Question

10. 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) would have a body surface area of approximately 2 m². What is the net amount of heat this person could radiate per second into a room at 18°C (about 65°F) if the surface temperature of the skin is 30°C?

A 140 W
B 120 W
C 110 W
D 130 W

Answer 1

B. 120 W

Step-by-step explanation:

To calculate the net amount of heat this person could radiate per second, you can use the Stefan-Boltzmann law for radiative heat transfer:

\[ \text{Heat Radiated per second} = \text{Stefan-Boltzmann Constant} \times \text{Surface Area} \times \text{Emissivity} \times \left( \text{Skin Temperature}^4 - \text{Room Temperature}^4 \right) \]

Given values:

- Stefan-Boltzmann Constant (\(\sigma\)) ≈ \(5.67 \times 10^{-8}\) W/(m²·K⁴)

- Surface Area (\(A\)) ≈ 2 m²

- Emissivity (\(e\)) is typically around 0.60 for human skin

- Skin Temperature (\(T_{\text{skin}}\)) = 30°C + 273.15 = 303.15 K

- Room Temperature (\(T_{\text{room}}\)) = 18°C + 273.15 = 291.15 K

Now, plug in the values:

\[ \text{Heat Radiated per second} = 5.67 \times 10^{-8} \, \text{W/(m²·K⁴)} \times 2 \, \text{m²} \times 0.60 \times \left( 303.15 \, \text{K}^4 - 291.15 \, \text{K}^4 \right) \]

\[ \text{Heat Radiated per second} \approx 123.55 \, \text{W} \]

So, the net amount of heat this person could radiate per second into a room at 18°C is approximately 123.55 W. The closest answer choice is B: 120 W.