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 1.56 m2, which has a temperature of 34 °C and an emissivity of 0.673. Suppose that metabolic processes are producing energy at a rate of 137 J/s. What is the temperature in kelvins of the coldest room in which this person could stand and not experience a drop in body temperature?

2 answers

Net radiative heat loss
= sigma*A*[(307)^4 - T^4]
= 137 watts

"sigma" is the Stefan Boltzmann constant. Look it up if you have never heared of it. A is the body area.

An emissivity of 1 has been assumed. Convective heat loss has been ignored. These may not be accurate assumptions.
Solve for T, the room temperature in Kelvin.
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