A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose a person is unclothed and energy is being lost via radiation from a body surface area of 1.42 m2, which has a temperature of 34 °C and an emissivity of 0.621. Suppose that metabolic processes are producing energy at a rate of 138 J/s. What is the temperature in Kelvin of the coldest room in which this person could stand and not experience a drop in body temperature? Thank you.
2007-06-13 04:17:16 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
We are assuming no heat transfer by conduction or convection.
Net irradiance (j) = emissivity (epsilon) * Stefan's constant (sigma) * (Tbody^4 - Troom^4)
j = Power / area
Troom = (Tbody^4 - Power / (epsilon sigma area))^(1/4)
They give you body temperature (convert to K!), power, emissivity, and the body area. Look up Stefan's constant. Plugnchug.
2007-06-13 04:32:33 · answer #1 · answered by Anonymous · 1⤊ 1⤋
Convective transfer would greatly dominate radiation.
2007-06-16 12:54:14 · answer #2 · answered by Dr. R 7 · 0⤊ 0⤋