHYSICS - a.A person with skin area of 3 m2 is nude in a room of still air at 22...

physics problem - At body temperature the latent heat of vaporization of water ...

Physics ptoblem - At body temperature the latent heat of vaporization of water ...

physics - a. A person with skin area 2 m2 and 0.97 radiation efficiency is at ...

physics - a. A person with skin area 2 m2 and 0.97 radiation efficiency is at ...