Question

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 that a person is unclothed and energy is being lost via radiation from a body surface area of 1.36 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 122 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature

Answer 1

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

`P = \sigma Ae\Delta T^4`

Where

A = Surface area of the Object

`\sigma =` Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

`A = 1.36m^2`

`\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4`

`\sigma = 5.67*10^(-8) J/(s m^2 K^4)`

`P = 122J/s`

`e = 0.7`

Replacing at our equation and solving to find the temperature 1 we have,

`P = \sigma Ae\Delta T^4`

`P = \sigma Ae (T_2^4 -T_1^4)`

`122 = (5.67*10^(-8))(1.36)(0.7)(307^4-T_1^4)`

`T_1 = 285.272K = 12.122\°C`

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C