Question

A person with a surface area of 1.20 m2, and a skin temperature of 27 oC, is in a room that is at a temperature of 18.8 °C. The emissivity of the skin is 0.895. The Stefan-Boltzmann constant is 5.67 x 10-8 W/(m2·K4).(a) How much power (power is the energy per second) is radiated by the person? Keep 2 decimal places.b. How much power is absorbed by the person from the surroundings? Keep 2 decimal places.c. What is his NET Power loss? Report as a Positive value. Keep 2 decimal places.

Answer 1

(a) 493.256 watt

(b) 441.5 Watt

(c) 51.76 Watt

Step-by-step explanation:

Area, A = 1.20 m^2

T = 27°C = 300 K

To = 18.8°C = 291.8 K

e = 0.895

Stefan's constant, σ = 5.67 x 10^-8 W/m^2K^4

(a) Power emitted by the person

P1 = A e σ T^4

P1 = 1.2 x 0.895 x 5.67 x 10^-8 x (300)^4

P1 = 493.256 watt

(b) Pwer absorbed by the person, P2

P2 = A e σ To^4

P2 = 1.2 x 0.895 x 5.67 x 10^-8 x (291.8)^4

P2 = 441.5 watt

(c) Net power loss, P = P1 - P2

P = 493.256 - 441.5 = 51.76 Watt

Answer 2

494.24 W

442.41 W

51.83 W

Step-by-step explanation:

A = Area of person = 1.2 m²

`T_1` = Person's temperature = 27+273.15 = 300.15 K

`T_2` = Room temperature = 18.8+273.15 = 291.95 K

`\epsilon` = Emissivity of skin = 0.895

`\sigma` = Stefan-Boltzmann constant =
`5.67* 10^(-8)\ W/m^2K^4`

Radiated power of the person

`P_1=\sigma\epsilon AT_1^4\\\Rightarrow P_1=5.67* 10^(-8)* 0.895* 1.2* 300.15^4\\\Rightarrow P_1=494.24\ W`

Power radiated by the person is 494.24 W

Radiated power absorbed by the person

`P_2=\sigma\epsilon AT_2^4\\\Rightarrow P_2=5.67* 10^(-8)* 0.895* 1.2* 291.95^4\\\Rightarrow P_2=442.41\ W`

Radiated power absorbed by the person is 442.41 W

Net power would be

`\Delta P=P_1-P_2\\\Rightarrow \Delta P=494.24-442.41\\\Rightarrow \Delta P=51.83\ W`

The net power loss is 51.83 W