Transfer of Heat: Temperature in Kelvins

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A person's body generates energy through metabolic processes, and if energy loss exceeds this production, body temperature can drop dangerously. In the scenario presented, the body loses energy through radiation from a surface area of 1.24 m² at a temperature of 34 °C with an emissivity of 0.614. The metabolic energy production rate is 98.8 J/s, and the discussion involves calculating the coldest room temperature in Kelvin where the individual would not experience a temperature drop. The Stefan-Boltzmann law is referenced to analyze radiative heat transfer, indicating the need to consider both the body and room temperatures in the calculations. Accurate application of these principles is crucial for determining safe environmental conditions for the individual.
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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 1.24 m2, which has a temperature of 34 °C and an emissivity of 0.614. Suppose that metabolic processes are producing energy at a rate of 98.8 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?


I used the formula Q= e*sigma*(T kelvins ^4) times(area) I get 384.21 after changing the temperature to 307.15 Kelvins. I am not sure where to place the rate of energy within this problem.
 
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The Stefan-Boltzmann law is based on the radiative heat transfer between hot and cold - i.e. (TH4 - TC4).
 
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