Answer:
494.24 W
442.41 W
51.83 W
Explanation:
A = Area of person = 1.2 m²
[tex]T_1[/tex] = Person's temperature = 27+273.15 = 300.15 K
[tex]T_2[/tex] = Room temperature = 18.8+273.15 = 291.95 K
[tex]\epsilon[/tex] = Emissivity of skin = 0.895
[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]
Radiated power of the person
[tex]P_1=\sigma\epsilon AT_1^4\\\Rightarrow P_1=5.67\times 10^{-8}\times 0.895\times 1.2\times 300.15^4\\\Rightarrow P_1=494.24\ W[/tex]
Power radiated by the person is 494.24 W
Radiated power absorbed by the person
[tex]P_2=\sigma\epsilon AT_2^4\\\Rightarrow P_2=5.67\times 10^{-8}\times 0.895\times 1.2\times 291.95^4\\\Rightarrow P_2=442.41\ W[/tex]
Radiated power absorbed by the person is 442.41 W
Net power would be
[tex]\Delta P=P_1-P_2\\\Rightarrow \Delta P=494.24-442.41\\\Rightarrow \Delta P=51.83\ W[/tex]
The net power loss is 51.83 W
Answer:
(a) 493.256 watt
(b) 441.5 Watt
(c) 51.76 Watt
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
