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kumpel [21]
3 years ago
6

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

s 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
Physics
1 answer:
Allushta [10]3 years ago
8 0

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

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