Question

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 75-kg (165-lb) person of height 1.83 m (6 ft) has a body surface area of approximately 2.0 m2. (a) What is the net amount of heat this person could radiate per second into a room at 18°C (about 65°F) if his skin’s surface temperature is 30°C? (At such temperatures, nearly all the heat is infrared radiation, for which the body’s emissivity is 1.0, regardless of the amount of pigment.) (b) Normally, 80% of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part (a) to find this person’s basal metabolic rate.

Answer 1

Answer:

(a) Q = 142.67 W

(b) Basal Metabolic Rate = 178.33 W

Explanation:

(a)

We can find the heat radiated by the person by using Stefan-Boltzman's law:

$Q = \sigma A (T^4 - T_{s}^4)$

where,

Q = heat radiated per second = ?

σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

A = Surface Area = 2 m²

T = Temperature of Skin = 30° C + 273 = 303 k

Ts = Temperature of room = 18° C +273 = 291 k

Therefore,

$Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4]$

Q = 142.67 W

(b)

Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,

$Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}$

Basal Metabolic Rate = 178.33 W