The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature and approximate the human body by a rectangular block with a height of 1.63 m, a width of 33.5 cm and a length of 31.0 cm. Calculate the power emitted by the human body. What is the wavelength of the peak in the spectral distribution for this temperature?

P = 1145.75W

λ = 4.0*10⁻²⁸m

Explanation:

Temperature T = 35°C = 308.15K

h = 1.63m

width (w) = 33.5cm = 0.335m

Length (L) = 31cm = 0.31m

Surface area of the human body =

2hw + 2hL + 2Lw

2*(1.63*0.335) + 2*(1.63*0.31) + 2*(0.31*0.335)

Area = 1.0921 + 1.0106 + 0.2077

Area = 2.3104m²

Using Stefan-Boltzmann equation,

E = εσΤ⁴

Ε = 0.97 * 5.67*10⁻⁸ * (308.15)⁴

E = 495.91 w/m²

Power = energy * surface area

Power = 495.91 * 2.3104

P = 1145.75W

b)

Applying Energy-Wavelength equation

E = hc / λ

λ = hc / E

λ = (6.626*10⁻³⁴ * 3.0⁸) / 495.91

λ = 4.0*10⁻²⁸m

E = 1.143 KW

(λ)max = 9.4 μm = 9.4 x 10⁻⁶ m

Explanation:

a)

The total emissive power of human body can be given by Stefan-Boltzman Law:

E = AεσΤ⁴

where,

E = Total emitted power

ε = emissivity = 97% = 0.97

σ = Stefan Boltzman Constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature = 35°C +273 = 308 K

A = Total Surface Area of rectangular approximation = 2(1.63 m)(0.335 m) + 2(1.63 m)(0.31 m) + 2(0.335 m)(0.31 m)

A = 2.31 m²

Therefore,

E = (2.31 m²)(0.97)(5.67 x 10⁻⁸ W/m².K⁴)(308 K)⁴

E = 1143.52 W = 1.143 KW

b)

The peak wavelength or the maximum wavelength can be found out by using wein's displacement law:

[(λ)max][T] = 2.8978 x 10⁻³ m.K

(λ)max = 2897.8 μm.K/308 K

(λ)max = 9.4 μm = 9.4 x 10⁻⁶ m

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.

truck y has a greater velocity than truck x.