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 a person is unclothed and energy is being lost via radiation from a body surface area of 1.38 m2, which has a temperature of 34 °C and an emissivity of 0.557. Suppose that metabolic processes are producing energy at a rate of 120 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

Answer 1

Answer:

$T_s = 6.8 degree C$

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

$\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)$

here we know that

T = 34 degree C = 307 K

$A = 1.38 m^2$

e = 0.557

$\sigma = 5.67 \times 10^{-8} W/m^2K^4$

$\frac{dQ}{dt} = 120 J/s$

now we have

$120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)$

$120 = (4.36 \times 10^{-8})(307^4 - T_s^4)$

$T_s = 279.8 K$

$T_s = 6.8 degree C$