Suppose the skin temperature of a naked person is 34 degree celsius when the person is standing inside a room whose temperature is 23 celsius. The skin area of the individual is 1.5m^2.

Question

(a) Assuming the emissivity is 0.80, find the net loss of radint power from the body.
(b) Determine the number of food Calories of energy(1 food calorie=4186J) that is lost in one hour due to the net loss rate obtained in part (a0. Metabolic conversion of food into energy replaces this loss.

I will be happy to critique your thinking.

Ignore convection losses, which in fact are probably greater than radiation.

Answer 1

Yeahh Go UBC